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 (4,-6,4,-1)
FORMULA
G.f.: x*(1 + 18*x + 17*x^2)/(1 - x)^4.
E.g.f.: x*(1 + 10*x + 6*x^2)*exp(x).
a(n) = n*A080859(n+1).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), for n>3.
See page 7 in Brent's paper:
A272379(n) = n^2*a(n) - n*(n-1)*a(n-1).
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*(6*x^2 - 8*x + 3) = Sum_{n >= 0} ((n+1)^7 + n^7)*a(n,x) and x*(6*x^2 - 8*x + 3) = Sum_{n >= 0} ((n+1)^6 - n^6)*a(n,x). Both identities are also valid for complex x in the half-plane Re(x) > 7/2. See the Bala link in A036970. Cf. A272379. (End)
MATHEMATICA
Table[n (6 n^2 - 8 n + 3), {n, 0, 50}]
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 22, 99}, 40] (* Harvey P. Dale, Dec 29 2017 *)
PROG
(Magma) [n*(6*n^2 - 8*n + 3): n in [0..50]];
(PARI) vector(100, n, n--; n*(6*n^2 - 8*n + 3)) \\ Altug Alkan, Apr 29 2016
(Python) for n in range(0, 10**3):print(n*(6*n**2-8*n+3), end=", ") # Soumil Mandal, Apr 30 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Apr 29 2016
STATUS
approved