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a(n) = Sum_{k=0..n} 3^(n-k)* A123125(n, k).
+20
18
1, 1, 4, 22, 160, 1456, 15904, 202672, 2951680, 48361216, 880405504, 17630351872, 385148108800, 9114999832576, 232311251144704, 6343764407375872, 184778982658539520, 5718564661248065536, 187389427488113557504, 6481629887083387420672, 235993351028007334051840
COMMENTS
a(n+1) = [1,4,22,160,1456,...] is the first Eulerian transform of A000244 (powers of 3), it is also the Stirling transform of A080599(n+1) = [1,3,12,66,450,...].
FORMULA
O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1-2*k*x). - Paul D. Hanna, Jul 20 2011 a(n) = A_{n}(3) where A_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010 E.g.f.: (exp(x) - 2*cosh(x))/(2*exp(x) - 3*cosh(x)) =1 + x/(U(0)-x) where U(k)= 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/U(k+1)))); (continued fraction, 4-step). - Sergei N. Gladkovskii, Nov 08 2012 G.f.: 1 + x/G(0) where G(k) = 1 - x*2*(2*k+2) + x^2*(k+1)*(k+2)*(1-2^2)/G(k+1); (continued fraction due to T. J. Stieltjes). - Sergei N. Gladkovskii, Jan 11 2013 G.f.: 1/Q(0), where Q(k) = 1 - x*(4*k+1) - 3*x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 30 2013 a(n) = 2^n*log(3)* Integral_{x >= 0} (floor(x))^n * 3^(-x) dx. - Peter Bala, Feb 14 2015 E.g.f.: 2/(3-exp(2*x)).
Special values of the generalized hypergeometric function n_F_(n-1):
a(n) = (2^(n+1)/9) * hypergeom([2,2,..2],[1,1,..1],1/3), where the sequence in the first square bracket ("upper" parameters) has n elements all equal to 2 whereas the sequence in the second square bracket ("lower" parameters) has n-1 elements all equal to 1.
Example: a(4) = (2^5/9) * hypergeom([2,2,2,2],[1,1,1],1/3) = 16. (End)
a(n) = (-1)^(n+1)*(Li_{-n}(sqrt(3)) + Li_{-n}(-sqrt(3)))/3, where Li_n(x) is the polylogarithm. - Vladimir Reshetnikov, Oct 31 2015 a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * 2^(k - 1) * a(n-k). - Ilya Gutkovskiy, Jan 16 2020 a(n) = 2^n*F_{n}(1/2), where F_{n}(x) is the Fubini polynomial. This is another way to state the above formula from Ilya Gutkovskiy. - Peter Luschny, May 21 2021 a(n+1) = -2*a(n) + 3*Sum_{k=0..n} binomial(n, k)*a(k)*a(n-k). - Michael Somos, Jun 05 2021 a(n) = (-2)^(n + 1)*PolyLog(-n, 3)/3. - Peter Luschny, Aug 20 2021
EXAMPLE
G.f. = 1 + x + 4*x^2 + 22*x^3 + 160*x^4 + 1456*x^5 + 15904*x^6 + ... - Michael Somos, Jun 05 2021
MAPLE
seq(subs(x=3, add(combinat:-eulerian1(n, k)*x^k, k=0..n)), n=0..20);
# Alternative:
gf := -2/(exp(2*x) - 3): series(gf, x, 21): seq(n!*coeff(%, x, n), n=0..20);
# (End)
# Or via the recurrence of the Fubini polynomials:
F := proc(n) option remember; if n = 0 then return 1 fi;
expand(add(binomial(n, k)*F(n - k)*x, k = 1..n)) end:
a := n -> 2^n*subs(x = 1/2, F(n)):
# next Maple program:
a:= proc(n) option remember; `if`(n=0, 1, add(
a(n-j)*binomial(n, j)*2^(j-1), j=1..n))
end:
MATHEMATICA
CoefficientList[Series[(Exp[x]-2*Cosh[x])/(2*Exp[x]-3*Cosh[x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 24 2013 *) Table[Sum[2^(n+1)*k^n/3^(k+1), {k, 0, Infinity}], {n, 0, 20}] (* Vaclav Kotesovec, Nov 28 2013 *) Round@Table[(-1)^(n+1) (PolyLog[-n, Sqrt[3]] + PolyLog[-n, -Sqrt[3]])/3, {n, 0, 20}] (* Vladimir Reshetnikov, Oct 31 2015 *) Table[Sum[StirlingS2[n, k]*2^(n-k)*k!, {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Jul 13 2018 *) Eulerian1[0, 0] = 1; Eulerian1[n_, k_] := Sum[(-1)^j (k-j+1)^n Binomial[ n+1, j], {j, 0, k+1}]; Table[Sum[Eulerian1[n, k]*3^k, {k, 0, n}], {n, 0, 20}] (* Jean-Fran�ois Alcover, Jul 13 2019, after Peter Luschny *) a[n_] := (-2)^(n + 1) PolyLog[-n, 3] / 3;
PROG
(PARI) {a(n)=polcoeff(sum(m=0, n, m!*x^m/prod(k=1, m, 1-2*k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */ (PARI) {a(n) = if(n<0, 0, n!*polcoeff( 2/(3 - exp(2*x + x*O(x^n))), n))}; /* Michael Somos, Jun 05 2021 */
EXTENSIONS
a(7) corrected (was 206672) and more terms from Peter Luschny, Aug 03 2010
T(n, k) = A343277(n)*[x^k] p(n, x) where p(n, x) = (1/(n+1))*Sum_{k=0..n} (-1)^k*E1(n, k)*x^(n - k) / binomial(n, k), and E1(n, k) are the Eulerian numbers A123125. Triangle read by rows, for 0 <= k <= n.
+20
2
1, 0, 1, 0, -1, 2, 0, 1, -4, 3, 0, -3, 22, -33, 12, 0, 1, -13, 33, -26, 5, 0, -5, 114, -453, 604, -285, 30, 0, 5, -200, 1191, -2416, 1985, -600, 35, 0, -35, 2470, -21465, 62476, -78095, 42930, -8645, 280, 0, 14, -1757, 21912, -88234, 156190, -132351, 51128, -7028, 126
COMMENTS
Conjecture: For even n >= 6 p(n, x)/x and for odd n >= 3 p(n, x)/(x^2 - x) is irreducible.
FORMULA
An alternative representation of the sequence of rational polynomials is:
p(n, x) = Sum_{k=1..n} x^k*k!*Sum_{j=0..k} (-1)^(n-j)*Stirling2(n, j)/((k - j)!(n - j + 1)*binomial(n + 1, j)) for n >= 1 and p(0, x) = 1.
(Sum_{k = 0..n} T(n, k)) / A343277(n) = Bernoulli(n, 1).
EXAMPLE
Triangle starts:
----------------------------------------------------------
[0] 1; [1]
[1] 0, 1; [2]
[2] 0, -1, 2; [6]
[3] 0, 1, -4, 3; [12]
[4] 0, -3, 22, -33, 12; [60]
[5] 0, 1, -13, 33, -26, 5; [30]
[6] 0, -5, 114, -453, 604, -285, 30; [210]
[7] 0, 5, -200, 1191, -2416, 1985, -600, 35; [280]
.
The coefficients of the polynomials p(n, x) = (Sum_{k = 0..n} T(n, k) x^k) / A343277(n) for the first few n: [0] 1;
[1] 0, 1/2;
[2] 0, -1/6, 1/3;
[3] 0, 1/12, -1/3, 1/4;
[4] 0, -1/20, 11/30, -11/20, 1/5;
[5] 0, 1/30, -13/30, 11/10, -13/15, 1/6.
