A003328 - OEIS

A003328

Numbers that are the sum of 5 positive cubes.

5, 12, 19, 26, 31, 33, 38, 40, 45, 52, 57, 59, 64, 68, 71, 75, 78, 82, 83, 89, 90, 94, 96, 97, 101, 108, 109, 115, 116, 120, 127, 129, 131, 134, 135, 136, 138, 143, 145, 146, 150, 152, 153, 155, 157, 162, 164, 169, 171, 172, 176, 181, 183, 188, 190, 192, 194, 195, 199

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

As the order of addition doesn't matter we can assume terms are in increasing order. - David A. Corneth, Aug 01 2020

It seems only a finite number N of positive integers are not in this sequence, and thus a(n) = n - N for all sufficiently large n. Is it true that 2243453, last term of A048927, is sufficiently large in that sense? - M. F. Hasler, Jan 04 2023

LINKS

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)

Eric Weisstein's World of Mathematics, Cubic Number.

EXAMPLE

From David A. Corneth, Aug 01 2020: (Start)

3084 is in the sequence as 3084 = 5^3 + 5^3 + 5^3 + 8^3 + 13^3.

4385 is in the sequence as 4385 = 4^3 + 4^3 + 9^3 + 11^3 + 13^3.

5426 is in the sequence as 5426 = 8^3 + 9^3 + 9^3 + 12^3 + 12^3. (End)

PROG

(PARI) select( {is_A003328(n, k=5, m=3, L=sqrtnint(abs(n-k+1), m))=if( n>k*L^m || n<k, 0, n<k*L^m, forstep(r=min(k-1, n\L^m), 0, -1, self()(n-r*L^m, k-r, m, L-1) && return(1)), 1)}, [1..200]) \\ M. F. Hasler, Aug 02 2020

A003328_upto(N, k=5, m=3)=[i|i<-[1..#N=sum(n=1, sqrtnint(N, m), 'x^n^m, O('x^N))^k], polcoef(N, i)] \\ M. F. Hasler, Aug 02 2020

(Python)

from collections import Counter

from itertools import combinations_with_replacement as combs_w_rep

def aupto(lim):

s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))

s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_w_rep(s, 5)))

s2counts = Counter(s2)

return sorted(k for k in s2counts)

print(aupto(200)) # Michael S. Branicky, May 12 2021

CROSSREFS

Cf. A003328, A048926, A048297.