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Zero-one sequence based on pentagonal numbers: a( A000325(k))=a(k); a( A183217(k))=1-a(k); a(1)=0.
+20
4
0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1
FORMULA
Let u= A000217 and v= A014132, so that u(n)=n(3n-1)/2 and v=complement(u) for n>=1. Then a is a self-generating zero-one sequence with initial value a(1)=0 and a(u(k))=a(k); a(v(k))=1-a(k).
MATHEMATICA
a[1] = 0; h = 128;
c = (u[#1] &) /@ Range[h];
d = (Complement[Range[Max[#1]], #1] &)[c]; (* A183217*) Table[a[d[[n]]] = 1 - a[n], {n, 1, h - 1}];
Table[a[c[[n]]] = a[n], {n, 1, h}] (* A189014*) Flatten[Position[%, 0]] (* A189015*) Flatten[Position[%%, 1]] (* A189016*)
2, 2, 4, 32, 4096, 134217728, 288230376151711744, 2658455991569831745807614120560689152, 452312848583266388373324160190187140051835877600158453279131187530910662656
PROG
(PARI) {a(n) = 2^(2^n-n)}
(Magma) [2^(2^n-n): n in [1..10]]; // Altug Alkan, Mar 04 2018
a(n) = 2^n - 1. (Sometimes called Mersenne numbers, although that name is usually reserved for A001348.)
(Formerly M2655 N1059)
+10
1292
0, 1, 3, 7, 15, 31, 63, 127, 255, 511, 1023, 2047, 4095, 8191, 16383, 32767, 65535, 131071, 262143, 524287, 1048575, 2097151, 4194303, 8388607, 16777215, 33554431, 67108863, 134217727, 268435455, 536870911, 1073741823, 2147483647, 4294967295, 8589934591
COMMENTS
This is the Gaussian binomial coefficient [n,1] for q=2.
Number of rank-1 matroids over S_n.
Numbers k such that the k-th central binomial coefficient is odd: A001405(k) mod 2 = 1. - Labos Elemer, Mar 12 2003 Also solutions (with minimum number of moves) for the problem of Benares Temple, i.e., three diamond needles with n discs ordered by decreasing size on the first needle to place in the same order on the third one, without ever moving more than one disc at a time and without ever placing one disc at the top of a smaller one. - Xavier Acloque, Oct 18 2003
a(0) = 0, a(1) = 1; a(n) = smallest number such that a(n)-a(m) == 0 (mod (n-m+1)), for all m. - Amarnath Murthy, Oct 23 2003 Binomial transform of [1, 1/2, 1/3, ...] = [1/1, 3/2, 7/3, ...]; (2^n - 1)/n, n=1,2,3, ... - Gary W. Adamson, Apr 28 2005 Numbers whose binary representation is 111...1. E.g., the 7th term is (2^7) - 1 = 127 = 1111111 (in base 2). - Alexandre Wajnberg, Jun 08 2005 Number of nonempty subsets of a set with n elements. - Michael Somos, Sep 03 2006 For n >= 2, a(n) is the least Fibonacci n-step number that is not a power of 2. - Rick L. Shepherd, Nov 19 2007 Let P(A) be the power set of an n-element set A. Then a(n+1) = the number of pairs of elements {x,y} of P(A) for which x and y are disjoint and for which either x is a subset of y or y is a subset of x. - Ross La Haye, Jan 10 2008 A simpler way to state this is that it is the number of pairs (x,y) where at least one of x and y is the empty set. - Franklin T. Adams-Watters, Oct 28 2011 2^n-1 is the sum of the elements in a Pascal triangle of depth n. - Brian Lewis (bsl04(AT)uark.edu), Feb 26 2008
Starting with offset 1 = the Jacobsthal sequence, A001045, (1, 1, 3, 5, 11, 21, ...) convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 23 2009 If n is even a(n) mod 3 = 0. This follows from the congruences 2^(2k) - 1 ~ 2*2*...*2 - 1 ~ 4*4*...*4 - 1 ~ 1*1*...*1 - 1 ~ 0 (mod 3). (Note that 2*2*...*2 has an even number of terms.) - Washington Bomfim, Oct 31 2009 Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n)=det(A). - Milan Janjic, Jan 26 2010 This is the sequence A(0,1;1,2;2) = A(0,1;3,-2;0) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010 a(n) = S(n+1,2), a Stirling number of the second kind. See the example below. - Dennis P. Walsh, Mar 29 2011 Entries of row a(n) in Pascal's triangle are all odd, while entries of row a(n)-1 have alternating parities of the form odd, even, odd, even, ..., odd.
Define the bar operation as an operation on signed permutations that flips the sign of each entry. Then a(n+1) is the number of signed permutations of length 2n that are equal to the bar of their reverse-complements and avoid the set of patterns {(-2,-1), (-1,+2), (+2,+1)}. (See the Hardt and Troyka reference.) - Justin M. Troyka, Aug 13 2011 This sequence is also the number of proper subsets of a set with n elements. - Mohammad K. Azarian, Oct 27 2011 a(n) is the number k such that the number of iterations of the map k -> (3k +1)/2 == 1 (mod 2) until reaching (3k +1)/2 == 0 (mod 2) equals n. (see the Collatz problem). - Michel Lagneau, Jan 18 2012 For integers a, b, denote by a<+>b the least c >= a such that Hd(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then a(n+1)=a(n)<+>1. Thus this sequence is the Hamming analog of nonnegative integers. - Vladimir Shevelev, Feb 13 2012 Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... apparently A007733. - R. J. Mathar, Aug 10 2012 Start with n. Each n generates a sublist {n-1,n-2,...,1}. Each element of each sublist also generates a sublist. Take the sum of all. E.g., 3->{2,1} and 2->{1}, so a(3)=3+2+1+1=7. - Jon Perry, Sep 02 2012 This is the Lucas U(P=3,Q=2) sequence. - R. J. Mathar, Oct 24 2012 The Mersenne numbers >= 7 are all Brazilian numbers, as repunits in base two. See Proposition 1 & 5.2 in Links: "Les nombres br�siliens". - Bernard Schott, Dec 26 2012 Number of line segments after n-th stage in the H tree. - Omar E. Pol, Feb 16 2013 a(n) is the highest power of 2 such that 2^a(n) divides (2^n)!. - Ivan N. Ianakiev, Aug 17 2013 In computer programming, these are the only unsigned numbers such that k&(k+1)=0, where & is the bitwise AND operator and numbers are expressed in binary. - Stanislav Sykora, Nov 29 2013 Minimal number of moves needed to interchange n frogs in the frogs problem (see for example the NRICH 1246 link or the Britton link below). - N. J. A. Sloane, Jan 04 2014 a(n) !