A345154 Numbers that are the sum of four third powers in exactly nine ways.

42120, 46683, 50806, 50904, 51408, 51480, 51688, 52208, 53865, 54971, 56385, 57113, 60515, 60984, 62433, 65303, 66276, 66339, 66430, 67158, 69048, 69832, 69930, 71162, 72072, 72520, 72576, 72800, 73017, 77714, 77903, 79345, 79667, 79849, 80066, 80073, 81207

Offset

1,1

Comments

Differs from A345146 at term 1 because 21896 = 1^3 + 11^3 + 19^3 + 22^3 = 2^3 + 2^3 + 12^3 + 26^3 = 2^3 + 3^3 + 19^3 + 23^3 = 2^3 + 5^3 + 15^3 + 25^3 = 3^3 + 10^3 + 16^3 + 24^3 = 3^3 + 17^3 + 19^3 + 19^3 = 4^3 + 6^3 + 20^3 + 22^3 = 5^3 + 8^3 + 14^3 + 25^3 = 7^3 + 11^3 + 17^3 + 23^3 = 8^3 + 9^3 + 19^3 + 22^3.

Links

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Example

42120 is a term because 42120 = 1^3 + 19^3 + 22^3 + 27^3  = 2^3 + 3^3 + 13^3 + 33^3  = 2^3 + 6^3 + 17^3 + 32^3  = 3^3 + 3^3 + 20^3 + 31^3  = 3^3 + 17^3 + 20^3 + 29^3  = 3^3 + 13^3 + 14^3 + 32^3  = 6^3 + 15^3 + 16^3 + 31^3  = 7^3 + 17^3 + 23^3 + 27^3  = 11^3 + 13^3 + 21^3 + 29^3.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 4):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 9])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A025365, A344927, A345120, A345146, A345153, A345156, A345186.

Sequence in context: A151818 A374368 A186604 * A103155 A271749 A031670

Adjacent sequences: A345151 A345152 A345153 * A345155 A345156 A345157

Keyword

nonn

Author

David Consiglio, Jr., Jun 09 2021

Status

approved