A303824 Number of ways of writing n as a sum of powers of 6, each power being used at most six times.

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2

Offset: 0

Author

Seiichi Manyama, May 01 2018

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..10000

Crossrefs

Number of ways of writing n as a sum of powers of b, each power being used at most b times: A054390 (b=3), A277872 (b=4), A277873 (b=5), this sequence (b=6), A303825 (b=7).

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      add(b(n-j*6^i, i-1), j=0..min(6, n/6^i))))
    end:
a:= n-> b(n, ilog[6](n)):
seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018

Mathematica

m = 100; A[_] = 1;
Do[A[x_] = Total[x^Range[0, 6]] A[x^6] + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 19 2019 *)

Ruby

def A(k, n)
  ary = [1]
  (1..n).each{|i|
    s = ary[i / k]
    s += ary[i / k - 1] if i % k == 0
    ary << s
  }
  ary
end
p A(6, 100)

Formula

G.f.: Product_{k>=0} (1-x^(7*6^k))/(1-x^(6^k)).

a(0)=1; for k>0, a(6*k) = a(k)+a(k-1) and a(6*k+r) = a(k) with r=1,2,3,4,5.

G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6) * A(x^6). - Ilya Gutkovskiy, Jul 09 2019