== 4 (mod 5); a(n) !== 10 (mod 11); a(n) !== 2, 4, 5, 6 (mod 7). - Carmine Suriano, Apr 06 2014 After 0, antidiagonal sums of the array formed by partial sums of integers (1, 2, 3, 4, ...). - Luciano Ancora, Apr 24 2015 a(n+1) equals the number of ternary words of length n avoiding 01,02. - Milan Janjic, Dec 16 2015 With offset 0 and another initial 0, the n-th term of 0, 0, 1, 3, 7, 15, ... is the number of commas required in the fully-expanded von Neumann definition of the ordinal number n. For example, 4 := {0, 1, 2, 3} := {{}, {{}}, {{}, {{}}}, {{}, {{}}, {{}, {{}}}}}, which uses seven commas. Also, for n>0, a(n) is the total number of symbols required in the fully-expanded von Neumann definition of ordinal n - 1, where a single symbol (as usual) is always used to represent the empty set and spaces are ignored. E.g., a(5) = 31, the total such symbols for the ordinal 4. - Rick L. Shepherd, May 07 2016 With the quantum integers defined by [n+1]_q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Mersenne numbers are a(n+1) = q^n [n+1]_q with q = sqrt(2), whereas the signed Jacobsthal numbers A001045 are given by q = i * sqrt(2) for i^2 = -1. Cf. A239473. - Tom Copeland, Sep 05 2016 Except for the initial terms, the decimal representation of the x-axis of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 659", "Rule 721" and "Rule 734", based on the 5-celled von Neumann neighborhood initialized with a single on cell. - Robert Price, Mar 14 2017 a(n), n > 1, is the number of maximal subsemigroups of the monoid of order-preserving partial injective mappings on a set with n elements. - James Mitchell and Wilf A. Wilson, Jul 21 2017 Also the number of independent vertex sets and vertex covers in the complete bipartite graph K_{n-1,n-1}. - Eric W. Weisstein, Sep 21 2017 Sum_{k=0..n} p^k is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*p + binomial(i+j-1, i), in this case p=2 (empirical observation). - Tony Foster III, May 11 2019 The rational numbers r(n) = a(n+1)/2^(n+1) = a(n+1)/ A000079(n+1) appear also as root of the n-th iteration f^{[n]}(c; x) = 2^(n+1)*x - a(n+1)*c of f(c; x) = f^{[0]}(c; x) = 2*x - c as r(n)*c. This entry is motivated by a riddle of Johann Peter Hebel (1760 - 1826): Erstes Rechnungsexempel(Ein merkw�rdiges Rechnungs-Exempel) from 1803, with c = 24 and n = 2, leading to the root r(2)*24 = 21 as solution. See the link and reference. For the second problem, also involving the present sequence, see a comment in A130330. - Wolfdieter Lang, Oct 28 2019 a(n) is the sum of the smallest elements of all subsets of {1,2,..,n} that contain n. For example, a(3)=7; the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 7. - Enrique Navarrete, Aug 21 2020 a(n-1) is the number of nonempty subsets of {1,2,..,n} which don't have an element that is the size of the set. For example, for n = 4, a(3) = 7 and the subsets are {2}, {3}, {4}, {1,3}, {1,4}, {3,4}, {1,2,4}. - Enrique Navarrete, Nov 21 2020 Also the number of dominating sets in the complete graph K_n.
Also the number of minimum dominating sets in the n-helm graph for n >= 3. (End)
Conjecture: except for a(2)=3, numbers m such that 2^(m+1) - 2^j - 2^k - 1 is composite for all 0 <= j < k <= m. - Chai Wah Wu, Sep 08 2021 a(n) is the number of three-in-a-rows passing through a corner cell in n-dimensional tic-tac-toe. - Ben Orlin, Mar 15 2022 a(n) == 1 (mod 30) for n == 1 (mod 4);
a(n) == 7 (mod 120) for n == 3 (mod 4);
(a(n) - 1)/30 = (a(n+2) - 7)/120 for n odd;
(a(n) - 1)/30 = (a(n+2) - 7)/120 = A131865(m) for n == 1 (mod 4) and m >= 0 with A131865(0) = 0. (End) a(n) is the number of n-digit numbers whose smallest decimal digit is 8. - Stefano Spezia, Nov 15 2023 Also, number of nodes in a perfect binary tree of height n-1, or: number of squares (or triangles) after the n-th step of the construction of a Pythagorean tree: Start with a segment. At each step, construct squares having the most recent segment(s) as base, and isosceles right triangles having the opposite side of the squares as hypotenuse ("on top" of each square). The legs of these triangles will serve as the segments which are the bases of the squares in the next step. - M. F. Hasler, Mar 11 2024 a(n) is the length of the longest path in the n-dimensional hypercube. - Christian Barrientos, Apr 13 2024 a(n) is the diameter of the n-Hanoi graph. Equivalently, a(n) is the largest minimum number of moves between any two states of the Towers of Hanoi problem (aka problem of Benares Temple described above). - Allan Bickle, Aug 09 2024
REFERENCES
P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 75.
Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134.
Johann Peter Hebel, Gesammelte Werke in sechs B�nden, Herausgeber: Jan Knopf, Franz Littmann und Hansgeorg Schmidt-Bergmann unter Mitarbeit von Ester Stern, Wallstein Verlag, 2019. Band 3, S. 20-21, Loesung, S. 36-37. See also the link below.
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).
D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, "Tower of Hanoi", Penguin Books, 1987, pp. 112-113.
LINKS
John Brillhart et al., Cunningham Project [Factorizations of b^n +- 1, b = 2, 3, 5, 6, 7, 10, 11, 12 up to high powers] [Subscription required]. P. Catarino, H. Campos, and P. Vasco, On the Mersenne sequence, Annales Mathematicae et Informaticae, 46 (2016) pp. 37-53. A. Hardt and J. M. Troyka, Slides (associated with the Hardt and Troyka reference above). Bernard Schott, Les nombres brésiliens, Reprinted from Quadrature, no. 76, avril-juin 2010, pages 30-38, included here with permission from the editors of Quadrature. Eric Weisstein's World of Mathematics, Complete
FORMULA
G.f.: x/((1-2*x)*(1-x)).
E.g.f.: exp(2*x) - exp(x).
