OFFSET
0,3
COMMENTS
I
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).
FORMULA
a(n) = (4*n^4+8*n^3+2*n^2+4*n+3*(1-(-1)^n))/24. Therefore :
a(2*k) = k*(k+1)*(8*k^2+1)/3, a(2*k+1) = (k+1)*(8*k^3+16*k^2+9*k+3)/3.
From Colin Barker, Nov 11 2016: (Start)
G.f.: x*(1 + 2*x + 5*x^2) / ((1 - x)^5 * (1 + x)).
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6) for n>5.
(End)
EXAMPLE
a(0) = 0, a(1) = 1, a(2) = 6, a(3) = 24, a(4) = 66.
MATHEMATICA
LinearRecurrence[{4, -5, 0, 5, -4, 1}, {0, 1, 6, 24, 66, 149}, 60] (* Harvey P. Dale, Jun 19 2021 *)
PROG
(PARI) concat(0, Vec(x*(1 + 2*x + 5*x^2) / ((1 - x)^5 * (1 + x)) + O(x^50))) \\ Colin Barker, Nov 11 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Luce ETIENNE, Nov 11 2016
STATUS
approved