%I #15 Oct 08 2022 22:13:50
%S 1,242,1463,4464,10045,19006,32147,50268,74169,104650,142511,188552,
%T 243573,308374,383755,470516,569457,681378,807079,947360,1103021,
%U 1274862,1463683,1670284,1895465,2140026,2404767,2690488,2997989,3328070,3681531,4059172,4461793
%N Number of unordered ways of making change for n dollars using coins of denominations 1, 5, 10, and 25.
%C a(n) is the number of distinct quadruplets (p, k, d, q) of nonnegative integers satisfying p + 5k + 10d + 25q = 100n.
%H Andrew Howroyd, <a href="/A160551/b160551.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,6,4,1).
%F a(n) = [x^(100*n)] 1/((1x)*(1x^5)*(1x^10)*(1x^25)).
%F a(n) = (3 + 53*n + 270*n^2 + 400*n^3) / 3.
%F From _Alois P. Heinz_, Oct 08 2022: (Start)
%F a(n) = A001299(100*n).
%F G.f.: (60*x^3+501*x^2+238*x+1)/(x1)^4. (End)
%e There are four ways to make $0.10: (1) 10 pennies, (2) 5 pennies and 1 nickel, (3) 2 nickels, and (4) 1 dime.
%p f := 1/(1x)/(1x^5)/(1x^10)/(1x^25); a := n > (convert(series(f,x,100*n+1),polynom)convert(series(f,x,100*n),polynom)) /x^(100*n);
%p a := n > (3 + 53*n + 270*n^2 + 400*n^3) / 3;
%o (PARI) a(n) = {(3 + 53*n + 270*n^2 + 400*n^3) / 3} \\ _Andrew Howroyd_, Feb 02 2020
%Y Cf. A001299.
%K nonn,easy
%O 0,2
%A _Lee A. Newberg_, May 18 2009, Jun 15 2009
%E Terms a(21) and beyond from _Andrew Howroyd_, Feb 02 2020