MAPLE
CoeffList := p -> op(PolynomialTools:-CoefficientList(p, x)):
E1 := (n, k) -> combinat:-eulerian1(n, k):
poly := n -> (1/(n+1))*add((-1)^k*E1(n, k)*x^(n-k)/binomial(n, k), k=0..n):
Trow := n -> denom(poly(n))*CoeffList(poly(n)): seq(Trow(n), n = 0..9);
MATHEMATICA
Poly342321[n_, x_] := If[n == 0, 1, Sum[x^k*k!*Sum[(-1)^(n - j)*StirlingS2[n, j] /((k - j)!(n - j + 1) Binomial[n + 1, j]), {j, 0, k}], {k, 1, n}]];
Table[ A343277[n] CoefficientList[Poly342321[n, x], x][[k+1]], {n, 0, 9}, {k, 0, n}] // Flatten
CROSSREFS
Sequences of rational polynomials p(n, x) with p(n, 1) = Bernoulli(n, 1):
Triangle, read by rows, T(n, k) = 2* A123125(n-1, k), for n >= 2, otherwise T(n, 0) = T(n, n) = -1, with T(0, 0) = T(1, 0) = 1.
+20
1
1, 1, -1, -1, 2, -1, -1, 2, 2, -1, -1, 2, 8, 2, -1, -1, 2, 22, 22, 2, -1, -1, 2, 52, 132, 52, 2, -1, -1, 2, 114, 604, 604, 114, 2, -1, -1, 2, 240, 2382, 4832, 2382, 240, 2, -1, -1, 2, 494, 8586, 31238, 31238, 8586, 494, 2, -1, -1, 2, 1004, 29216, 176468, 312380, 176468, 29216, 1004, 2, -1, -1, 2, 2026, 95680, 910384, 2620708, 2620708, 910384, 95680, 2026, 2, -1
REFERENCES
Douglas C. Montgomery and Lynwood A. Johnson, Forecasting and Time Series Analysis, McGraw-Hill, New York, 1976, page 91.
FORMULA
T(n, k) = 2* A123125(n-1, k), with T(0, 0) = T(1, 0) = 1, otherwise T(n, 0) = T(n, n) = -1. Sum_{k=0..n} T(n, k) = 2*033312(n), for n >= 1, otherwise 1 (n=0).
T(n, k) = 2* A008292(n, k) for n >= 2, 1 <= k <= n-1, with T(n, 0) = T(n, n) = -1, T(0, 0) = T(1, 0) = 1. T(n, n-k) = T(n, k) for n >= 2. (End)
EXAMPLE
Triangle begins as:
1;
1, -1;
-1, 2, -1;
-1, 2, 2, -1;
-1, 2, 8, 2, -1;
-1, 2, 22, 22, 2, -1;
-1, 2, 52, 132, 52, 2, -1;
-1, 2, 114, 604, 604, 114, 2, -1;
-1, 2, 240, 2382, 4832, 2382, 240, 2, -1;
-1, 2, 494, 8586, 31238, 31238, 8586, 494, 2, -1;
-1, 2, 1004, 29216, 176468, 312380, 176468, 29216, 1004, 2, -1;
MATHEMATICA
(* First program *)
f[x_, n_]:= f[x, n]= (1-x)^(n+1)*Sum[k^n*x^k, {k, 0, Infinity}];
Table[Simplify[f[x, n]], {n, 0, 10}];
Join[{{1}}, Table[Join[CoefficientList[2*f[x, n] -1, x], {-1}], {n, 0, 10}]]//Flatten
(* Second program *)
Eulerian[n_, k_]:= Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}]; (* A008292 *) A141591[n_, k_]:= If[k==0 || k==n, -1, 2*Eulerian[n-1, k]] +2*Boole[n==0 || n ==1 && k==0];
PROG
(Magma)
Eulerian:= func< n, k | (&+[(-1)^j*Binomial(n+1, j)*(k-j)^n: j in [0..k]]) >; // A008292 if n eq 0 then return 1;
elif k eq 0 and n eq 1 then return 1;
elif k eq 0 or k eq n then return -1;
else return 2*Eulerian(n-1, k);
end if;
end function;
(SageMath)
@CachedFunction
def A008292(n, k): return sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in range(k+1)) if (k==0 and n==0): return 1
elif (k==0 and n==1): return 1
elif (k==0 or k==n): return -1
An extended triangle of Eulerian coefficients A123125: f(x,n)=x^(n+1)+1+ A123125(x,n).
+20
0
1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 11, 11, 1, 1, 1, 1, 26, 66, 26, 1, 1, 1, 1, 57, 302, 302, 57, 1, 1, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1, 1, 1013
COMMENTS
Not entirely symmetrical, the x^(n+1)+1 polynomials was added to remove zeros and make the triangle more symmetrical.
Row sums are:
{1, 3, 3, 4, 8, 26, 122, 722, 5042, 40322, 362882, 3628802}.
MATHEMATICA
lear[f, x, n, a] f[x_, n_] := f[x, n] = x^(n + 1) + (1 - x)^(n + 1)*Sum[k^n*x^k, {k, 0, Infinity}] + 1; Table[FullSimplify[ExpandAll[f[x, n]]], {n, 0, 10}]; a = Join[{{1}}, Table[CoefficientList[FullSimplify[ExpandAll[f[x, n]]], x], {n, 0, 10}]]; Flatten[a]
Triangle read by rows: T(n,k) = 2^k* A123125(n,k).
+20
0
1, 0, 2, 0, 2, 4, 0, 2, 16, 8, 0, 2, 44, 88, 16, 0, 2, 104, 528, 416, 32, 0, 2, 228, 2416, 4832, 1824, 64, 0, 2, 480, 9528, 38656, 38112, 7680, 128, 0, 2, 988, 34344, 249904, 499808, 274752, 31616, 256, 0, 2, 2008, 116864, 1411744, 4998080, 5646976, 1869824
COMMENTS
Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,...] DELTA [2,0,4,0,6,0,8,0,10,0,...] where DELTA is the operator defined in A084938. - Philippe Del�ham, Sep 15 2008
EXAMPLE
{1},
{0, 2},
{0, 2, 4},
{0, 2, 16, 8},
{0, 2, 44, 88, 16},
{0, 2, 104, 528, 416, 32},
{0, 2, 228, 2416, 4832, 1824, 64},
{0, 2, 480, 9528, 38656, 38112, 7680, 128}, ...
EXTENSIONS
New name using formula by Philippe Del�ham, Joerg Arndt, Jan 25 2020
Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.
+10
406
1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
COMMENTS
The indexing used here follows that given in the classic books by Riordan and Comtet. For two other versions see A173018 and A123125. - N. J. A. Sloane, Nov 21 2010 Coefficients of Eulerian polynomials. Number of permutations of n objects with k-1 rises. Number of increasing rooted trees with n+1 nodes and k leaves.
