%I #17 Sep 08 2022 08:46:16
%S 1,1641,26465,134953,427905,1046441,2172001,4026345,6871553,11010025,
%T 16784481,24577961,34813825,47955753,64507745,85014121,110059521,
%U 140268905,176307553,218881065,268735361,326656681,393471585,470046953,557289985,656148201,767609441
%N a(n) = 1680*n^4 - 168*n^2 + 128*n + 1.
%C This is the polynomial Qbar(4,n) in Brent. See A160485 for the triangle of coefficients (with signs) of the Qbar polynomials. - _Peter Bala_, Jan 21 2019
%H Vincenzo Librandi, <a href="/A272127/b272127.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_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F O.g.f.: (1+1636*x+18270*x^2+19028*x^3+1385*x^4)/(1-x)^5.
%F E.g.f.: (1+1640*x+11592*x^2+10080*x^3+1680*x^4)*exp(x).
%F a(n) = (2*n+1)*(840*n^3-420*n^2+126*n+1).
%F a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), for n>4
%F See page 7 in Brent's paper:
%F a(n) = (2*n+1)^2*A272126(n) - 2*n*(2*n+1)*A272126(n-1).
%F A272128(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
%F From _Peter Bala_, Jan 22 2019: (Start)
%F a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^8 * binomial(2*n + 1, n - k).
%F a(n-1) = 2/4^n * binomial(2*n,n) * ( 1 + 3^8*(n - 1)/(n + 1) + 5^8*(n - 1)*(n - 2)/((n + 1)*(n + 2)) + 7^8*(n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)) + ... ). (End)
%t Table[1680 n^4 - 168 n^2 + 128 n + 1, {n, 0, 30}]
%t LinearRecurrence[{5,-10,10,-5,1},{1,1641,26465,134953,427905},30] (* _Harvey P. Dale_, Nov 27 2017 *)
%o (Magma) [1680*n^4-168*n^2+128*n+1: n in [0..50]];
%o (PARI) a(n) = 1680*n^4 - 168*n^2 + 128*n + 1; \\ _Altug Alkan_, Apr 30 2016
%Y Cf. A014641, A272126, A160485, A245244, A272129.
%K nonn,easy
%O 0,2
%A _Vincenzo Librandi_, Apr 25 2016