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%I #30 Oct 19 2017 10:46:16
%S 5,25,125,625,3125,15625,78005,384550,1829850,8209410,34219650,
%T 131875900,470597480,1562441800,4855374080,14208711350,39381411950,
%U 103917328350,262270328730,635683810740,1484963848500,3353799866500
%N a(n) = the number of sequences of n integers such that each integer is in the range 0..4 and the sum of the integers is in the range 0..24.
%C Changing 4 to 2 and 24 to 3 gives A105163. - _Don Reble_, Aug 14 2012
%H Colin Barker, <a href="/A102169/b102169.txt">Table of n, a(n) for n = 1..1000</a>
%F From _Michael David Hirschhorn_, Aug 10 2012: (Start)
%F a(n) is the sum of the coefficients of 1, x, x^2, ..., x^24 in (1+x+x^2+x^3+x^4)^n = (1-x^5)^n/(1-x)^n.
%F But this is equal to the coefficient of x^24 in (1-x^5)^n/(1-x)^(n+1) = Sum_{k=0..n} (-1)^k binomial(n,k) x^5k times Sum_{m>=0} binomial(n+m,m) x^m.
%F Hence a(n) = Sum_{k=0..4} (-1)^k binomial(n,k) binomial(n+24-5k,n).
%F For example, if n=2, a(2) = 325-420+120 = 25. (End)
%F G.f.: -x*(x^24 -25*x^23 +300*x^22 -2300*x^21 +12650*x^20 -53060*x^19 +175980*x^18 -472300*x^17 +1042375*x^16 -1915575*x^15 +2962780*x^14 -3894200*x^13 +4384980*x^12 -4251000*x^11 +3547700*x^10 -2533840*x^9 +1532975*x^8 -776575*x^7 +325880*x^6 -111900*x^5 +30750*x^4 -6500*x^3 +1000*x^2 -100*x +5) / (x-1)^25. - _Colin Barker_, Nov 01 2014
%e a(2)=25 because there are five choices for either integer.
%t Table[Sum[(-1)^k * Binomial[n,k] * Binomial[n+24-5k,n],{k,0,4}],{n,1,20}] (* _Vaclav Kotesovec_, Nov 01 2014 after _Michael David Hirschhorn_ *)
%Y Cf. A105163.
%K nonn,easy
%O 1,1
%A Tony Berard (TheMathDude(AT)worldnet.att.net), Feb 16 2005
%E Edited by _Don Reble_, Mar 19 2007
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