E.g.f. if offset 1: ((exp(x)-1)^2)/2.
a(n) = Sum_{k=0..n-1} 2^k. - Paul Barry, May 26 2003 a(n) = a(n-1) + 2*a(n-2) + 2, a(0)=0, a(1)=1. - Paul Barry, Jun 06 2003 Let b(n) = (-1)^(n-1)*a(n). Then b(n) = Sum_{i=1..n} i!*i*Stirling2(n,i)*(-1)^(i-1). E.g.f. of b(n): (exp(x)-1)/exp(2x). - Mario Catalani (mario.catalani(AT)unito.it), Dec 19 2003
a(n+1) = 2*a(n) + 1, a(0) = 0.
a(n) = Sum_{k=1..n} binomial(n, k).
a(n+1) = (n+1)*Sum_{k=0..n} binomial(n, k)/(k+1). - Paul Barry, Aug 06 2004 a(n+1) = Sum_{k=0..n} binomial(n+1, k+1). - Paul Barry, Aug 23 2004 Inverse binomial transform of A001047. Also U sequence of Lucas sequence L(3, 2). - Ross La Haye, Feb 07 2005 a(n) = J_n(2), where J_n is the n-th Jordan Totient function: ( A007434, is J_2). a(n) = Sum_{d|2} d^n*mu(2/d). (End)
a(n) = det(|s(i+2,j+1)|, 1 <= i,j <= n-1), where s(n,k) are Stirling numbers of the first kind. - Mircea Merca, Apr 06 2013 G.f.: Q(0), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 - 1/(2*4^k - 8*x*16^k/(4*x*4^k - 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013 E.g.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - (k+1)/Q(k+1))); (continued fraction).
G.f.: Q(0), where Q(k) = 1 - 1/(2^k - 2*x*4^k/(2*x*2^k - 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 23 2013 a(n) = Sum_{t_1+2*t_2+...+n*t_n=n} n*multinomial(t_1+t_2 +...+t_n,t_1,t_2,...,t_n)/(t_1+t_2 +...+t_n). - Mircea Merca, Dec 06 2013 Sum_{n>=0} a(n)/n! = A090142. (End) Convolution of binomial coefficient C(n,a(k)) with itself is C(n,a(k+1)) for all k >= 3. - Anton Zakharov, Sep 05 2016 a(n) = n + Sum_{j=1..n-1} (n-j)*2^(j-1). See a Jun 14 2017 formula for A000918(n+1) with an interpretation. - Wolfdieter Lang, Jun 14 2017 a(n+m) = a(n)*a(m) + a(n) + a(m). - Yuchun Ji, Jul 27 2018 a(n+m) = a(n+1)*a(m) - 2*a(n)*a(m-1). - Taras Goy, Dec 23 2018 a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(i + j - 1, j)*2 + binomial(i+j-1, i) (empirical observation). - Tony Foster III, May 11 2019
EXAMPLE
For n=3, a(3)=S(4,2)=7, a Stirling number of the second kind, since there are 7 ways to partition {a,b,c,d} into 2 nonempty subsets, namely,
{a}U{b,c,d}, {b}U{a,c,d}, {c}U{a,b,d}, {d}U{a,b,c}, {a,b}U{c,d}, {a,c}U{b,d}, and {a,d}U{b,c}. - Dennis P. Walsh, Mar 29 2011 Since a(3) = 7, there are 7 signed permutations of 4 that are equal to the bar of their reverse-complements and avoid {(-2,-1), (-1,+2), (+2,+1)}. These are:
(+1,+2,-3,-4),
(+1,+3,-2,-4),
(+1,-3,+2,-4),
(+2,+4,-1,-3),
(+3,+4,-1,-2),
(-3,+1,-4,+2),
(-3,-4,+1,+2). (End)
G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 63*x^6 + 127*x^7 + ...
For the Towers of Hanoi problem with 2 disks, the moves are as follows, so a(2) = 3.
12|_|_ -> 2|1|_ -> _|1|2 -> _|_|12 - Allan Bickle, Aug 07 2024
MAPLE
A000225 := n->2^n-1; [ seq(2^n-1, n=0..50) ];
MATHEMATICA
Array[2^# - 1 &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
CoefficientList[Series[1/(1 - 3 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
PROG
(Haskell)
a000225 = (subtract 1) . (2 ^)
a000225_list = iterate ((+ 1) . (* 2)) 0
(PARI) concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015 (SageMath)
def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019 (Python)
CROSSREFS
Cf. A001348 (Mersenne numbers with n prime).
Eulerian numbers (Euler's triangle: column k=2 of A008292, column k=1 of A173018).
(Formerly M3416 N1382)
+10
204
0, 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558
COMMENTS
There are 2 versions of Euler's triangle:
* A008292 Classic version of Euler's triangle used by Comtet (1974). * A173018 Version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990). Euler's triangle rows and columns indexing conventions:
* A008292 The rows and columns of the Eulerian triangle are both indexed starting from 1. (Classic version: used in the classic books by Riordan and Comtet.) * A173018 The rows and columns of the Eulerian triangle are both indexed starting from 0. (Graham et al.) Number of Dyck paths of semilength n having exactly one long ascent (i.e., ascent of length at least two). Example: a(4)=11 because among the 14 Dyck paths of semilength 4, the paths that do not have exactly one long ascent are UDUDUDUD (no long ascent), UUDDUUDD and UUDUUDDD (two long ascents). Here U=(1,1) and D=(1,-1). Also number of ordered trees with n edges having exactly one branch node (i.e., vertex of outdegree at least two). - Emeric Deutsch, Feb 22 2004 Number of permutations of {1,2,...,n} with exactly one descent (i.e., permutations (p(1),p(2),...,p(n)) such that #{i: p(i)>p(i+1)}=1). E.g., a(3)=4 because the permutations of {1,2,3} with one descent are 132, 213, 231 and 312.
Number of partitions of an n-set having exactly one block of size > 1. Example: a(4)=11 because, if the partitioned set is {1,2,3,4}, then we have 1234, 123|4, 124|3, 134|2, 1|234, 12|3|4, 13|2|4, 14|2|3, 1|23|4, 1|24|3 and 1|2|34. - Emeric Deutsch, Oct 28 2006 n divides a(n+1) for n = A014741(n) = {1, 2, 6, 18, 42, 54, 126, 162, 294, 342, 378, 486, 882, 1026, ...}. - Alexander Adamchuk, Nov 03 2006 (Number of permutations avoiding patterns 321, 2413, 3412, 21534) minus one. - Jean-Luc Baril, Nov 01 2007, Mar 21 2008 The chromatic invariant of the prism graph P_n for n >= 3. - Jonathan Vos Post, Aug 29 2008 Decimal integer corresponding to the result of XORing the binary representation of 2^n - 1 and the binary representation of n with leading zeros. This sequence and a few others are syntactically similar. For n > 0, let D(n) denote the decimal integer corresponding to the binary number having n consecutive 1's. Then D(n).OP.n represents the n-th term of a sequence when .OP. stands for a binary operator such as '+', '-', '*', 'quotentof', 'mod', 'choose'. We then get the various sequences A136556, A082495, A082482, A066524, A000295, A052944. Another syntactically similar sequence results when we take the n-th term as f(D(n)).OP.f(n). For example if f='factorial' and .OP.='/', we get ( A136556)( A000295) ; if f='squaring' and .OP.='-', we get ( A000295)( A052944). - K.V.Iyer, Mar 30 2009 Chromatic invariant of the prism graph Y_n.
