%I #38 Feb 03 2022 00:33:10
%S 1,2,4,6,9,13,18,24,31,39,49,60,73,87,103,121,141,163,187,213,242,273,
%T 307,343,382,424,469,517,568,622,680,741,806,874,946,1022,1102,1186,
%U 1274,1366,1463,1564,1670,1780,1895,2015,2140,2270,2405,2545,2691,2842
%N Expansion of 1/((1-x)^2*(1-x^2)*(1-x^5)).
%C Ways of making change for n cents using coins of 1, 2 and 5 cents, if two different kinds of 1-cent coin are counted as different. - _Matthew Vandermast_, Feb 27 2003
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 113, Example (2), D(n; 1,2,4,10).
%H Vincenzo Librandi, <a href="/A001304/b001304.txt">Table of n, a(n) for n = 0..1000</a>
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=198">Encyclopedia of Combinatorial Structures 198</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1,1,-2,0,2,-1).
%F G.f.: 1/((1-x)^2*(1-x^2)*(1-x^5)) = 1 / ((1+x)*(x^4+x^3+x^2+x+1)*(x-1)^4).
%F a(n) = floor((n+8)*(2*n^2+11*n+18)/120). - _Tani Akinari_, May 14 2014
%p a:= proc(n) local m, r; m:= iquo(n, 10, 'r'); r:= r+1; (53+ (135+ 100*m) *m) *m/6+ [1, 2, 4, 6, 9, 13, 18, 24, 31, 39][r]+ [0, 5, 11, 18, 26, 35, 45, 56, 68, 81][r]*m+ (r-1)*5 *m^2 end: seq(a(n), n=0..100); # _Alois P. Heinz_, Oct 05 2008
%t CoefficientList[Series[1/((1-x)^2*(1-x^2)*(1-x^5)),{x,0,50}],x] (* _Vincenzo Librandi_, Feb 24 2012 *)
%t LinearRecurrence[{2,0,-2,1,1,-2,0,2,-1},{1,2,4,6,9,13,18,24,31},60] (* _Harvey P. Dale_, Oct 03 2018 *)
%o (PARI) a(n)=floor((n+8)*(2*n^2+11*n+18)/120) \\ _Tani Akinari_, May 14 2014
%Y First differences are in A000115.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_