A160551 - OEIS

%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/((1-x)*(1-x^5)*(1-x^10)*(1-x^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)/(x-1)^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/(1-x)/(1-x^5)/(1-x^10)/(1-x^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