OFFSET
0,3
LINKS
J. Conrad, Table of n, a(n) for n = 0..260
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
Conjectures: a(n) = n*(11*n^3+6*n^2+n+6)/24. G.f.: x*(1+5*x+5*x^2)/(1-x)^5. - Colin Barker, Jan 09 2012
Comment from Doron Zeilberger, Feb 18 2016 (Start):
The geometric series 1+x+x^2+..+x^(n-1) = (1-x^n)/(1-x).
Hence for a fixed k (in the above cases k=5..9, but the argument holds in general)
the coefficient of x^(2*n-2) in (1+x+...+x^(n-1))^k =
coefficient of x^(2*n-2) in (1-x^n)^k*(1-x)^(-k) =
coefficient of x^(2*n-2) in (1-k*x^n + Sum of powers higher than x^(2*n-2)..)
= coefficient of x^(2*n-2) in (1-x)^(-k) -k*(the coefficient of x^(n-2) in (1-x)^(-k))
= (-1)^(2*n-2)*binomial(-k,2*n-2)- k* (-1)^(n-2)*binomial(-k,n-2)=
Using (-1)^m *binomial(-m,k)= binomial(m+k-1,k-1) this is
binomial(k+2*n-3,k-1) - k *binomial(k+n-3,k-1)
and this agrees with the conjectures for k=5..9 (End)
E.g.f.: exp(x)*x*(24 + 96*x + 72*x^2 + 11*x^3)/24. - Stefano Spezia, Mar 28 2023
MATHEMATICA
f[n_] := CoefficientList[ Series[ Sum[x^k, {k, 0, n - 1}]^5, {x, 0, 2 n + 3}], x][[2 n - 1]]; Array[f, 36] (* Robert G. Wilson v, Jul 30 2010 *)
PROG
(PARI) a(n) = polcoeff(((x^n-1)/(x-1))^5, 2*n-2); \\ Michel Marcus, Feb 17 2016
(PARI) A179095(n)=n*(11*n^3+6*n^2+n+6)\24 \\ M. F. Hasler, Feb 19 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Michael A. Jackson, Jun 29 2010
EXTENSIONS
More terms from Robert G. Wilson v, Jul 30 2010
STATUS
approved