A057788 Expansion of (1+x)/(1-x)^12.

1, 13, 90, 442, 1729, 5733, 16744, 44200, 107406, 243542, 520676, 1058148, 2057510, 3848222, 6953544, 12183560, 20764055, 34512075, 56071470, 89224590, 139299615, 213696795, 322561200, 479634480, 703323660, 1018031196, 1455797448, 2058314440, 2879378332

Offset: 0

Author

N. J. A. Sloane, Nov 04 2000

Keywords

• nonn
• easy

Comments

1/2^10 of twelfth unsigned column of triangle A053120 (T-Chebyshev, rising powers, zeros omitted).

If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-12) is the number of 12-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

11-dimensional square numbers, tenth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = sum_{i=0..n} C(n+10,i+10)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

2*a(n) is number of ways to place 10 queens on an (n+10) X (n+10) chessboard so that they diagonally attack each other exactly 45 times. The maximal possible attack number, p=binomial(k,2) =45 for k=10 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form the corresponding complete graph. - Antal Pinter, Dec 27 2015

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000
• Milan Janjic, Two Enumerative Functions.
• Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
• Index entries for linear recurrences with constant coefficients, signature (12,-66, 220,-495,792,-924,792,-495,220,-66,12,-1).
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

Cf. A053120, A054334, A054333, A053347, A002415, A005585, A040977, A050486.

Partial sums of A054334.

Sixth column of A111125 (s=5, without leading zeros). - Wolfdieter Lang, Oct 18 2012

Programs

GAP

List([0..30], n -> (2*n+11)*Binomial(n+10, 10)/11); # G. C. Greubel, Dec 02 2018

Magma

[Binomial(n+10,10)*(2*n+11)/11: n in [0..40]]; // Vincenzo Librandi, Feb 14 2016

Maple

A057788 := proc(n)
        1/39916800*(2*n+11) *(n+10) *(n+9) *(n+8) *(n+7) *(n+6) *(n+5) *(n+4) *(n+3) *(n+2) *(n+ 1) ; end proc: # R. J. Mathar, Mar 22 2011

Mathematica

Table[(2*n+11)*Binomial[n+10, 10]/11, {n,0,40}] (* G. C. Greubel, Dec 02 2018 *)
CoefficientList[Series[(1 + x) / (1 - x)^12, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 14 2016 *)
LinearRecurrence[{12,-66,220,-495,792,-924,792,-495,220,-66,12,-1},{1,13,90,442,1729,5733,16744,44200,107406,243542,520676,1058148},30] (* Harvey P. Dale, Sep 07 2022 *)

PARI

Vec((1+x)/(1-x)^12+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

Sage

[(2*n+11)*binomial(n+10, 10)/11 for n in range(40)] # G. C. Greubel, Dec 02 2018

Formula

a(n) = 2*C(n+11, 11) - C(n+10, 10). - Paul Barry, Mar 04 2003

a(n) = C(n+10,10) + 2*C(n+10,11). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

a(n) = C(n+10,10)*(2n+11)/11. - Antal Pinter, Dec 27 2015

a(n) = 12*a(n-1)-66*a(n-2)+220*a(n-3)-495*a(n-4)+792*a(n-5)-924*a(n-6)+792*a(n-7)-495*a(n-8)+220*a(n-9)-66*a(n-10)+12*a(n-11)-a(n-12) for n >11. - Vincenzo Librandi, Feb 14 2016

a(n) = (2*n+11)*binomial(n+10, 10)/11. - G. C. Greubel, Dec 02 2018

From Amiram Eldar, Jan 26 2022: (Start)

Sum_{n>=0} 1/a(n) = 419751541/13230 - 2883584*log(2)/63.

Sum_{n>=0} (-1)^n/a(n) = 720896*Pi/63 - 237793798/6615. (End)