Number of labelings of a full binary tree of height n-1, such that each path from root to any leaf contains each label from {1,2,...,n-1} exactly once. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009
Also number of nontrivial equivalence classes generated by the weak associative law X((YZ)T)=(X(YZ))T on words with n open and n closed parentheses. Also the number of join (resp. meet)-irreducible elements in the pruning-grafting lattice of binary trees with n leaves. - Jean Pallo, Jan 08 2010
Nonzero terms of this sequence can be found from the row sums of the third sub-triangle extracted from Pascal's triangle as indicated below by braces:
1;
1, 1;
{1}, 2, 1;
{1, 3}, 3, 1;
{1, 4, 6}, 4, 1;
{1, 5, 10, 10}, 5, 1;
{1, 6, 15, 20, 15}, 6, 1;
For integers a, b, denote by a<+>b the least c >= a, such that the Hamming distance D(a,c) = b (note that, generally speaking, a<+>b differs from b<+>a). Then for n >= 3, a(n) = n<+>n. This has a simple explanation: for n >= 3 in binary we have a(n) = (2^n-1)-n = "anti n". - Vladimir Shevelev, Feb 14 2012 a(n) is the number of binary sequences of length n having at least one pair 01. - Branko Curgus, May 23 2012 Nonzero terms are those integers k for which there exists a perfect (Hamming) error-correcting code. - L. Edson Jeffery, Nov 28 2012 a(n) is the number of length n binary words constructed in the following manner: Select two positions in which to place the first two 0's of the word. Fill in all (possibly none) of the positions before the second 0 with 1's and then complete the word with an arbitrary string of 0's or 1's. So a(n) = Sum_{k=2..n} (k-1)*2^(n-k). - Geoffrey Critzer, Dec 12 2013 Without first 0: a(n)/2^n equals Sum_{k=0..n} k/2^k. For example: a(5)=57, 57/32 = 0/1 + 1/2 + 2/4 + 3/8 + 4/16 + 5/32. - Bob Selcoe, Feb 25 2014 The first barycentric coordinate of the centroid of the first n rows of Pascal's triangle, assuming the numbers are weights, is A000295(n+1)/ A000337(n). See attached figure. - C�sar Eliud Lozada, Nov 14 2014 Starting (0, 1, 4, 11, ...), this is the binomial transform of (0, 1, 2, 2, 2, ...). - Gary W. Adamson, Jul 27 2015 Also the number of (non-null) connected induced subgraphs in the n-triangular honeycomb rook graph. - Eric W. Weisstein, Aug 27 2017 a(n) is the number of swaps needed in the worst case to transform a binary tree with n full levels into a heap, using (bottom-up) heapify. - Rudy van Vliet, Sep 19 2017 The utility of large networks, particularly social networks, with n participants is given by the terms a(n) of this sequence. This assertion is known as Reed's Law, see the Wikipedia link. - Johannes W. Meijer, Jun 03 2019 a(n-1) is the number of subsets of {1..n} in which the largest element of the set exceeds by at least 2 the next largest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,2,4}, {1,2,5}, {1,3,5}, {2,3,5}, {1,2,3,5}. - Enrique Navarrete, Apr 08 2020 a(n-1) is also the number of subsets of {1..n} in which the second smallest element of the set exceeds by at least 2 the smallest element. For example, for n = 5, a(4) = 11 and the 11 sets are {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}, {1,3,4}, {1,3,5}, {1,4,5}, {2,4,5}, {1,3,4,5}. - Enrique Navarrete, Apr 09 2020 a(n+1) is the sum of the smallest elements of all subsets of {1..n}. For example, for n=3, a(4)=11; the subsets of {1,2,3} are {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}, and the sum of smallest elements is 11. - Enrique Navarrete, Aug 20 2020 Number of subsets of an n-set that have more than one element. - Eric M. Schmidt, Mar 13 2021 Number of individual bets in a "full cover" bet on n-1 horses, dogs, etc. in different races. Each horse, etc. can be bet on or not, giving 2^n bets. But, by convention, singles (a bet on only one race) are not included, reducing the total number bets by n. It is also impossible to bet on no horses at all, reducing the number of bets by another 1. A full cover on 4 horses, dogs, etc. is therefore 6 doubles, 4 trebles and 1 four-horse etc. accumulator. In British betting, such a bet on 4 horses etc. is a Yankee; on 5, a super-Yankee. - Paul Duckett, Nov 17 2021 Number of binary sequences of length n with at least two 1's.
a(n-1) is the number of ways to choose an odd number of elements greater than or equal to 3 out of n elements.
a(n+1) is the number of ways to split [n] = {1,2,...,n} into two (possibly empty) complementary intervals {1,2,...,i} and {i+1,i+2,...,n} and then select a subset from the first interval (2^i choices, 0 <= i <= n), and one block/cell (i.e., subinterval) from the second interval (n-i choices, 0 <= i <= n).
(End)
Number of possible conjunctions in a system of n planets; for example, there can be 0 conjunctions with one planet, one with two planets, four with three planets (three pairs of planets plus one with all three) and so on. - Wendy Appleby, Jan 02 2023 Largest exponent m such that 2^m divides (2^n-1)!. - Franz Vrabec, Aug 18 2023 It seems that a(n-1) is the number of odd r with 0 < r < 2^n for which there exist u,v,w in the x-independent beginning of the Collatz trajectory of 2^n x + r with u+v = w+1, as detailed in the link "Collatz iteration and Euler numbers?". A better understanding of this might also give a formula for A374527. - Markus Sigg, Aug 02 2024
REFERENCES
O. Bottema, Problem #562, Nieuw Archief voor Wiskunde, 28 (1980) 115.
L. Comtet, "Permutations by Number of Rises; Eulerian Numbers." Section 6.5 in Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed. Dordrecht, Netherlands: Reidel, pp. 51 and 240-246, 1974.
F. N. David and D. E. Barton, Combinatorial Chance. Hafner, NY, 1962, p. 151.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, p. 34.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
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).
LINKS
P. J. Cameron, M. Gadouleau, J. D. Mitchell, and Y. Peresse, Chains of subsemigroups, arXiv preprint arXiv:1501.06394 [math.GR], 2015. See Table 4. 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
a(n) = 2^n - n - 1.
