OFFSET
0,2
COMMENTS
This is the polynomial Qbar(5,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - Peter Bala, Feb 01 2019
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16).
Richard P. Brent, Generalising Tuenter's binomial sums, Journal of Integer Sequences, 18 (2015), Article 15.3.2.
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
O.g.f.: (1 + 14757*x + 540242*x^2 + 1949762*x^3 + 1073517*x^4 + 50521*x^5)/(1-x)^6.
E.g.f.: (1 + 14762*x + 299640*x^2 + 609840*x^3 + 277200*x^4 + 30240*x^5)*exp(x).
a(n) = (2*n+1)*(15120*n^4-20160*n^3+12600*n^2-2640*n+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).
From Peter Bala, Feb 01 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^10 * binomial(2*n + 1, n - k).
a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^10*(n - 1)/(n + 1) + 5^10*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^10*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)
MATHEMATICA
Table[30240 n^5 - 25200 n^4 + 5040 n^3 + 7320 n^2 - 2638 n + 1, {n, 0, 30}]
PROG
(Magma) [30240*n^5-25200*n^4+5040*n^3+7320*n^2-2638*n+1: n in [0..40]];
(PARI) a(n)=30240*n^5-25200*n^4+5040*n^3+7320*n^2-2638*n+1 \\ Charles R Greathouse IV, Apr 30 2016
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Vincenzo Librandi, Apr 25 2016
STATUS
approved