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A179095 Rectified 5-cell numbers: the coefficient of x^{2n-2} in (1+x+x^2+ ... + x^{n-1})^5.

%I A179095 #36 Mar 29 2023 09:02:39
%S A179095 0,1,10,45,135,320,651,1190,2010,3195,4840,7051,9945,13650,18305,
%T A179095 24060,31076,39525,49590,61465,75355,91476,110055,131330,155550,
%U A179095 182975,213876,248535,287245,330310,378045,430776,488840,552585,622370,698565,781551,871720,969475
%N A179095 Rectified 5-cell numbers: the coefficient of x^{2n-2} in (1+x+x^2+ ... + x^{n-1})^5.
%H A179095 J. Conrad, <a href="/A179095/b179095.txt">Table of n, a(n) for n = 0..260</a>
%H A179095 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A179095 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
%F A179095 Comment from _Doron Zeilberger_, Feb 18 2016 (Start):
%F A179095 The conjectures in A179095-A179099 are true. Proof:
%F A179095 The geometric series 1+x+x^2+..+x^(n-1) = (1-x^n)/(1-x).
%F A179095 Hence for a fixed k (in the above cases k=5..9, but the argument holds in general)
%F A179095 the coefficient of  x^(2*n-2) in (1+x+...+x^(n-1))^k =
%F A179095 coefficient of  x^(2*n-2) in (1-x^n)^k*(1-x)^(-k) =
%F A179095 coefficient of  x^(2*n-2) in (1-k*x^n + Sum of powers higher than x^(2*n-2)..)
%F A179095 = coefficient of x^(2*n-2) in (1-x)^(-k) -k*(the coefficient of x^(n-2) in (1-x)^(-k))
%F A179095 = (-1)^(2*n-2)*binomial(-k,2*n-2)- k* (-1)^(n-2)*binomial(-k,n-2)=
%F A179095 Using  (-1)^m *binomial(-m,k)= binomial(m+k-1,k-1) this is
%F A179095 binomial(k+2*n-3,k-1) - k *binomial(k+n-3,k-1)
%F A179095 and this agrees with the conjectures for k=5..9 (End)
%F A179095 E.g.f.: exp(x)*x*(24 + 96*x + 72*x^2 + 11*x^3)/24. - _Stefano Spezia_, Mar 28 2023
%t A179095 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 *)
%o A179095 (PARI) a(n) = polcoeff(((x^n-1)/(x-1))^5, 2*n-2); \\ _Michel Marcus_, Feb 17 2016
%o A179095 (PARI) A179095(n)=n*(11*n^3+6*n^2+n+6)\24 \\ _M. F. Hasler_, Feb 19 2016
%Y A179095 Cf. A179096, A179097, A179098, A179099.
%K A179095 nonn,easy
%O A179095 0,3
%A A179095 _Michael A. Jackson_, Jun 29 2010
%E A179095 More terms from _Robert G. Wilson v_, Jul 30 2010