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A297199
a(n) = number of partitions of n into consecutive positive cubes.
3
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
1,216
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..65537 (terms 1..10000 from Robert Israel)
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
EXAMPLE
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.
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
PROG
(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
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Jan 15 2018
STATUS
approved