T(n,k) = number of permutations of [n] with k runs. T(n,k) = number of permutations of [n] requiring k readings (see the Knuth reference). T(n,k) = number of permutations of [n] having k distinct entries in its inversion table. - Emeric Deutsch, Jun 09 2004 T(n,k) = number of ways to write the Coxeter element s_{e1}s_{e1-e2}s_{e2-e3}s_{e3-e4}...s_{e_{n-1}-e_n} of the reflection group of type B_n, using s_{e_k} and as few reflections of the form s_{e_i+e_j}, where i = 1, 2, ..., n and j is not equal to i, as possible. - Pramook Khungurn (pramook(AT)mit.edu), Jul 07 2004
T(n,k)/n! also represents the n-dimensional volume of the portion of the n-dimensional hypercube cut by the (n-1)-dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k-1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k-1 and k. - Stefano Zunino, Oct 25 2006
[E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [-P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials and P(n,t) are the polynomials in A131758. - Tom Copeland, Sep 30 2007 G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1*x + (2+t)*x^2/2! + (6+6*t+t^2)*x^3/3! + ... gives row polynomials for A090582, the reverse f-polynomials for the permutohedra (see A019538). G(x,t-1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra (Postnikov et al.). G((t+1)*x, -1/(t+1)) = 1 + (1+t)*x + (1+3*t+2*t^2)*x^2/2! + ... gives row polynomials for A028246. (End)
A subexceedant function f on [n] is a map f:[n] -> [n] such that 1 <= f(i) <= i for all i, 1 <= i <= n. T(n,k) equals the number of subexceedant functions f of [n] such that the image of f has cardinality k [Mantaci & Rakotondrajao]. Example T(3,2) = 4: if we identify a subexceedant function f with the word f(1)f(2)...f(n) then the subexceedant functions on [3] are 111, 112, 113, 121, 122 and 123 and four of these functions have an image set of cardinality 2. - Peter Bala, Oct 21 2008 Further to the comments of Tom Copeland above, the n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type A_(n-1). The corresponding f-vectors are the rows of A019538. For example, 1 + 4*x + x^2 = y^2 + 6*y + 6 and 1 + 11*x + 11*x^2 + x^3 = y^3 + 14*y^2 + 36*y + 24, where x = y + 1, give [1,6,6] and [1,14,36,24] as the third and fourth rows of A019538. The Hilbert transform of this triangle (see A145905 for the definition) is A047969. See A060187 for the triangle of Eulerian numbers of type B (the h-vectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of h-vectors of type D. For tables of restricted Eulerian numbers see A144696 - A144699. - Peter Bala, Oct 26 2008 For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - Tom Copeland, Nov 06 2008 The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1)) for n >=1 lead directly to the triangle of Eulerian numbers. - Johannes W. Meijer, May 24 2009 From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
The (Eulerian) polynomials e(n,x) = Sum_{k=0..n-1} T(n,k+1)*x^k turn out to be also the numerators of the closed-form expressions of the infinite sums:
S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
(Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
If n is an odd prime, then all numbers of the (n-2)-th and (n-1)-th rows are in the progression k*n+1. - Vladimir Shevelev, Jul 01 2011 The Eulerian triangle is an element of the formula for the r-th successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k-1) * binomial(j-k+n+r, j+r). - Gary Detlefs, Nov 11 2011 Li and Wong show that T(n,k) counts the combinatorially inequivalent star polygons with n+1 vertices and sum of angles (2*k-n-1)*Pi. An equivalent formulation is: define the total sign change S(p) of a permutation p in the symmetric group S_n to be equal to Sum_{i=1..n} sign(p(i)-p(i+1)), where we take p(n+1) = p(1). T(n,k) gives the number of permutations q in S_(n+1) with q(1) = 1 and S(q) = 2*k-n-1. For example, T(3,2) = 4 since in S_4 the permutations (1243), (1324), (1342) and (1423) have total sign change 0. - Peter Bala, Dec 27 2011 Xiong, Hall and Tsao refer to Riordan and mention that a traditional Eulerian number A(n,k) is the number of permutations of (1,2...n) with k weak exceedances. - Susanne Wienand, Aug 25 2014 Connections to algebraic geometry/topology and characteristic classes are discussed in the Buchstaber and Bunkova, the Copeland, the Hirzebruch, the Lenart and Zainoulline, the Losev and Manin, and the Sheppeard links; to the Grassmannian, in the Copeland, the Farber and Postnikov, the Sheppeard, and the Williams links; and to compositional inversion and differential operators, in the Copeland and the Parker links. - Tom Copeland, Oct 20 2015 The bivariate e.g.f. noted in the formulas is related to multiplying edges in certain graphs discussed in the Aluffi-Marcolli link. See p. 42. - Tom Copeland, Dec 18 2016 Distribution of left children in treeshelves is given by a shift of the Eulerian numbers. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. See A278677, A278678 or A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016 The row polynomial P(n, x) = Sum_{k=1..n} T(n, k)*x^k appears in the numerator of the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} P(n, x)/(1 - x)^(n+2) for n >= 0 (with 0^0=1). See also triangle A131689 with a Mar 31 2017 comment for a rewritten form. For the e.g.f see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017 For relations to Ehrhart polynomials, volumes of polytopes, polylogarithms, the Todd operator, and other special functions, polynomials, and sequences, see A131758 and the references therein. - Tom Copeland, Jun 20 2017 For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - Tom Copeland, Mar 19 2018 Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - Tom Copeland, Jun 25 2018
REFERENCES
Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 106.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.[Worpitzky's identity (6.37)]
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
Meng Li and Ron Goldman. "Limits of sums for binomial and Eulerian numbers and their associated distributions." Discrete Mathematics 343.7 (2020): 111870.
Anthony Mendes and Jeffrey Remmel, Generating functions from symmetric functions, Preliminary version of book, available from Jeffrey Remmel's home page http://math.ucsd.edu/~remmel/
K. Mittelstaedt, A stochastic approach to Eulerian numbers, Amer. Math. Mnthly, 127:7 (2020), 618-628.
T. K. Petersen, Eulerian Numbers, Birkhauser, 2015.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Figure M3416, Academic Press, 1995.
H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, 1973, see p. 208.
D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 101.
LINKS
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48. David Callan, Problem 498, The College Mathematics Journal, Vol. 24, No. 2 (Mar., 1993), pp. 183-190 (8 pages). P. A. MacMahon, The divisors of numbers, Proc. London Math. Soc., (2) 19 (1920), 305-340; Coll. Papers II, pp. 267-302. O. J. Munch, Om potensproduktsummer [Norwegian, English summary], Nordisk Matematisk Tidskrift, 7 (1959), 5-19. [Annotated scanned copy] O. J. Munch, Om potensproduktsummer [ Norwegian, English summary ], Nordisk Matematisk Tidskrift, 7 (1959), 5-19. Vincent Pilaud and V. Pons, Permutrees, arXiv preprint arXiv:1606.09643 [math.CO], 2016-2017. P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257-260. P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257-260. [Annotated scanned copy]
FORMULA
T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
E.g.f. A(x, q) = Sum_{n>0} (Sum_{k=1..n} T(n, k) * q^k) * x^n / n! = q * ( e^(q*x) - e^x ) / ( q*e^x - e^(q*x) ) satisfies dA / dx = (A + 1) * (A + q). - Michael Somos, Mar 17 2011 For a column listing, n-th term: T(c, n) = c^(n+c-1) + Sum_{i=1..c-1} (-1)^i/i!*(c-i)^(n+c-1)*Product_{j=1..i} (n+c+1-j). - Randall L Rathbun, Jan 23 2002 From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
Four characterizations of Eulerian numbers T(i, n):
1. T(0, n)=1 for n>=1, T(i, 1)=0 for i>=1, T(i, n) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).
2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
3. Let C_n be the unit cube in R^n with vertices (e_1, e_2, ..., e_n) where each e_i is 0 or 1 and all 2^n combinations are used. Then T(i, n)/n! is the volume of C_n between the hyperplanes x_1 + x_2 + ... + x_n = i and x_1 + x_2 + ... + x_n = i+1. Hence T(i, n)/n! is the probability that i <= X_1 + X_2 + ... + X_n < i+1 where the X_j are independent uniform [0, 1] distributions. - See Ehrenborg & Readdy reference.
