%I #17 Sep 08 2022 08:46:16

%S 1,14763,628805,5501167,24943689,79549811,203823373,449807415,

%T 890712977,1624547899,2777745621,4508793983,7011864025,10520438787,

%U 15310942109,21706367431,30079906593,40858578635,54526858597,71630306319,92779195241,118652141203

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

%C 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

%H Vincenzo Librandi, <a href="/A272128/b272128.txt">Table of n, a(n) for n = 0..1000</a>

%H Richard P. Brent, <a href="http://arxiv.org/abs/1407.3533">Generalising Tuenter's binomial sums</a>, arXiv:1407.3533 [math.CO], 2014. (page 16).

%H Richard P. Brent, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Brent/brent5.html">Generalising Tuenter's binomial sums</a>, Journal of Integer Sequences, 18 (2015), Article 15.3.2.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F O.g.f.: (1 + 14757*x + 540242*x^2 + 1949762*x^3 + 1073517*x^4 + 50521*x^5)/(1-x)^6.

%F E.g.f.: (1 + 14762*x + 299640*x^2 + 609840*x^3 + 277200*x^4 + 30240*x^5)*exp(x).

%F a(n) = (2*n+1)*(15120*n^4-20160*n^3+12600*n^2-2640*n+1).

%F 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).

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

%F From _Peter Bala_, Feb 01 2019: (Start)

%F a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^10 * binomial(2*n + 1, n - k).

%F 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)

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

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

%o (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

%Y Cf. A014641, A272126, A272127.

%K easy,nonn

%O 0,2

%A _Vincenzo Librandi_, Apr 25 2016