A272380 a(n) = n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155).

0, 1, 342, 6315, 40492, 157125, 456546, 1099567, 2321880, 4448457, 7907950, 13247091, 21145092, 32428045, 48083322, 69273975, 97353136, 133878417, 180626310, 239606587, 313076700, 403556181, 513841042, 647018175, 806479752, 995937625, 1219437726, 1481374467

Offset

0,3

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.: x*(1 + 336*x + 4278*x^2 + 7712*x^3 + 2073*x^4)/(1-x)^6.

E.g.f.: x*(1 + 170*x + 882*x^2 + 720*x^3 + 120*x^4)*exp(x).

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), for n>5.

a(n) = n^2*A272379(n) - n*(n-1)*A272379(n-1), see page 7 in Brent's paper.

From Peter Bala, Jan 30 2019: (Start)

Let a(n,x) = Product_{k = 0..n} (x - k)/(x + k). Then for positive integer x we have x^2*(120*x^4 - 480*x^3 + 762*x^2 - 556*x + 155) = Sum_{n >= 0} ((n+1)^11 + n^11)*a(n,x) and x*(120*x^4 - 480*x^3 + 762*x^2 - 556*x + 155) = Sum_{n >= 0} ((n+1)^10 - n^10)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 11/2. See the Bala link in A036970. Cf. A272378 and A272379. (End)

Mathematica

Table[n (120 n^4 - 480 n^3 + 762 n^2 - 556 n + 155), {n, 0, 50}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 342, 6315, 40492, 157125}, 40] (* Harvey P. Dale, Mar 15 2018 *)

Prog

(Magma) [n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155): n in [0..50]];
(PARI) vector(100, n, n--; n*(120*n^4 - 480*n^3 + 762*n^2 - 556*n + 155)) \\ Altug Alkan, Apr 29 2016

Crossrefs

Cf. A272378, A272379.

Sequence in context: A158595 A231267 A252235 * A023909 A035844 A028681

Adjacent sequences: A272377 A272378 A272379 * A272381 A272382 A272383

Keyword

nonn,easy

Author

Vincenzo Librandi, Apr 29 2016

Status

approved