G.f.: x^2/((1-2*x)*(1-x)^2).
a(0)=0, a(1)=0, a(n) = 3*a(n-1) - 2*a(n-2) + 1. - Miklos Kristof, Mar 09 2005 a(0)=0, a(n) = 2*a(n-1) + n - 1 for all n in Z.
a(n) = Sum_{k=2..n} binomial(n, k). - Paul Barry, Jun 05 2003 a(n+1) = Sum_{i=1..n} Sum_{j=1..i} C(i, j). - Benoit Cloitre, Sep 07 2003 a(0)=0, a(1)=0, a(n) = Sum_{i=0..n-1} i+a(i) for i > 1. - Gerald McGarvey, Jun 12 2004 a(n+1) = Sum_{k=0..n} (n-k)*2^k. - Paul Barry, Jul 29 2004 a(n) = Sum_{k=0..n} binomial(n, k+2); a(n+2) = Sum_{k=0..n} binomial(n+2, k+2). - Paul Barry, Aug 23 2004 a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n-k-1, k+1)*2^(n-k-2)*(-1/2)^k. - Paul Barry, Oct 25 2004 a(0) = 0, a(n) = Sum_{k=0..n-1} 2^k - 1. - Doug Bell, Jan 19 2009 Column k=1 of A173018 starts a'(n) = 0, 1, 4, 11, ... and has the hypergeometric representation n*hypergeom([1, -n+1], [-n], 2). This can be seen as a formal argument to prefer Euler's A173018 over A008292. - Peter Luschny, Sep 19 2014 E.g.f.: exp(x)*(exp(x)-1-x); this is U(0) where U(k) = 1 - x/(2^k - 2^k/(x + 1 - x^2*2^(k+1)/(x*2^(k+1) - (k+1)/U(k+1)))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012
EXAMPLE
G.f. = x^2 + 4*x^3 + 11*x^4 + 26*x^5 + 57*x^6 + 120*x^7 + 247*x^8 + 502*x^9 + ...
MAPLE
[ seq(2^n-n-1, n=1..50) ];
# Grammar specification:
spec := [S, { B = Set(Z, 1 <= card), C = Sequence(B, 2 <= card), S = Prod(B, C) }, unlabeled]:
struct := n -> combstruct[count](spec, size = n+1);
MATHEMATICA
a[n_] = If[n==0, 0, n*(HypergeometricPFQ[{1, 1-n}, {2}, -1] - 1)];
PROG
(Magma) [EulerianNumber(n, 1): n in [0..40]]; // G. C. Greubel, Oct 02 2024 (SageMath) [2^n -(n+1) for n in range(41)] # G. C. Greubel, Oct 02 2024
CROSSREFS
Cf. A008292 (classic version of Euler's triangle used by Comtet (1974)). Cf. A173018 (version of Euler's triangle used by Graham, Knuth and Patashnik in Concrete Math. (1990)).
a(n) = 2^n - 2.
(Formerly M1599 N0625)
+10
154
-1, 0, 2, 6, 14, 30, 62, 126, 254, 510, 1022, 2046, 4094, 8190, 16382, 32766, 65534, 131070, 262142, 524286, 1048574, 2097150, 4194302, 8388606, 16777214, 33554430, 67108862, 134217726, 268435454, 536870910, 1073741822, 2147483646, 4294967294, 8589934590, 17179869182, 34359738366, 68719476734, 137438953470
COMMENTS
For n > 1, a(n) is the expected number of tosses of a fair coin to get n-1 consecutive heads. - Pratik Poddar, Feb 04 2011 For n > 0, the number of nonempty proper subsets of an n-element set. - Ross La Haye, Feb 07 2004 Numbers j such that abs( Sum_{k=0..j} (-1)^binomial(j, k)*binomial(j + k, j - k) ) = 1. - Benoit Cloitre, Jul 03 2004 For n > 2 this formula also counts edge-rooted forests in a cycle of length n. - Woong Kook (andrewk(AT)math.uri.edu), Sep 08 2004
For n >= 1, conjectured to be the number of integers from 0 to (10^n)-1 that lack 0, 1, 2, 3, 4, 5, 6 and 7 as a digit. - Alexandre Wajnberg, Apr 25 2005 Beginning with a(2) = 2, these are the partial sums of the subsequence of A000079 = 2^n beginning with A000079(1) = 2. Hence for n >= 2 a(n) is the smallest possible sum of exactly one prime, one semiprime, one triprime, ... and one product of exactly n-1 primes. A060389 (partial sums of the primorials, A002110, beginning with A002110(1)=2) is the analog when all the almost primes must also be squarefree. - Rick L. Shepherd, May 20 2005 From the second term on (n >= 1), the binary representation of these numbers is a 0 preceded by (n - 1) 1's. This pattern (0)111...1110 is the "opposite" of the binary 2^n+1: 1000...0001 (cf. A000051). - Alexandre Wajnberg, May 31 2005 The numbers 2^n - 2 (n >= 2) give the positions of 0's in A110146. Also numbers k such that k^(k + 1) = 0 mod (k + 2). - Zak Seidov, Feb 20 2006 Number of surjections from an n-element set onto a two-element set, with n >= 2. - Mohamed Bouhamida, Dec 15 2007 It appears that these are the numbers n such that 3* A135013(n) = n*(n + 1), thus answering Problem 2 on the Mathematical Olympiad Foundation of Japan, Final Round Problems, Feb 11 1993 (see link Japanese Mathematical Olympiad). Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x is a proper subset of y or y is a proper subset of x and x and y are disjoint. Then a(n+1) = |R|. - Ross La Haye, Mar 19 2009 The permutohedron Pi_n has 2^n - 2 facets [Pashkovich]. - Jonathan Vos Post, Dec 17 2009 a(n) is the number of branches of a complete binary tree of n levels. - Denis Lorrain, Dec 16 2011 For n>=1, a(n) is the number of length-n words on alphabet {1,2,3} such that the gap(w)=1. For a word w the gap g(w) is the number of parts missing between the minimal and maximal elements of w. Generally for words on alphabet {1,2,...,m} with g(w)=g>0 the e.g.f. is Sum_{k=g+2..m} (m - k + 1)*binomial((k - 2),g)*(exp(x) - 1)^(k - g). a(3)=6 because we have: 113, 131, 133, 311, 313, 331. Cf. A240506. See the Heubach/Mansour reference. - Geoffrey Critzer, Apr 13 2014 For n > 0, a(n) is the minimal number of internal nodes of a red-black tree of height 2*n-2. See the Oct 02 2015 comment under A027383. - Herbert Eberle, Oct 02 2015 Actually this follows from the procedure for determining the multiplicity of prime p in C(n) given in A000108 by Franklin T. Adams-Watters: For p=2, the multiplicity is the number of 1 digits minus 1 in the binary representation of n+1. Obviously, the smallest k achieving "number of 1 digits" = k is 2^k-1. Therefore C(2^k-2) is divisible by 2^(k-1) for k > 0 and there is no smaller m for which 2^(k-1) divides C(m) proving the conjecture. - Peter Schorn, Feb 16 2020 For n >= 0, a(n) is the largest number you can write in bijective base-2 (a.k.a. the dyadic system, A007931) with n digits. - Harald Korneliussen, May 18 2019 The terms of this sequence are also the sum of the terms in each row of Pascal's triangle other than the ones. - Harvey P. Dale, Apr 19 2020 For n > 1, binomial(a(n),k) is odd if and only if k is even. - Charlie Marion, Dec 22 2020 For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 2. - David desJardins, Oct 27 2022 For n >= 1, a(n+1) is the number of roots of unity in the unique degree-n unramified extension of the 2-adic field Q_2. Note that for each p, the unique degree-n unramified extension of Q_p is the splitting field of x^(p^n) - x, hence containing p^n - 1 roots of unity for p > 2 and 2*(2^n - 1) for p = 2. - Jianing Song, Nov 08 2022
REFERENCES
H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, Addison-Wesley, 2004, p. 134. [From Mohammad K. Azarian, October 27 2011] S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2009 page 86, Exercise 3.16.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
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).