4. Let f(i, n) = T(i, n)/n!. The f(i, n) are the unique coefficients so that (1/(r-1)^(n+1)) Sum_{i=0..n-1} f(i, n) r^{i+1} = Sum_{j>=0} (j^n)/(r^j) whenever n>=1 and abs(r)>1. (End)
O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - Vladeta Jovovic, Sep 02 2002 Triangle T(n, k), n>0 and k>0, read by rows; given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] (positive integers interspersed with 0's) where DELTA is Del�ham's operator defined in A084938. Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n} E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)). (Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
From the relations between the h- and f-polynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t-1)((t-1)d/dx)^n 1/(t-exp(x)) evaluated at x=0 gives the n-th Eulerian row polynomial in t and the n-th row polynomial in (t-1) of A019538 and A090582. From the Comtet and Copeland references in A139605: ((t+exp(x)-1)d/dx)^(n+1) x gives pairs of the Eulerian polynomials in t as the coefficients of x^0 and x^1 in its Taylor series expansion in x. - Tom Copeland, Oct 05 2008 G.f: 1/(1-x/(1-x*y/1-2*x/(1-2*x*y/(1-3*x/(1-3*x*y/(1-... (continued fraction). - Paul Barry, Mar 24 2010 If n is odd prime, then the following consecutive 2*n+1 terms are 1 modulo n: a((n-1)*(n-2)/2+i), i=0..2*n. This chain of terms is maximal in the sense that neither the previous term nor the following one are 1 modulo n. - _Vladimir Shevelev, Jul 01 2011
For k = 0,1,2,... put G(k,x,t) := x -(1+2^k*t)*x^2/2 +(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 0 and for A008517 when k = 1. The e.g.f. B(x,t) := compositional inverse with respect to x of G(0,x,t) = (exp(x)-exp(x*t))/(exp(x*t)-t*exp(x)) = x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... satisfies the autonomous differential equation dB/dx = (1+B)*(1+t*B) = 1 + (1+t)*B + t*B^2.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the Eulerian polynomials: A(n,t) counts plane increasing trees on n vertices where each vertex has outdegree <= 2, the vertices of outdegree 1 come in 1+t colors and the vertices of outdegree 2 come in t colors. An example is given below. Cf. A008517. Applying [Dominici, Theorem 4.1] gives the following method for calculating the Eulerian polynomials: Let f(x,t) = (1+x)*(1+t*x) and let D be the operator f(x,t)*d/dx. Then A(n+1,t) = D^n(f(x,t)) evaluated at x = 0. (End)
With e.g.f. A(x,t) = G[x,(t-1)]-1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t) = log(t-(t-1)/(1+x))/(t-1). - Tom Copeland, Oct 11 2011 T(2*n+1,n+1) = (2*n+2)*T(2*n,n). (E.g., 66 = 6*11, 2416 = 8*302, ...) - Gary Detlefs, Nov 11 2011 Let {A(n,x)} n>=1 denote the sequence of Eulerian polynomials beginning [1, 1 + x, 1 + 4*x + x^2, ...]. Given two complex numbers a and b, the polynomial sequence defined by R(n,x) := (x+b)^n*A(n+1,(x+a)/(x+b)), n >= 0, satisfies the recurrence equation R(n+1,x) = d/dx((x+a)*(x+b)*R(n,x)). These polynomials give the row generating polynomials for several triangles in the database including A019538 (a = 0, b = 1), A156992 (a = 1, b = 1), A185421 (a = (1+i)/2, b = (1-i)/2), A185423 (a = exp(i*Pi/3), b = exp(-i*Pi/3)) and A185896 (a = i, b = -i). (End)
E.g.f.: 1 + x/(T(0) - x*y), where T(k) = 1 + x*(y-1)/(1 + (k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 07 2013 A) Bivariate e.g.f. A(x,a,b)= (e^(ax)-e^(bx))/(a*e^(bx)-b*e^(ax)) = x + (a+b)*x^2/2! + (a^2+4ab+b^2)*x^3/3! + (a^3+11a^2b+11ab^2+b^3)x^4/4! + ...
B) B(x,a,b)= log((1+ax)/(1+bx))/(a-b) = x - (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 - (a^3+a^2b+ab^2+b^3)x^4/4 + ... = log(1+u.*x), with (u.)^n = u_n = h_(n-1)(a,b) a complete homogeneous polynomial, is the compositional inverse of A(x,a,b) in x (see Drake, p. 56).
C) A(x) satisfies dA/dx = (1+a*A)(1+b*A) and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova).
D) The bivariate Eulerian row polynomials are generated by the iterated derivative ((1+ax)(1+bx)d/dx)^n x evaluated at x=0 (see A145271). E) A(x,a,b)= -(e^(-ax)-e^(-bx))/(a*e^(-ax)-b*e^(-bx)), A(x,-1,-1) = x/(1+x), and B(x,-1,-1) = x/(1-x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1-ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the n-th Eulerian polynomial A(n,x) = (x - 1)^n * log(x) * Integral_{u>=0} (ceiling(u))^n * x^(-u) du. - Peter Bala, Feb 06 2015 Sum_{j>=0} j^n/e^j, for n>=0, equals Sum_{k=1..n} T(n,k)e^k/(e-1)^(n+1), a rational function in the variable "e" which evaluates, approximately, to n! when e = A001113 = 2.71828... - Richard R. Forberg, Feb 15 2015 For a fixed k, T(n,k) ~ k^n, proved by induction. - Ran Pan, Oct 12 2015 From A145271, multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n (1+a*x)*(1+b*x) evaluated at x= 0, i.e., g_0 = 1, g_1 = (a+b), g_2 = 2ab, and g_n = 0 otherwise, to obtain the tridiagonal matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then the m-th bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^(m-1) (1, a+b, 2ab, 0, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. Also, P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^m (0, 1, 0, ...)^T. - Tom Copeland, Aug 02 2016 Cumulatively summing a row generates the n starting terms of the n-th differences of the n-th powers. Applying the finite difference method to x^n, these terms correspond to those before constant n! in the lowest difference row. E.g., T(4,k) is summed as 0+1=1, 1+11=12, 12+11=23, 23+1=4!. See A101101, A101104, A101100, A179457. - Andy Nicol, May 25 2024
EXAMPLE
The triangle T(n, k) begins:
n\k 1 2 3 4 5 6 7 8 9 10 ...
1: 1
2: 1 1
3: 1 4 1
4: 1 11 11 1
5: 1 26 66 26 1
6: 1 57 302 302 57 1
7: 1 120 1191 2416 1191 120 1
8: 1 247 4293 15619 15619 4293 247 1
9: 1 502 14608 88234 156190 88234 14608 502 1
10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013 1
-----------------------------------------------------------------
E.g.f. = (y) * x^1 / 1! + (y + y^2) * x^2 / 2! + (y + 4*y^2 + y^3) * x^3 / 3! + ... - Michael Somos, Mar 17 2011 Let n=7. Then the following 2*7+1=15 consecutive terms are 1(mod 7): a(15+i), i=0..14. - Vladimir Shevelev, Jul 01 2011 Row 3: The plane increasing 0-1-2 trees on 3 vertices (with the number of colored vertices shown to the right of a vertex) are
.
