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%I A034839 #127 Aug 04 2024 13:40:42
%S A034839 1,1,1,1,1,3,1,6,1,1,10,5,1,15,15,1,1,21,35,7,1,28,70,28,1,1,36,126,
%T A034839 84,9,1,45,210,210,45,1,1,55,330,462,165,11,1,66,495,924,495,66,1,1,
%U A034839 78,715,1716,1287,286,13
%N A034839 Triangular array formed by taking every other term of each row of Pascal's triangle.
%C A034839 Number of compositions of n having k parts greater than 1. Example: T(5,2)=5 because we have 3+2, 2+3, 2+2+1, 2+1+2 and 1+2+2. Number of binary words of length n-1 having k runs of consecutive 1's. Example: T(5,2)=5 because we have 1010, 1001, 0101, 1101 and 1011. - _Emeric Deutsch_, Mar 30 2005
%C A034839 From _Gary W. Adamson_, Oct 17 2008: (Start)
%C A034839 Received from _Herb Conn_:
%C A034839 Let T = tan x, then
%C A034839 tan x = T
%C A034839 tan 2x = 2T / (1 - T^2)
%C A034839 tan 3x = (3T - T^3) / (1 - 3T^2)
%C A034839 tan 4x = (4T - 4T^3) / (1 - 6T^2 + T^4)
%C A034839 tan 5x = (5T - 10T^3 + T^5) / (1 - 10T^2 + 5T^4)
%C A034839 tan 6x = (6T - 20T^3 + 6T^5) / (1 - 15T^2 + 15T^4 - T^6)
%C A034839 tan 7x = (7T - 35T^3 + 21T^5 - T^7) / (1 - 21T^2 + 35T^4 - 7T^6)
%C A034839 tan 8x = (8T - 56T^3 + 56T^5 - 8T^7) / (1 - 28T^2 + 70T^4 - 28T^6 + T^8)
%C A034839 tan 9x = (9T - 84T^3 + 126T^5 - 36T^7 + T^9) / (1 - 36 T^2 + 126T^4 - 84T^6 + 9T^8)
%C A034839 ... To get the next one in the series, (tan 10x), for the numerator add:
%C A034839 9....84....126....36....1 previous numerator +
%C A034839 1....36....126....84....9 previous denominator =
%C A034839 10..120....252...120...10 = new numerator
%C A034839 For the denominator add:
%C A034839 ......9.....84...126...36...1 = previous numerator +
%C A034839 1....36....126....84....9.... = previous denominator =
%C A034839 1....45....210...210...45...1 = new denominator
%C A034839 ...where numerators = A034867, denominators = A034839
%C A034839 (End)
%C A034839 Triangle, with zeros omitted, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Dec 12 2011
%C A034839 The row (1,66,495,924,495,66,1) plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534. - _Tom Copeland_, Dec 12 2016
%C A034839 Binomial(n,2k) is also the number of permutations avoiding both 123 and 132 with k ascents, i.e., positions with w[i]<w[i+1]. - _Lara Pudwell_, Dec 19 2018
%C A034839 Coefficients in expansion of ((x-1)^n+(x+1)^n)/2 or ((x-i)^n+(x+i)^n)/2 with alternating sign. - _Eugeniy Sokol_, Sep 20 2020
%C A034839 Number of permutations of length n avoiding simultaneously the patterns 213 and 312 with the maximum number of non-overlapping descents equal k (equivalently, with the maximum number of non-overlapping ascents equal k). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i) < a(i+1) (resp., a(i) > a(i+1)). - _Tian Han_, Nov 16 2023
%H A034839 G. C. Greubel, <a href="/A034839/b034839.txt">Table of n, a(n) for the first 101 rows, flattened</a>
%H A034839 M. Bukata, R. Kulwicki, N. Lewandowski, L. Pudwell, J. Roth, and T. Wheeland, <a href="https://arxiv.org/abs/1812.07112">Distributions of Statistics over Pattern-Avoiding Permutations</a>, arXiv preprint arXiv:1812.07112 [math.CO], 2018.
%H A034839 H. Chan, S. Cooper, and P. Toh, <a href="http://dx.doi.org/10.1016/j.aim.2005.12.003">The 26th power of Dedekind's eta function</a> Advances in Mathematics, 207 (2006) 532-543.
%H A034839 Tom Copeland, <a href="https://tcjpn.wordpress.com/2020/07/15/juggling-zeros-in-the-matrix-example-ii/">Juggling Zeros in the Matrix: Example II</a>, 2020.
%H A034839 C. Corsani, D. Merlini, and R. Sprugnoli, <a href="http://dx.doi.org/10.1016/S0012-365X(97)00110-6">Left-inversion of combinatorial sums</a>, Discrete Mathematics, 180 (1998) 107-122.