A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
LINKS
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
G.f.: 1/(1 - 2*x) - 2/(1 - x), e.g.f.: (e^x - 1)^2 - 1. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
G.f.: (3*x - 1)/((2*x - 1)*(x - 1)). - Simon Plouffe in his 1992 dissertation for the sequence without the leading -1 G.f.: U(0) - 1, where U(k) = 1 + x/(2^k + 2^k/(x - 1 - x^2*2^(k + 1)/(x*2^(k + 1) - (k + 1)/U(k + 1) ))); (continued fraction, 3rd kind, 4-step). - Sergei N. Gladkovskii, Dec 01 2012 a(n) = Sum_{k=1..n-1} binomial(n, k) for n > 0. - Dan McCandless, Nov 14 2015 a(n+1) = 2*(n + Sum_{j=1..n-1} (n-j)*2^(j-1)), n >= 1. This is the number of the rationals k/2, k = 1..2*n for n >= 1 and (2*k+1)/2^j for j = 2..n, n >= 2, and 2*k+1 < n-(j-1). See the example for n = 3 below. Motivated by the proposal A287012 by Mark Rickert. - Wolfdieter Lang, Jun 14 2017
EXAMPLE
a(4) = 14 because the 14 = 6 + 4 + 4 rationals (in lowest terms) for n = 3 are (see the Jun 14 2017 formula above): 1/2, 1, 3/2, 2, 5/2, 3; 1/4, 3/4, 5/4, 7/4; 1/8, 3/8, 5/8, 7/8. - Wolfdieter Lang, Jun 14 2017
MAPLE
seq(2^n-2, n=0..20) ;
ZL := [S, {S=Prod(B, B), B=Set(Z, 1 <= card)}, labeled]: [-1, seq(combstruct[count](ZL, size=n), n=1..28)]; # Zerinvary Lajos, Mar 13 2007
PROG
(Sage) [gaussian_binomial(n, 1, 2)-1 for n in range(0, 29)] # Zerinvary Lajos, May 31 2009 (Haskell)
a000918 = (subtract 2) . (2 ^)
a000918_list = iterate ((subtract 2) . (* 2) . (+ 2)) (- 1)
Chernoff sequence: a(n) = Product_{k=1..n} prime(k)^(n-k+1).
(Formerly M2050)
+10
110
1, 2, 12, 360, 75600, 174636000, 5244319080000, 2677277333530800000, 25968760179275365452000000, 5793445238736255798985527240000000, 37481813439427687898244906452608585200000000, 7517370874372838151564668004911177464757864076000000000, 55784440720968513813368002533861454979548176771615744085560000000000
COMMENTS
Product of first n primorials: a(n) = Product_{i=1..n} A002110(i). Superprimorials, from primorials by analogy with superfactorials.
Smallest number k with n distinct exponents in its prime factorization, i.e., A071625(k) = n. This might be a good place to explain the name "Chernoff sequence" since his name does not appear in the References or Links as of Mar 22 2014. - Jonathan Sondow, Mar 22 2014 Pickover (1992) named this sequence after Paul Chernoff of California, who contributed this sequence to his book. He was possibly referring to American mathematician Paul Robert Chernoff (1942 - 2017), a professor at the University of California. - Amiram Eldar, Jul 27 2020
REFERENCES
Clifford A. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 351.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James K. Strayer, Elementary number theory, Waveland Press, Inc., Long Grove, IL, 1994. See p. 37.
FORMULA
a(n) = m(1)*m(2)*m(3)*...*m(n), where m(n) = n-th primorial number. - N. J. A. Sloane, Feb 20 2005 a(0) = 1, a(n) = a(n - 1)p(n)#, where p(n)# is the n-th primorial A002110(n) (the product of the first n primes). - Alonso del Arte, Sep 30 2011
EXAMPLE
a(4) = 360 because 2^3 * 3^2 * 5 = 1 * 2 * 6 * 30 = 360.
a(5) = 75600 because 2^4 * 3^3 * 5^2 * 7 = 1 * 2 * 6 * 30 * 210 = 75600.
MAPLE
a := []; printlevel := -1; for k from 0 to 20 do a := [op(a), product(ithprime(i)^(k-i+1), i=1..k)] od; print(a);
MATHEMATICA
Rest[FoldList[Times, 1, FoldList[Times, 1, Prime[Range[15]]]]] (* Harvey P. Dale, Jul 07 2011 *) Table[Times@@Table[Prime[i]^(n - i + 1), {i, n}], {n, 12}] (* Alonso del Arte, Sep 30 2011 *)
PROG
(Haskell)
a006939 n = a006939_list !! n
(Magma) [1] cat [(&*[NthPrime(k)^(n-k+1): k in [1..n]]): n in [1..15]]; // G. C. Greubel, Oct 14 2018
CROSSREFS
Cf. A000178 (product of first n factorials), A007489 (sum of first n factorials), A060389 (sum of first n primorials). A000142 counts divisors of superprimorials.
A000325 counts uniform divisors of superprimorials.
A008302 counts divisors of superprimorials by bigomega.
A022915 counts permutations of prime indices of superprimorials.
A118914 has row a(n) equal to {1..n}.
A124010 has row a(n) equal to {n..1}.
A130091 lists numbers with distinct prime multiplicities.
A317829 counts factorizations of superprimorials.
