A345088 Numbers that are the sum of three third powers in exactly eight ways.

2562624, 5618250, 6525000, 6755328, 7374375, 12742920, 13581352, 14027112, 14288373, 18443160, 20500992, 22783032, 23113728, 25305048, 26936064, 27131840, 29515968, 30205440, 32835375, 38269440, 39317832, 39339000, 40189248, 42144192, 42183504, 43077952

Offset

1,1

Comments

Differs from A345087 at term 10 because 14926248 = 2^3 + 33^3 + 245^3 = 11^3 + 185^3 + 203^3 = 14^3 + 32^3 + 245^3 = 50^3 + 113^3 + 236^3 = 71^3 + 89^3 + 239^3 = 74^3 + 189^3 + 196^3 = 89^3 + 185^3 + 197^3 = 98^3 + 148^3 + 219^3 = 105^3 + 149^3 + 217^3.

Links

David Consiglio, Jr., Table of n, a(n) for n = 1..100

Example

2562624 is a term because 2562624 = 7^3 + 35^3 + 135^3  = 7^3 + 63^3 + 131^3  = 11^3 + 99^3 + 115^3  = 16^3 + 45^3 + 134^3  = 29^3 + 102^3 + 112^3  = 35^3 + 59^3 + 131^3  = 50^3 + 84^3 + 121^3  = 68^3 + 71^3 + 122^3.

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, 3):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 8])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A025328, A344738, A345085, A345087, A345120, A345153.

Sequence in context: A251857 A096557 A345087 * A234664 A225321 A259563

Adjacent sequences: A345085 A345086 A345087 * A345089 A345090 A345091

Keyword

nonn

Author

David Consiglio, Jr., Jun 07 2021

Status

approved