. 1o (1+t) 1o t 1o t
. | / \ / \
. | / \ / \
. 2o (1+t) 2o 3o 3o 2o
. |
. |
. 3o
.
The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.
MAPLE
A008292 := proc(n, k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1, k)+(n-k+1)*procname(n-1, k-1) ; end if; end proc:
MATHEMATICA
t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *) Table[Tally[
Count[#, x_ /; x > 0] & /@ (Differences /@
Permutations[Range[n]])][[;; , 2]], {n, 10}] (* Li Han, Oct 11 2020 *)
PROG
(PARI) {T(n, k) = if( k<1 || k>n, 0, if( n==1, 1, k * T(n-1, k) + (n-k+1) * T(n-1, k-1)))}; /* Michael Somos, Jul 19 1999 */ (PARI) {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */ { A008292(c, n)=c^(n+c-1)+sum(i=1, c-1, (-1)^i/i!*(c-i)^(n+c-1)*prod(j=1, i, n+c+1-j))} (Haskell)
import Data.List (genericLength)
a008292 n k = a008292_tabl !! (n-1) !! (k-1)
a008292_row n = a008292_tabl !! (n-1)
a008292_tabl = iterate f [1] where
f xs = zipWith (+)
(zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
where ks = [1 .. 1 + genericLength xs]
(Python)
from sympy import binomial
def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 08 2017 (R)
T <- function(n, k) {
S <- numeric()
for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
return(sum(S))
}
for (n in 1:10){
for (k in 1:n) print(T(n, k))
(GAP) Flat(List([1..10], n->List([1..n], k->Sum([0..k], j->(-1)^j*(k-j)^n*Binomial(n+1, j))))); # Muniru A Asiru, Jun 29 2018 (Sage) [[sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in (0..k)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Feb 23 2019 (Magma) Eulerian:= func< n, k | (&+[(-1)^j*Binomial(n+1, j)*(k-j+1)^n: j in [0..k+1]]) >; [[Eulerian(n, k): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Apr 15 2019
CROSSREFS
Cf. A019538, A028246, A048993, A048994, A049019, A086885, A090582, A129185, A131758, A139605, A173018.
Sixth powers: a(n) = n^6.
(Formerly M5330 N2318)
+10
202
0, 1, 64, 729, 4096, 15625, 46656, 117649, 262144, 531441, 1000000, 1771561, 2985984, 4826809, 7529536, 11390625, 16777216, 24137569, 34012224, 47045881, 64000000, 85766121, 113379904, 148035889, 191102976, 244140625, 308915776, 387420489, 481890304
COMMENTS
Totally multiplicative sequence with a(p) = p^6 for prime p. - Jaroslav Krizek, Nov 01 2009 Numbers n for which the order of the torsion subgroup of the elliptic curve y^2 = x^3 + n is t = 6, cf. Gebel link. - Artur Jasinski, Jun 30 2010 The binomial transform yields A056468. The inverse binomial transform yields the (finite) 0, 1, 62, 540, ..., 720, the 6th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013 For n > 0, a(n) is the largest number k such that k + n^3 divides k^2 + n^3. - Derek Orr, Oct 01 2014
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity, eq. (6.37).
Granino A. Korn and Theresa M.Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
G.f.: -x*(1+x)*(x^4+56*x^3+246*x^2+56*x+1) / (x-1)^7. - Simon Plouffe in his 1992 dissertation E.g.f.: (x + 31x^2 + 90x^3 + 65x^4 + 15x^5 + x^6)*exp(x). Generally, the e.g.f. for n^m is Sum_{k=1..m} A008277(m,k)*x^k*exp(x). - Geoffrey Critzer, Aug 25 2013 Signature {7, -21, 35, -35, 21, -7, 1}.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) + 720. (End)
a(n) == 1 (mod 7) if gcd(n, 7) = 1, otherwise a(n) == 0 (mod 7). See A109720. - Jake Lawrence, May 28 2016 Dirichlet g.f.: zeta(s-6).
Sum_{n>=1} 1/a(n) = Pi^6/945 = A013664. (End) a(n) = Sum_{k=1..6} Eulerian(6, k)*binomial(n+6-k, 6), with Eulerian(6, k) = A008292(6, k) (the numbers are 1, 57, 302, 302, 57, 1) for n >= 0. Worpitzki's identity for powers of 6. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019 Sum_{n>=1} (-1)^(n+1)/a(n) = 31*zeta(6)/32 = 31*Pi^6/30240 ( A275703). - Amiram Eldar, Oct 08 2020 Product_{n>=1} (1 + 1/a(n)) = (cosh(Pi)-cos(sqrt(3)*Pi))*sinh(Pi)/(2*Pi^3).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)^2/(6*Pi^2). (End)
EXAMPLE
The 6th powers of the first few integers are: 0^6 = 0 = a(0), 1^6 = 1 = a(1), 2^6 = 64 = a(2), 3^6 = 9^3 = 729 = a(3), 4^6 = 2^12 = 4096 = a(4), 5^6 = 25^3 = 15625 = a(5), etc.
PROG
(Haskell)
a001014 n = a001014_list !! n
EXTENSIONS
Comments from 2010 - 2011 edited by M. F. Hasler, Jul 05 2024
Fifth powers: a(n) = n^5.
(Formerly M5231 N2277)
+10
184
0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049, 100000, 161051, 248832, 371293, 537824, 759375, 1048576, 1419857, 1889568, 2476099, 3200000, 4084101, 5153632, 6436343, 7962624, 9765625, 11881376, 14348907, 17210368, 20511149
COMMENTS
Totally multiplicative sequence with a(p) = p^5 for prime p. - Jaroslav Krizek, Nov 01 2009 The binomial transform yields A059338. The inverse binomial transform yields the (finite) 0, 1, 30, 150, 240, 120, the 5th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
G.f.: x*(1+26*x+66*x^2+26*x^3+x^4) / (x-1)^6. [ Simon Plouffe in his 1992 dissertation] a(n) = 5*a(n-1) - 10* a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) + 120. - Ant King, Sep 23 2013 a(n) = n + Sum_{j=0..n-1}{k=1..4}binomial(5,k)*j^(5-k). - Patrick J. McNab, Mar 28 2016 a(n) = Sum_{k=0..n-1} A300656(n,k). (End) a(n) = Sum_{k=1..5} Eulerian(5, k)*binomial(n+5-k, 5), with Eulerian(5, k) = A008292(5, k), the numbers 1, 26, 66, 26, 1, for n >= 0. Worpitzki's identity for powers of 5. See. e.g., Graham et al., eq. (6, 37) (using A173018, the row reversed version of A123125). - Wolfdieter Lang, Jul 17 2019 Sum_{n>=1} 1/a(n) = zeta(5) ( A013663). Sum_{n>=1} (-1)^(n+1)/a(n) = 15*zeta(5)/16 ( A267316). (End)
PROG
(Haskell)
(Maxima) makelist(n^5, n, 0, 30); /* Martin Ettl, Nov 12 2012 */
Euler's triangle: triangle of Eulerian numbers T(n,k) (n>=0, 0 <= k <= n) read by rows.
