OFFSET
0,2
COMMENTS
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
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 (5,-10,10,-5,1).
FORMULA
O.g.f.: (1+1636*x+18270*x^2+19028*x^3+1385*x^4)/(1-x)^5.
E.g.f.: (1+1640*x+11592*x^2+10080*x^3+1680*x^4)*exp(x).
a(n) = (2*n+1)*(840*n^3-420*n^2+126*n+1).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5), for n>4
See page 7 in Brent's paper:
A272128(n) = (2*n+1)^2*a(n) - 2*n*(2*n+1)*a(n-1).
From Peter Bala, Jan 22 2019: (Start)
a(n) = 1/4^n * Sum_{k = 0..n} (2*k + 1)^8 * binomial(2*n + 1, n - k).
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)
MATHEMATICA
Table[1680 n^4 - 168 n^2 + 128 n + 1, {n, 0, 30}]
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 1641, 26465, 134953, 427905}, 30] (* Harvey P. Dale, Nov 27 2017 *)
PROG
(Magma) [1680*n^4-168*n^2+128*n+1: n in [0..50]];
(PARI) a(n) = 1680*n^4 - 168*n^2 + 128*n + 1; \\ Altug Alkan, Apr 30 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Apr 25 2016
STATUS
approved