A277872 Number of ways of writing n as a sum of powers of 4, each power being used at most four times.
1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 3, 3, 3, 5, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 5, 3, 3, 3, 4, 1
Offset: 0
Keywords
Examples
a(72) = 4 because 72 = 64+4+4 = 64+4+1+1+1+1 = 16+16+16+16+4+4 = 16+16+16+16+4+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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Mathematica
n:=250; r:=3; (* To get up to n-th term, need r such that 4^r < n < 4^(r+1) *) h4 := CoefficientList[ Series[ Product[ (1 - q^(5*4^i))/(1 - q^(4^i)) , {i, 0, r}], {q, 0, n} ], q]
Formula
G.f.: Product_{j>=0} (1-x^(5*4^j))/(1-x^(4^j)).
G.f.: Product_{j>=0} (1+x^(4^j)+x^(2*4^j)+x^(3*4^j)+x^(4*4^j)).
a(0)=1 and for n>0, a(4n)=a(n)+a(n-1), a(4n+r)=a(n) for r=1,2,3.
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4) * A(x^4). - Ilya Gutkovskiy, Jul 09 2019
Comments