+10
128
1, 1, 0, 1, 1, 0, 1, 4, 1, 0, 1, 11, 11, 1, 0, 1, 26, 66, 26, 1, 0, 1, 57, 302, 302, 57, 1, 0, 1, 120, 1191, 2416, 1191, 120, 1, 0, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 0, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 0, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1, 0
COMMENTS
This version indexes the Eulerian numbers in the same way as Graham et al.'s Concrete Mathematics (see references section). The traditional indexing, used by Riordan, Comtet and others, is given in A008292, which is the main entry for the Eulerian numbers. Triangle T(n,k), read by rows, given by [1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...] DELTA [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,8,0,9,...] where DELTA is the operator defined in A084938. - Philippe Del�ham Sep 30 2011 [ E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [ -P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials (e.g., E(2,t)= 1+t) and P(n,t) are the polynomials related to polylogarithms in A131758. - Tom Copeland, Oct 03 2014 See A131758 for connections of the evaluation of these polynomials at -1 (alternating row sum) to the Euler, Genocchi, Bernoulli, and zag/tangent numbers and values of the Riemann zeta function and polylogarithms. See also A119879 for the Swiss-knife polynomials. - Tom Copeland, Oct 20 2015
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, table 254.
See A008292 for additional references and links.
FORMULA
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(n+1, j)*(k+1-j)^n. - G. C. Greubel, Feb 25 2019 T(n, k) = (-1)^n*(n+1)!*[x^k][t^n](1 + log((x*exp(-t) - exp(-t*x))/(x-1))/(t*x)). - Peter Luschny, Aug 12 2022
EXAMPLE
Triangle begins:
[ 0] 1,
[ 1] 1, 0,
[ 2] 1, 1, 0,
[ 3] 1, 4, 1, 0,
[ 4] 1, 11, 11, 1, 0,
[ 5] 1, 26, 66, 26, 1, 0,
[ 6] 1, 57, 302, 302, 57, 1, 0,
[ 7] 1, 120, 1191, 2416, 1191, 120, 1, 0,
[ 8] 1, 247, 4293, 15619, 15619, 4293, 247, 1, 0,
[ 9] 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 0,
[10] 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1, 0.
MAPLE
T:= proc(n, k) option remember;
if k=0 and n>=0 then 1
elif k<0 or k>n then 0
else (n-k) * T(n-1, k-1) + (k+1) * T(n-1, k)
fi
end:
# Maple since version 13:
# Or:
egf := 1 + log((x*exp(-t) - exp(-t*x))/(x-1))/(t*x):
ser := series(egf, t, 12): ct := n -> coeff(ser, t, n):
seq(print(seq((-1)^n*(n+1)!*coeff(ct(n), x, k), k=0..n)), n=0..8); # Peter Luschny, Aug 12 2022
MATHEMATICA
t[n_ /; n >= 0, 0] = 1; t[n_, k_] /; k < 0 || k > n = 0;
t[n_, k_] := t[n, k] = (n-k)*t[n-1, k-1] + (k+1)*t[n-1, k]; Flatten[Table[t[n, k], {n, 0, 11}, {k, 0, n}]][[1 ;; 60]]
<< Combinatorica`
Flatten[Table[Eulerian[n, k], {n, 0, 20}, {k, 0, n}]]
(* To generate the table of the numbers T(n, k) *)
RecurrenceTable[{T[n + 1, k + 1] == (n - k) T[n, k] + (k + 2) T[n, k + 1], T[0, k] == KroneckerDelta[k]}, T, {n, 0, 12}, {k, 0, 12}] (* Emanuele Munarini, Jan 03 2018 *) Table[If[n==0, 1, Sum[(-1)^j*Binomial[n+1, j]*(k+1-j)^n, {j, 0, k+1}]], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 25 2019 *)
PROG
(Sage)
@CachedFunction
def eulerian1(n, k):
if k==0: return 1
if k==n: return 0
return eulerian1(n-1, k)*(k+1)+eulerian1(n-1, k-1)*(n-k)
for n in (0..9): [eulerian1(n, k) for k in(0..n)] # Peter Luschny, Nov 11 2012 (Sage) [1] + [[sum((-1)^(k-j+1)*binomial(n+1, k-j+1)*j^n for j in (0..k+1)) for k in (0..n)] for n in (1..12)] # G. C. Greubel, Feb 25 2019 (Haskell)
a173018 n k = a173018_tabl !! n !! k
a173018_row n = a173018_tabl !! n
a173018_tabl = map reverse a123125_tabl
(Magma) [[n le 0 select 1 else (&+[(-1)^j*Binomial(n+1, j)*(k-j+1)^n: j in [0..k+1]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 25 2019 (Magma) T:= func< n, k | n eq 0 select 1 else &+[(-1)^(k-j+1)*Binomial(n+1, k-j+1)*j^n: j in [0..k+1]] >;
[T(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 28 2020 (PARI) T(n, k) = if(n==0, 1, sum(j=0, k+1, (-1)^(k-j+1)*binomial(n+1, k-j+1)*j^n)); \\ G. C. Greubel, Feb 28 2020
CROSSREFS
Cf. A123125 (row reversed version). For this triangle read mod m for m=2 through 10 see A290452- A290460. See also A047999 for the mod 2 version.
Triangular array a(n,k) = (1/k)*Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.
+10
63
1, 1, 1, 1, 3, 2, 1, 7, 12, 6, 1, 15, 50, 60, 24, 1, 31, 180, 390, 360, 120, 1, 63, 602, 2100, 3360, 2520, 720, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 362880
COMMENTS
Let M = n X n matrix with (i,j)-th entry a(n+1-j, n+1-i), e.g., if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n-1)], let b = [b(0)..b(n-1)] be its inverse binomial transform and let c = [c(0)..c(n-1)] = M^(-1)*transpose(b). Then s(k) = Sum_{i=0..n-1} b(i)*binomial(k,i) = Sum_{i=0..n-1} c(i)*k^i, k=0..n-1. - Gary W. Adamson, Nov 11 2001 Julius Worpitzky's 1883 algorithm generates Bernoulli numbers.
By way of example [Wikipedia]:
B0 = 1;
B1 = 1/1 - 1/2;
B2 = 1/1 - 3/2 + 2/3;
B3 = 1/1 - 7/2 + 12/3 - 6/4;
B4 = 1/1 - 15/2 + 50/3 - 60/4 + 24/5;
B5 = 1/1 - 31/2 + 180/3 - 390/4 + 360/5 - 120/6;
B6 = 1/1 - 63/2 + 602/3 - 2100/4 + 3360/5 - 2520/6 + 720/7;
...
