A272128 a(n) = 30240*n^5-25200*n^4+5040*n^3+7320*n^2-2638*n+1.

1, 14763, 628805, 5501167, 24943689, 79549811, 203823373, 449807415, 890712977, 1624547899, 2777745621, 4508793983, 7011864025, 10520438787, 15310942109, 21706367431, 30079906593, 40858578635, 54526858597, 71630306319, 92779195241, 118652141203

Offset: 0

Author

Vincenzo Librandi, Apr 25 2016

Keywords

• easy
• nonn

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).

Crossrefs

Cf. A014641, A272126, A272127.

Programs

Magma

[30240*n^5-25200*n^4+5040*n^3+7320*n^2-2638*n+1: n in [0..40]];

Mathematica

Table[30240 n^5 - 25200 n^4 + 5040 n^3 + 7320 n^2 - 2638 n + 1, {n, 0, 30}]

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

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).

See page 7 in Brent's paper: a(n) = (2*n+1)^2*A272127(n) - 2*n*(2*n+1)*A272127(n-1).

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)