%H A034839 Tian Han and Sergey Kitaev, <a href="https://arxiv.org/abs/2311.02974">Joint distributions of statistics over permutations avoiding two patterns of length 3</a>, arXiv:2311.02974 [math.CO], 2023.
%H A034839 S.-M. Ma, <a href="http://arxiv.org/abs/1205.0735">On some binomial coefficients related to the evaluation of tan(nx)</a>, arXiv preprint arXiv:1205.0735 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 13 2012
%H A034839 K. Oliver and H. Prodinger, The continued fraction expansion of Gauss' hypergeometric function and a new application to the tangent function, Transactions of the Royal Society of South Africa, Vol. 76 (2012), 151-154, <a href="http://dx.doi.org/10.1080/0035919X.2012.727363">[DOI]</a>; <a href="http://math.sun.ac.za/~hproding/pdffiles/Avery-contribution-July-2012.pdf">[PDF]</a>. - From _N. J. A. Sloane_, Jan 03 2013
%H A034839 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Tangent.html">Tangent</a> [From _Eric W. Weisstein_, Oct 18 2008]
%F A034839 E.g.f.: exp(x)*cosh(x*sqrt(y)). - _Vladeta Jovovic_, Mar 20 2005
%F A034839 From _Emeric Deutsch_, Mar 30 2005: (Start)
%F A034839 T(n, k) = binomial(n, 2*k), for n >= 0 and k = 0, 1, ..., floor(n/2).
%F A034839 G.f.: (1-z)/((1-z)^2 - t*z^2). (End)
%F A034839 O.g.f. for column no. k is (1/(1-x))*(x/(1-x))^(2*k), k >= 0 [from the g.f. given in the preceding formula]. - _Wolfdieter Lang_, Jan 18 2013
%F A034839 From _Peter Bala_, Jul 14 2015: (Start)
%F A034839 Stretched Riordan array ( 1/(1 - x ), x^2/(1 - x)^2 ) in the terminology of Corsani et al.
%F A034839 Denote this array by P. Then P * A007318 = A201701.
%F A034839 P * transpose(P) is A119326 read as a square array.
%F A034839 Let Q denote the array ( (-1)^k*binomial(2*n,k) )n,k>=0. Q is a signed version of A034870. Then Q*P = the identity matrix, that is, Q is a left-inverse array of P (see Corsani et al., p. 111).
%F A034839 P * A034870 = A080928. (End)
%F A034839 Even rows are A086645. An aerated version of this array is A099174 with each diagonal divided by the first element of the diagonal, the double factorials A001147. - _Tom Copeland_, Dec 12 2015
%e A034839 Triangular array begins:
%e A034839   1
%e A034839   1
%e A034839   1  1
%e A034839   1  3
%e A034839   1  6  1
%e A034839   1 10  5
%e A034839   1 15 15 1
%e A034839   ...
%e A034839 cosh(4x) = (cosh x)^5 + 10 (cosh x)^3 (sinh x)^2 + 5 (cosh x) (sinh x)^4, so row 4 is (1,10,5). See Mathematica program. - _Clark Kimberling_, Aug 03 2024
%p A034839 for n from 0 to 13 do seq(binomial(n,2*k),k=0..floor(n/2)) od;# yields sequence in triangular form; # _Emeric Deutsch_, Mar 30 2005
%t A034839 u[1, x_] := 1; v[1, x_] := 1; z = 12;
%t A034839 u[n_, x_] := u[n - 1, x] + x*v[n - 1, x]
%t A034839 v[n_, x_] := u[n - 1, x] + v[n - 1, x]
%t A034839 cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t A034839 TableForm[cu]  (* A034839 as a triangle *)
%t A034839 cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t A034839 TableForm[cv]  (* A034867 as a triangle *)
%t A034839 (* _Clark Kimberling_, Feb 18 2012 *)
%t A034839 Table[Binomial[n, k], {n, 0, 13}, {k, 0, Floor[n, 2], 2}] // Flatten (* _Michael De Vlieger_, Dec 13 2016 *)
%t A034839 (* The triangle gives coefficients for cosh(nx) as a linear combination of products (cosh(x)^h)*(sinh(x)^k) *)
%t A034839 Column[Table[TrigExpand[Cosh[n  x]], {n, 0, 10}]]
%t A034839 (* _Clark Kimberling_, Aug 03 2024 *)
%o A034839 (PARI) for(n=0,15, for(k=0,floor(n/2), print1(binomial(n, 2*k), ", "))) \\ _G. C. Greubel_, Feb 23 2018
%o A034839 (Magma) /* As a triangle */ [[Binomial(n,2*k):k in [0..Floor(n/2)]] : n in [0..10]]; // _G. C. Greubel_, Feb 23 2018
%Y A034839 Cf. A007318, A034867, A034870, A080928, A119326, A201701.
%Y A034839 Cf. A008619 (row lengths), A086645.
%K A034839 nonn,easy,tabf
%O A034839 0,6
%A A034839 _N. J. A. Sloane_

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