Note that in this algorithm, odd n's for the Bernoulli numbers sum to 0, not 1, and the sum for B1 = 1/2 = (1/1 - 1/2). B3 = 0 = (1 - 7/2 + 13/3 - 6/4) = 0. The summation for B4 = -1/30. (End)
Pursuant to Worpitzky's algorithm and given M = A028246 as an infinite lower triangular matrix, M * [1/1, -1/2, 1/3, ...] (i.e., the Harmonic series with alternate signs) = the Bernoulli numbers starting [1/1, 1/2, 1/6, ...]. - Gary W. Adamson, Mar 22 2012 G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1 x + (2 + t)*x^2/2! + (6 + 6t + t^2)*x^3/3! + ... gives row polynomials for A090582, the f-polynomials for the permutohedra (see A019538). G(x,t-1) = 1 + 1*x + (1 + t)*x^2 / 2! + (1 + 4t + t^2)*x^3 / 3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra. G[(t+1)x,-1/(t+1)] = 1 + (1+ t) x + (1 + 3t + 2 t^2) x^2 / 2! + ... gives row polynomials for the present triangle. (End)
The Worpitzky triangle seems to be an apt name for this triangle. - Johannes W. Meijer, Jun 18 2009 If Pascal's triangle is written as a lower triangular matrix and multiplied by A028246 written as an upper triangular matrix, the product is a matrix where the (i,j)-th term is (i+1)^j. For example, 1,0,0,0 1,1,1, 1 1,1, 1, 1
1,1,0,0 * 0,1,3, 7 = 1,2, 4, 8
1,2,1,0 0,0,2,12 1,3, 9,27
1,3,3,1 0,0,0, 6 1,4,16,64
So, numbering all three matrices' rows and columns starting at 0, the (i,j) term of the product is (i+1)^j. - Jack A. Cohen (ProfCohen(AT)comcast.net), Aug 03 2010
The Fi1 and Fi2 triangle sums are both given by sequence A000670. For the definition of these triangle sums see A180662. The mirror image of the Worpitzky triangle is A130850. - Johannes W. Meijer, Apr 20 2011 Let S_n(m) = 1^m + 2^m + ... + n^m. Then, for n >= 0, we have the following representation of S_n(m) as a linear combination of the binomial coefficients:
S_n(m) = Sum_{i=1..n+1} a(i+n*(n+1)/2)*C(m,i). E.g., S_2(m) = a(4)*C(m,1) + a(5)*C(m,2) + a(6)*C(m,3) = C(m,1) + 3*C(m,2) + 2*C(m,3). - Vladimir Shevelev, Dec 21 2011 Given the set X = [1..n] and 1 <= k <= n, then a(n,k) is the number of sets T of size k of subset S of X such that S is either empty or else contains 1 and another element of X and such that any two elemements of T are either comparable or disjoint. - Michael Somos, Apr 20 2013 Working with the row and column indexing starting at -1, a(n,k) gives the number of k-dimensional faces in the first barycentric subdivision of the standard n-dimensional simplex (apply Brenti and Welker, Lemma 2.1). For example, the barycentric subdivision of the 2-simplex (a triangle) has 1 empty face, 7 vertices, 12 edges and 6 triangular faces giving row 4 of this triangle as (1,7,12,6). Cf. A053440. - Peter Bala, Jul 14 2014 See A074909 and above g.f.s for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014 An e.g.f. G(x,t) = exp[P(.,t)x] = 1/t - 1/[t+(1-t)(1-e^(-xt^2))] = (1-t) * x + (-2t + 3t^2 - t^3) * x^2/2! + (6t^2 - 12t^3 + 7t^4 - t^5) * x^3/3! + ... for the shifted, reverse, signed polynomials with the first element nulled, is generated by the infinitesimal generator g(u,t)d/du = [(1-u*t)(1-(1+u)t)]d/du, i.e., exp[x * g(u,t)d/du] u eval. at u=0 generates the polynomials. See A019538 and the G. Rzadkowski link below for connections to the Bernoulli and Eulerian numbers, a Ricatti differential equation, and a soliton solution to the KdV equation. The inverse in x of this e.g.f. is Ginv(x,t) = (-1/t^2)*log{[1-t(1+x)]/[(1-t)(1-tx)]} = [1/(1-t)]x + [(2t-t^2)/(1-t)^2]x^2/2 + [(3t^2-3t^3+t^4)/(1-t)^3]x^3/3 + [(4t^3-6t^4+4t^5-t^6)/(1-t)^4]x^4/4 + ... . The numerators are signed, shifted A135278 (reversed A074909), and the rational functions are the columns of A074909. Also, dG(x,t)/dx = g(G(x,t),t) (cf. A145271). (Analytic G(x,t) added, and Ginv corrected and expanded on Dec 28 2015.) - Tom Copeland, Nov 21 2014 The operator R = x + (1 + t) + t e^{-D} / [1 + t(1-e^(-D))] = x + (1+t) + t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... contains an e.g.f. of the reverse row polynomials of the present triangle, i.e., A123125 * A007318 (with row and column offset 1 and 1). Umbrally, R^n 1 = q_n(x;t) = (q.(0;t)+x)^n, with q_m(0;t) = (t+1)^(m+1) - t^(m+1), the row polynomials of A074909, and D = d/dx. In other words, R generates the Appell polynomials associated with the base sequence A074909. For example, R 1 = q_1(x;t) = (q.(0;t)+x) = q_1(0;t) + q__0(0;t)x = (1+2t) + x, and R^2 1 = q_2(x;t) = (q.(0;t)+x)^2 = q_2(0:t) + 2q_1(0;t)x + q_0(0;t)x^2 = 1+3t+3t^2 + 2(1+2t)x + x^2. Evaluating the polynomials at x=0 regenerates the base sequence. With a simple sign change in R, R generates the Appell polynomials associated with A248727. - Tom Copeland, Jan 23 2015 The e.g.f. E(n, x) for {S(n, m)}_{m>=0} with S(n, m) = Sum_{k=1..m} k^n, n >= 0, (with undefined sum put to 0) is exp(x)*R(n+1, x) with the exponential row polynomials R(n, x) = Sum_{k=1..n} a(n, k)*x^k/k!. E.g., e.g.f. for n = 2, A000330: exp(x)*(1*x/1!+3*x^2/2!+2*x^3/3!). The o.g.f. G(n, x) for {S(n, m)}_{m >=0} is then found by Laplace transform to be G(n, 1/p) = p*Sum_{k=1..n} a(n+1, k)/(p-1)^(2+k).
Hence G(n, x) = x/(1 - x)^(n+2)*Sum_{k=1..n} A008292(n,k)*x^(k-1). E.g., n=2: G(2, 1/p) = p*(1/(p-1)^2 + 3/(p-1)^3 + 2/(p-1)^4) = p^2*(1+p)/(p-1)^4; hence G(2, x) = x*(1+x)/(1-x)^4.
This works also backwards: from the o.g.f. to the e.g.f. of {S(n, m)}_{m>=0}. (End)
a(n,k) is the number of k-tuples of pairwise disjoint and nonempty subsets of a set of size n. - Dorian Guyot, May 21 2019 a(n-1,k) is the number of chains of length k in a partially ordered set formed from subsets of an n-element set ordered by inclusion such that the first term of the chains is either the empty set or an n-element set.
Also, a(n-1,k) is the number of distinct k-level rooted fuzzy subsets of an n-set ordered by set inclusion. (End)
The relations on p. 34 of Hasan (also p. 17 of Franco and Hasan) agree with the relation between A019538 and this entry given in the formula section. - Tom Copeland, May 14 2020 T(n,k) is the size of the Green's L-classes in the D-classes of rank (k-1) in the semigroup of partial transformations on an (n-1)-set. - Geoffrey Critzer, Jan 09 2023 T(n,k) is the number of strongly connected binary relations on [n] that have period k ( A367948) and index 1. See Theorem 5.4.25(6) in Ki Hang Kim reference. - Geoffrey Critzer, Dec 07 2023
REFERENCES
Ki Hang Kim, Boolean Matrix Theory and Applications, Marcel Dekker, New York and Basel (1982).
LINKS
A. Riskin and D. Beckwith, Problem 10231, Amer. Math. Monthly, 102 (1995), 175-176.
FORMULA
a(n, k) = S2(n, k)*(k-1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538. Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Del�ham's operator defined in A084938, but the notation is different. The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t) = t(t+1)(d/dt)P(n, t) for n >= 1 (see the Riskin and Beckwith reference). - Emeric Deutsch, Aug 09 2005 Delta-matrix as can be read from H. Hasse's proof of a connection between the zeta-function and Bernoulli numbers (see link below).
Let P = lower triangular matrix with entries P[row,col] = binomial(row,col).
Let J = unit matrix with alternating signs J[r,r]=(-1)^r.
Let N(m) = column matrix with N(m)(r) = (r+1)^m, N(1)--> natural numbers.
