A345120 Numbers that are the sum of three third powers in exactly nine ways.

14926248, 16819704, 20168784, 44946000, 45580536, 54042624, 59768064, 62099136, 66203136, 67956624, 69393024, 78008832, 78716448, 79539832, 80621568, 80996544, 89354448, 90757584, 99616392, 100088568, 101352168, 101943360, 112216896, 112720896, 114306984

Offset

1,1

Comments

Differs from A345119 at term 4 because 34012224 = 35^3 + 215^3 + 287^3 = 38^3 + 152^3 + 311^3 = 40^3 + 113^3 + 318^3 = 44^3 + 245^3 + 266^3 = 71^3 + 113^3 + 317^3 = 99^3 + 191^3 + 295^3 = 101^3 + 226^3 + 276^3 = 117^3 + 185^3 + 295^3 = 161^3 + 215^3 + 269^3 = 172^3 + 213^3 + 266^3.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..6359

Example

14926248 is a term 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.

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 == 9])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A025329, A344751, A345088, A345119, A345122, A345154.

Sequence in context: A015364 A081640 A345119 * A227286 A125565 A204313

Adjacent sequences: A345117 A345118 A345119 * A345121 A345122 A345123

Keyword

nonn

Author

David Consiglio, Jr., Jun 08 2021

Status

approved