A336417 counts perfect-power divisors of superprimorials.
A336426 gives non-products of superprimorials.
Cullen numbers: a(n) = n*2^n + 1.
(Formerly M2795 N1125)
+10
71
1, 3, 9, 25, 65, 161, 385, 897, 2049, 4609, 10241, 22529, 49153, 106497, 229377, 491521, 1048577, 2228225, 4718593, 9961473, 20971521, 44040193, 92274689, 192937985, 402653185, 838860801, 1744830465, 3623878657, 7516192769, 15569256449, 32212254721
COMMENTS
Let A be the Hessenberg matrix of order n defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1] =-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)= (-1)^(n-1)*coeff(charpoly(A,x),x). - Milan Janjic, Jan 26 2010 Add the list of fractions beginning with 1/2 + 3/4 + 7/8 + ... + (2^n - 1)/2^n and take the sums pairwise from left to right. For 1/2 + 3/4 = 5/4, 5 + 4 = 9 = a(2); for 5/4 + 7/8 = 17/8, 17 + 8 = 25 = a(3); for 17/8 + 15/16 = 49/16, 49 + 16 = 65 = a(4); for 49/16 + 31/32 = 129/32, 129 + 32 = 161 = a(5). For each pairwise sum a/b, a + b = n*2^(n+1). - J. M. Bergot, May 06 2015 Number of divisors of (2^n)^(2^n). - Gus Wiseman, May 03 2021 Named after the Irish Jesuit priest James Cullen (1867-1933), who checked the primality of the terms up to n=100. - Amiram Eldar, Jun 05 2021
REFERENCES
G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
R. K. Guy, Unsolved Problems in Number Theory, B20.
W. Sierpiński, Elementary Theory of Numbers. Państ. Wydaw. Nauk., Warsaw, 1964, p. 346.
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).
LINKS
James Cullen, Question 15897, Educational Times, Vol. 58 (December 1905), p. 534.
FORMULA
a(n) = 4a(n-1) - 4a(n-2) + 1. - Paul Barry, Jun 12 2003 a(0)=1, a(1)=3, a(2)=9, a(n) = 5*a(n-1)-8*a(n-2)+4*a(n-3). - Harvey P. Dale, Oct 13 2011
EXAMPLE
G.f. = 1 + 3*x + 9*x^2 + 25*x^3 + 65*x^4 + 161*x^5 + 385*x^6 + 897*x^7 + ... - Michael Somos, Jul 18 2018
MAPLE
A002064:=-(1-2*z+2*z**2)/((z-1)*(-1+2*z)**2); # conjectured by Simon Plouffe in his 1992 dissertation
MATHEMATICA
LinearRecurrence[{5, -8, 4}, {1, 3, 9}, 51] (* Harvey P. Dale, Oct 13 2011 *) CoefficientList[Series[(1 - 2 x + 2 x^2)/((1 - x) (2 x - 1)^2), {x, 0, 50}], x] (* Vincenzo Librandi, May 07 2015 *)
PROG
(Haskell)
a002064 n = n * 2 ^ n + 1
a002064_list = 1 : 3 : (map (+ 1) $ zipWith (-) (tail xs) xs)
where xs = map (* 4) a002064_list
CROSSREFS
A188385 gives the highest prime exponent in n^n.
a(n) = 2^n + n.
(Formerly M2547)
+10
56
1, 3, 6, 11, 20, 37, 70, 135, 264, 521, 1034, 2059, 4108, 8205, 16398, 32783, 65552, 131089, 262162, 524307, 1048596, 2097173, 4194326, 8388631, 16777240, 33554457, 67108890, 134217755, 268435484, 536870941, 1073741854, 2147483679, 4294967328, 8589934625
COMMENTS
For numbers m=n+2^n such that equation x=2^(m-x) has solution x=2^n, see A103354. - Zak Seidov, Mar 23 2005 Primes of the form x^x+1 must be of the form 2^2^(a(n))+1, that is, Fermat number F_(a(n)) (Sierpiński 1958). - David W. Wilson, May 08 2005 a(n) is also the number of all connected subtrees of a star tree, having n leaves. The star tree is a tree, where all n leaves are connected to one parent P. - Viktar Karatchenia, Feb 29 2016
REFERENCES
John H. Conway, R. K. Guy, The Book of Numbers, Copernicus Press, p. 84.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
EXAMPLE
a(0) = 1. There are n=0 leaves, it is a trivial tree consisting of a single parent node P.
a(1) = 3. There is n=1 leaf, the tree is P-A, the subtrees are: 2 singles: P, A; 1 pair: P-A; 2+1 = 3 subtrees in total.
a(2) = 6. When n=2, the tree is P-A P-B, the subtrees are: 3 singles: P, A, B; 2 pairs: P-A, P-B; 1 triple: A-P-B (the whole tree); 3+2+1 = 6.
a(3) = 11. For n=3 leaf nodes, the tree is P-A P-B P-C, the subtrees are: 4 singles: P, A, B, C; 3 pairs: P-A, P-B, P-C; 3 triples: A-P-B, A-P-C, B-P-C; 1 quad: P-A P-B P-C (the whole tree); 4+3+3+1 = 11 in total.
a(4) = 20. For n=4 leaves, the tree is P-A P-B P-C P-D, the subtrees are: 5 singles: P, A, B, C, D; 4 pairs: P-A, P-B, P-C, P-D; 6 triples: A-P-B, A-P-C, B-P-C, A-P-D, B-P-D, C-P-D; 4 quads: P-A P-B P-C, P-A P-B P-D, P-A P-C P-D, P-B P-C P-D; the whole tree counts as 1; 5+4+6+4+1 = 20.
In general, for n leaves, connected to the parent node P, there will be: (n+1) singles; (n, 1) pairs; (n, 2) triples; (n, 3) quads; ... ; (n, n-1) subtrees having (n-1) edges; 1 whole tree, having all n edges. Thus, the total number of all distinct subtrees is: (n+1) + (n, 1) + (n, 2) + (n, 3) + ... + (n, n-1) + 1 = (n + (n, 0)) + (n, 1) + (n, 2) + (n, 3) + ... + (n, n-1) + (n, n) = n + (sum of all binomial coefficients of size n) = n + (2^n). (End)
MATHEMATICA
Table[BitXOr(i, 2^i), {i, 1, 100}] (* Peter Luschny, Jun 01 2011 *) LinearRecurrence[{4, -5, 2}, {1, 3, 6}, 40] (* Harvey P. Dale, Feb 08 2023 *)
PROG
(Haskell)
a006127 n = a000079 n + n
a006127_list = s [1] where
s xs = last xs : (s $ zipWith (+) [1..] (xs ++ reverse xs))
(Python)
1 followed by {1, 2} repeated.