Let V = Vandermonde matrix with V[r,c] = (r+1)^c.
V is then also N(0)||N(1)||N(2)||N(3)... (indices r,c always beginning at 0).
Then Delta = P*J * V and B' = N(-1)' * Delta, where B is the column matrix of Bernoulli numbers and ' means transpose, or for the single k-th Bernoulli number B_k with the appropriate column of Delta,
B_k = N(-1)' * Delta[ *,k ] = N(-1)' * P*J * N(k).
Using a single column instead of V and assuming infinite dimension, H. Hasse showed that in x = N(-1) * P*J * N(s), where s can be any complex number and s*zeta(1-s) = x.
His theorem reads: s*zeta(1-s) = Sum_{n>=0..inf} (n+1)^-1*delta(n,s), where delta(n,s) = Sum_{j=0..n} (-1)^j * binomial(n,j) * (j+1)^s.
(End)
a(n,k) = k*a(n-1,k) + (k-1)*a(n-1,k-1) with a(n,1) = 1 and a(n,n) = (n-1)!. - Johannes W. Meijer, Jun 18 2009 Rephrasing the Meijer recurrence above: Let M be the (n+1)X(n+1) bidiagonal matrix with M(r,r) = M(r,r+1) = r, r >= 1, in the two diagonals and the rest zeros. The row a(n+1,.) of the triangle is row 1 of M^n. - Gary W. Adamson, Jun 24 2011 With e.g.f.. A(x,t) = G[(t+1)x,-1/(t+1)]-1 (from 2008 comment) = -1 + 1/[1-(1+t)(1-e^(-x))] = (1+t)x + (1+3t+2t^2)x^2/2! + ..., the comp. inverse in x is
B(x,t)= -log(t/(1+t)+1/((1+t)(1+x))) = (1/(1+t))x - ((1+2t)/(1+t)^2)x^2/2 + ((1+3t+3t^2)/(1+t)^3)x^3/3 + .... The numerators are the row polynomials of A074909, and the rational functions are (omitting the initial constants) signed columns of the re-indexed Pascal triangle A007318. Let h(x,t)= 1/(dB/dx) = (1+x)(1+t(1+x)), then the row polynomial P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(1,t)=1+t. (Series added Dec 29 2015.)(End)
Let <n,k> denote the Eulerian numbers A173018(n,k), then T(n,k) = Sum_{j=0..n} <n,j>*binomial(n-j,n-k). - Peter Luschny, Jul 12 2013 Matrix product A007318 * A131689. The n-th row polynomial R(n,x) = Sum_{k >= 1} k^(n-1)*(x/(1 + x))^k, valid for x in the open interval (-1/2, inf). Cf A038719. R(n,-1/2) = (-1)^(n-1)*(2^n - 1)*Bernoulli(n)/n. - Peter Bala, Jul 14 2014 n-th row polynomial R(n,x) = (1+x) o (1+x) o ... o (1+x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E11 in the Bala link. - Peter Bala, Jan 12 2018 Sum_{i=0..k} binomial(k,i) * a(n,i) = (k+1)^n.
Sum_{k=0..n} a(n,k) = 2* A000670(n). (End)
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry, A028246, are given by x^n * A_n(1 + 1/x;0). Other specializations of A_n(x;y) give A046802, A090582, A119879, A130850, and A248727. - Tom Copeland, Jan 24 2020 The row generating polynomials R(n,x) = Sum_{i=1..n} a(n,i) * x^i satisfy the recurrence equation R(n+1,x) = R(n,x) + Sum_{k=0..n-1} binomial(n-1,k) * R(k+1,x) * R(n-k,x) for n >= 1 with initial value R(1,x) = x. - Werner Schulte, Jun 17 2021
EXAMPLE
The triangle a(n, k) starts:
n\k 1 2 3 4 5 6 7 8 9
1: 1
2: 1 1
3: 1 3 2
4: 1 7 12 6
5: 1 15 50 60 24
6: 1 31 180 390 360 120
7: 1 63 602 2100 3360 2520 720
8: 1 127 1932 10206 25200 31920 20160 5040
9: 1 255 6050 46620 166824 317520 332640 181440 40320
-----------------------------------------------------
Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.
Also, for power sums, we have
S_0(n) = C(n,1);
S_1(n) = C(n,1) + C(n,2);
S_2(n) = C(n,1) + 3*C(n,2) + 2*C(n,3);
S_3(n) = C(n,1) + 7*C(n,2) + 12*C(n,3) + 6*C(n,4);
S_4(n) = C(n,1) + 15*C(n,2) + 50*C(n,3) + 60*C(n,4) + 24*C(n,5); etc.
(End)
For X = [1,2,3], the sets T are {{}}, {{},{1,2}}, {{},{1,3}}, {{},{1,2,3}}, {{},{1,2},{1,2,3}}, {{},{1,3},{1,2,3}} and so a(3,1)=1, a(3,2)=3, a(3,3)=2. - Michael Somos, Apr 20 2013
MAPLE
a := (n, k) -> add((-1)^(k-i)*binomial(k, i)*i^n, i=0..k)/k;
seq(print(seq(a(n, k), k=1..n)), n=1..10);
T := (n, k) -> add(eulerian1(n, j)*binomial(n-j, n-k), j=0..n):
seq(print(seq(T(n, k), k=0..n)), n=0..9); # Peter Luschny, Jul 12 2013
MATHEMATICA
a[n_, k_] = Sum[(-1)^(k-i) Binomial[k, i]*i^n, {i, 0, k}]/k; Flatten[Table[a[n, k], {n, 10}, {k, n}]] (* Jean-Fran�ois Alcover, May 02 2011 *)
PROG
(PARI) {T(n, k) = if( k<0 || k>n, 0, n! * polcoeff( (x / log(1 + x + x^2 * O(x^n) ))^(n+1), n-k))}; /* Michael Somos, Oct 02 2002 */ (PARI) {T(n, k) = stirling(n, k, 2)*(k-1)!}; \\ G. C. Greubel, May 31 2019 (Sage)
x = polygen(ZZ, 'x')
A = []
for m in range(0, n, 1) :
A.append((-x)^m)
for j in range(m, 0, -1):
A[j - 1] = j * (A[j - 1] - A[j])
return list(A[0])
(Sage) [[stirling_number2(n, k)*factorial(k-1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, May 30 2019 (Magma) [[StirlingSecond(n, k)*Factorial(k-1): k in [1..n]]: n in [1..10]]; // G. C. Greubel, May 30 2019 (GAP) Flat(List([1..10], n-> List([1..n], k-> Stirling2(n, k)* Factorial(k-1) ))) # G. C. Greubel, May 30 2019 (Python) # Assuming offset (n, k) = (0, 0).
def T(n, k):
if k > n: return 0
if k == 0: return 1
return k*T(n - 1, k - 1) + (k + 1)*T(n - 1, k)
for n in range(9):
print([T(n, k) for k in range(n + 1)]) # Peter Luschny, Apr 26 2022
CROSSREFS
Dropping the column of 1's gives A053440. Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277. Row sums give A000629(n-1) for n >= 1. Cf. A007318, A008292, A046802, A074909, A090582, A123125, A130850, A135278, A141618, A145271, A163626, A248727, A263634. Diagonal gives A000142(n-1), for n >=1. Next-to-last diagonal gives A001710,
EXTENSIONS
Definition corrected by Li Guo, Dec 16 2006
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![plots of exceedances for permutations of [4]](/proxy/https://oeis.org/wiki/File:Exceedances_4.png){kind=link}
