A040977 a(n) = binomial(n+5,5)*(n+3)/3.

1, 8, 35, 112, 294, 672, 1386, 2640, 4719, 8008, 13013, 20384, 30940, 45696, 65892, 93024, 128877, 175560, 235543, 311696, 407330, 526240, 672750, 851760, 1068795, 1330056, 1642473, 2013760, 2452472, 2968064, 3570952, 4272576, 5085465

Offset: 0

Author

Barry E. Williams, Dec 14 1999

Keywords

• easy
• nonn

Comments

Sequence is n^2*(n^2-1)*(n^2-4)/360 if offset 3.

If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-7) is the number of 7-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

6-dimensional square numbers, fifth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+5,i+5)*b(i), where b(i) = [1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

Sequence of the absolute values of the z^2 coefficients divided by 5 of the polynomials in the GF2 denominators of A156925. See A157703 for background information. - Johannes W. Meijer, Mar 07 2009

2*a(n) is number of ways to place 5 queens on an (n+5) X (n+5) chessboard so that they diagonally attack each other exactly 10 times. The maximal possible attack number, p=binomial(k,2)=10 for k=5 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form a corresponding complete graph. - Antal Pinter, Dec 27 2015

Ehrhart polynomial for the Chan-Robbins-Yuen polytope CRY_4. [De Loera et al.] - N. J. A. Sloane, Apr 16 2016

Coefficients in the terminating series identity 1 - 8*n/(n + 7) + 35*n*(n - 1)/((n + 7)*(n + 8)) - 112*n*(n - 1)*(n - 2)/((n + 7)*(n + 8)*(n + 9)) + ... = 0 for n = 1,2,3,.... Cf. A005585 and A050486. - Peter Bala, Feb 18 2019

References

• Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.
• Herbert John Ryser, Combinatorial Mathematics, "The Carus Mathematical Monographs", No. 14, John Wiley and Sons, 1963, pp. 1-16.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Jesús A. De Loera, Fu Liu and Ruriko Yoshida, A generating function for all semi-magic squares and the volume of the Birkhoff polytope, J. Algebraic Combin., Vol. 30, No. 1 (2009), pp. 113-139. See page 138, n=4 entry in table.
• 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 sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Partial sums of A005585.

Cf. A000292, A000579, A050486, A053120, A133111, A133112.

Cf. A156925, A157703. - Johannes W. Meijer, Mar 07 2009

Programs

Magma

[Binomial(n+5, 5) + 2*Binomial(n+5, 6): n in [0..35]]; // Vincenzo Librandi, Jun 09 2013

Maple

with(combinat); A040977 := n->binomial(n+5,5)*(n+3)/3;
a:=n->(sum((numbcomp(n,6)), j=4..n))/3:seq(a(n), n=6..38); # Zerinvary Lajos, Aug 26 2008
nmax:=34; for n from 0 to nmax do fz(n):=product((1-m*z)^(n+1-m),m=1..n); c(n):= abs(coeff(fz(n),z,2))/5; end do: a:=n-> c(n): seq(a(n), n=2..nmax); # Johannes W. Meijer, Mar 07 2009

Mathematica

CoefficientList[Series[(1 + x) / (1 - x)^7, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{1,8,35,112,294,672,1386},40] (* Harvey P. Dale, Feb 20 2016 *)

PARI

vector(20,n,n--;2*binomial(n+6,6)-binomial(n+5,5)) \\ Derek Orr, May 05 2015

PARI

Vec((1+x)/(1-x)^7 + O(x^100)) \\ Altug Alkan, Nov 29 2015

Formula

a(n) = (-1)^n*A053120(2*n+6, 6)/32, (1/32 of seventh unsigned column of Chebyshev T-triangle, zeros omitted).

G.f.: (1+x)/(1-x)^7.

a(n-3) = Sum_{i+j+k=n} i*j*k^2. - Benoit Cloitre, Nov 01 2002

a(n) = 2*binomial(n+6, 6) - binomial(n+5, 5). - Paul Barry, Mar 04 2003

a(n-3) = 1/(1!*2!*3!)*Sum_{1 <= x_1, x_2, x_3 <= n} |det V(x_1,x_2,x_3)| = 1/12*Sum_{1 <= i,j,k <= n} |(i-j)(i-k)(j-k)|, where V(x_1,x_2,x_3) is the Vandermonde matrix of order 3. - Peter Bala, Sep 13 2007

a(n) = binomial(n+5,5) + 2*binomial(n+5,6). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009

a(n) = (n+1)*(n+2)*(n+3)^2*(n+4)*(n+5)/360. - Wesley Ivan Hurt, May 05 2015

a(n) = A000579(n+5) + A000579(n+6). - R. J. Mathar, Nov 29 2015

Sum_{n>=0} 1/a(n) = 15*Pi^2 - 1175/8. - Jaume Oliver Lafont, Jul 11 2017

Sum_{n>=0} (-1)^n/a(n) = 15*Pi^2/2 - 585/8. - Amiram Eldar, Jan 24 2022