A277873 Number of ways of writing n as a sum of powers of 5, each power being used at most five times.
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 1, 1, 2
Offset: 0
Keywords
Examples
a(140) = 4 because 140 = 125+5+5+5 = 125+5+5+1+1+1+1+1 = 25+25+25+25+25+5+5+5 = 25+25+25+25+25+5+5+1+1+1+1+1.
Links
- Timothy B. Flowers, Table of n, a(n) for n = 0..10000
- K. Courtright and J. Sellers, Arithmetic properties for hyper m-ary partition functions, Integers, 4 (2004), A6.
- Timothy B. Flowers, Extending a Recent Result on Hyper m-ary Partition Sequences, Journal of Integer Sequences, Vol. 20 (2017), #17.6.7.
- T. B. Flowers and S. R. Lockard, Identifying an m-ary partition identity through an m-ary tree, Integers, 16 (2016), A10.
Programs
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Maple
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0, add(b(n-j*5^i, i-1), j=0..min(5, n/5^i)))) end: a:= n-> b(n, ilog[5](n)): seq(a(n), n=0..120); # Alois P. Heinz, May 01 2018 -
Mathematica
n:=250; r:=3; (* To get up to n-th term, need r such that 5^r < n < 5^(r+1) *) h5 := CoefficientList[ Series[ Product[ (1 - q^(6*5^i))/(1 - q^(5^i)) , {i, 0, r}], {q, 0, n} ], q]
Formula
G.f.: Product_{j >= 0} (1-x^(6*5^j))/(1-x^(5^j)).
G.f.: Product_{j >= 0} Sum_{k=0..5} x^(k*5^j).
a(0)=1; for k>0, a(5*k) = a(k)+a(k-1) and a(5*k+r) = a(k) with r=1,2,3,4.
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^5). - Ilya Gutkovskiy, Jul 09 2019
Comments