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

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

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

Examples

			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)
		

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.

Crossrefs

Cf. A003328, A048926, A048297.

Cf. A057906 (Complement)

Cf. A###### (x, y) = Numbers that are the sum of x nonzero y-th powers:

A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A003072 (3, 3), A003325 (3, 2), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3), A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4), A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5), A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6), A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9), A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10), A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11), A047700 (5, 2).

Programs

PARI

select( {is_A003328(n,k=5,m=3,L=sqrtnint(abs(n-k+1),m))=if( n>k*L^m || nM. 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