+10
54
1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2
COMMENTS
Continued fraction for sqrt(3).
Also coefficient of the highest power of q in the expansion of the polynomial nu(n) defined by: nu(0)=1, nu(1)=b and for n>=2, nu(n)=b*nu(n-1)+lambda*(n-1)_q*nu(n-2) with (b,lambda)=(1,1), where (n)_q=(1+q+...+q^(n-1)) and q is a root of unity. - Y. Kelly Itakura (yitkr(AT)mta.ca), Aug 21 2002
nu(0)=1 nu(1)=1; nu(2)=2; nu(3)=3+q; nu(4)=5+3q+2q^2; nu(5)=8+7q+6q^2+4q^3+q^4; nu(6)=13+15q+16q^2+14q^3+11q^4+5q^5+2q^6.
a(n) = denominators of arithmetic means of the first n positive integers for n >= 1.
See A026741(n+1) or A145051(n) - denominators of arithmetic means of the first n positive integers. (End) This is a prototype of multiplicative sequences defined by a(p^e)=1 for odd primes p, and a(2^e)=c with some constant c, here c=2. They have Dirichlet generating functions (1+(c-1)/2^s)*zeta(s).
Examples are A153284, A176040 (c=3), A040005 (c=4), A021070, A176260 (c=5), A040011, A176355 (c=6), A176415 (c=7), A040019, A021059 (c=8), A040029 (c=10), A040041 (c=12). (End) a(n) = p(-1) where p(x) is the unique degree-n polynomial such that p(k) = A000325(k) for k = 0, 1, ..., n. - Michael Somos, May 12 2012 For n > 0, a(n) is the minimal gap of distinct numbers coprime to n. Proof: denote the minimal gap by b(n). For odd n we have A058026(n) > 0, hence b(n) = 1. For even n, since 1 and -1 are both coprime to n we have b(n) <= 2, and that b(n) >= 2 is obvious. The maximal gap is given by A048669. (End)
FORMULA
Multiplicative with a(p^e) = 2 if p even; 1 if p odd. - David W. Wilson, Aug 01 2001 G.f.: (1 + x + x^2) / (1 - x^2). E.g.f.: (3*exp(x)-2*exp(0)+exp(-x))/2. - Paul Barry, Apr 27 2003 a(n) = (3-2*0^n +(-1)^n)/2. a(-n)=a(n). a(2n+1)=1, a(2n)=2, n nonzero.
a(n) = sum{k=0..n, F(n-k+1)*(-2+(1+(-1)^k)/2+C(2, k)+0^k)}. - Paul Barry, Jun 22 2007 Euler transform of length 3 sequence [ 1, 1, -1]. - Michael Somos, Aug 04 2009 Moebius transform is length 2 sequence [ 1, 1]. - Michael Somos, Aug 04 2009 a(n) = sign(n) + ((n+1) mod 2) = 1 + sign(n) - (n mod 2). - Wesley Ivan Hurt, Dec 13 2013
EXAMPLE
1.732050807568877293527446341... = 1 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...))))
G.f. = 1 + x + 2*x^2 + x^3 + 2*x^4 + x^5 + 2*x^6 + x^7 + 2*x^8 + x^9 + ...
MAPLE
Digits := 100: convert(evalf(sqrt(N)), confrac, 90, 'cvgts'):
PROG
(PARI) {a(n) = 2 - (n==0) - (n%2)} /* Michael Somos, Jun 11 2003 */ (PARI) { allocatemem(932245000); default(realprecision, 12000); x=contfrac(sqrt(3)); for (n=0, 20000, write("b040001.txt", n, " ", x[n+1])); } \\ Harry J. Smith, Jun 01 2009 (Haskell)
a040001 0 = 1; a040001 n = 2 - mod n 2
CROSSREFS
Apart from a(0) the same as A134451.
Second-order Eulerian numbers: a(n) = 2^n - 2*n.
(Formerly M1838)
+10
39
1, 0, 0, 2, 8, 22, 52, 114, 240, 494, 1004, 2026, 4072, 8166, 16356, 32738, 65504, 131038, 262108, 524250, 1048536, 2097110, 4194260, 8388562, 16777168, 33554382, 67108812, 134217674, 268435400, 536870854, 1073741764, 2147483586
COMMENTS
Starting with n=2, a(n) is the second-order Eulerian number <<n-1,1>> (see A008517). Also, number of 3 X n binary matrices avoiding simultaneously the right-angled numbered polyomino patterns (ranpp) (00;1), (01;0) and (01;1). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1<i2, j1<j2 and these elements are in same relative order as those in the triple (x,y,z). - Sergey Kitaev, Nov 11 2004 This is the number of target DNA sequences of the correct length present after each successive cycle of the Polymerase Chain Reaction (PCR). The first two cycles produce 0 target sequences, there are 2 target sequences present after the third cycle, then 8 after the fourth cycle, and so forth. - A. Timothy Royappa, Jun 16 2012
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
Equals binomial transform of [1, -1, 1, 1, 1, ...]. - Gary W. Adamson, Jul 14 2008 a(0) = 1 and a(n) = Sum_{k=0..n-3} ((-1)^(n+k+1)*binomial(2*n-1,k)*stirling1(2*n-k-3,n-k-2)), n=>1. - Johannes W. Meijer, Oct 16 2009 a(0)=1, a(1)=0, a(2)=0, a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3). - Harvey P. Dale, May 21 2011 a(n) = A000225(n+1) - A081494(n+1), n > 1. In other words, a(n) equals the sum of the elements in a Pascal triangle of depth n+1 minus the sum of the elements on its perimeter. - Ivan N. Ianakiev, Jun 01 2014
EXAMPLE
G.f. = 1 + 2*x^3 + 8*x^4 + 22*x^5 + 52*x^6 + 114*x^7 + 240*x^8 + 494*x^9 + ...
MAPLE
A005803:=-2*z/(2*z-1)/(z-1)**2; # conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence except for three leading terms
MATHEMATICA
Table[2^n-2n, {n, 0, 50}] (* or *) LinearRecurrence[{4, -5, 2}, {1, 0, 0}, 51] (* Harvey P. Dale, May 21 2011 *)
PROG
(PARI) {a(n) = if( n<0, 0, 2^n - 2*n)}; /* Michael Somos, Oct 13 2002 */ (Haskell)
a005803 n = 2 ^ n - 2 * n
a005803_list = 1 : f 1 [0, 2 ..] where
f x (z:zs@(z':_)) = y : f y zs where y = (x + z) * 2 - z'
CROSSREFS
Equivalent to second column of A008517. Equals for n =>3 the third right hand column of A163936.
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