A179095 Rectified 5-cell numbers: the coefficient of x^{2n-2} in (1+x+x^2+ ... + x^{n-1})^5.
0, 1, 10, 45, 135, 320, 651, 1190, 2010, 3195, 4840, 7051, 9945, 13650, 18305, 24060, 31076, 39525, 49590, 61465, 75355, 91476, 110055, 131330, 155550, 182975, 213876, 248535, 287245, 330310, 378045, 430776, 488840, 552585, 622370, 698565, 781551, 871720, 969475
Offset: 0
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).
Programs
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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 *) -
PARI
a(n) = polcoeff(((x^n-1)/(x-1))^5, 2*n-2); \\ Michel Marcus, Feb 17 2016
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PARI
A179095(n)=n*(11*n^3+6*n^2+n+6)\24 \\ M. F. Hasler, Feb 19 2016
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
Extensions
More terms from Robert G. Wilson v, Jul 30 2010