A297199 a(n) = number of partitions of n into consecutive positive cubes.

1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Seiichi Manyama, Jan 15 2018

Keywords

• nonn

Examples

			    1 = 1^3,                   so   a(1) = 1.
    8 = 2^3,                   so   a(8) = 1.
    9 = 1^3 + 2^3,             so   a(9) = 1.
   27 = 3^3,                   so  a(27) = 1.
   35 = 2^3 + 3^3,             so  a(35) = 1.
   36 = 1^3 + 2^3 + 3^3,       so  a(36) = 1.
   64 = 4^3,                   so  a(64) = 1.
   91 = 3^3 + 4^3,             so  a(91) = 1.
   99 = 2^3 + 3^3 + 4^3,       so  a(99) = 1.
  100 = 1^3 + 2^3 + 3^3 + 4^3, so a(100) = 1.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537 (terms 1..10000 from Robert Israel)

Crossrefs

Cf. A000578, A001227, A062682, A217843, A265845, A296338, A307609.

Programs

Maple

N:= 200: # to get a(1)..a(N)
F:= (a, b) -> (b^2*(b+1)^2-a^2*(a-1)^2)/4:
A:= Vector(N):
for b from 1 to floor(N^(1/3)) do
  for a from b to 1 by -1 do
     v:= F(a,b);
     if v > N then break fi;
     A[v]:= A[v]+1;
od od:
convert(A,list); # Robert Israel, Jan 15 2018, corrected Jan 29 2018

PARI

A297199(n) = { my(s=0, k=1, c); while((c=k^3) <= n, my(u=n-c, i=k); while(u>0, i++; c = i^3; u=u-c); s += (!u); k++); (s); }; \\ Antti Karttunen, Aug 22 2019

Formula

a(A217843(n)) >= 1 for n > 1.

a(n) >= 2 for n in A265845. - Robert Israel, Jan 15 2018

G.f.: Sum_{i>=1} Sum_{j>=i} Product_{k=i..j} x^(k^3). - Ilya Gutkovskiy, Apr 18 2019

a(A000578(n)) = A307609(n). - Antti Karttunen, Aug 22 2019