A345122 Numbers that are the sum of three third powers in exactly ten ways.

34012224, 58995000, 71319312, 72505152, 92853216, 94118760, 95331816, 139755888, 147545280, 150506000, 157464000, 159874560, 161023680, 164186352, 171904032, 182393856, 184909824, 188224128, 189771336, 191260224, 199108125, 201342240, 202440384, 217054720

Offset

1,1

Comments

Differs from A345121 at term 3 because 69190848 = 23^3 + 107^3 + 407^3 = 23^3 + 191^3 + 395^3 = 33^3 + 271^3 + 365^3 = 35^3 + 299^3 + 347^3 = 50^3 + 137^3 + 404^3 = 89^3 + 308^3 + 338^3 = 95^3 + 178^3 + 396^3 = 107^3 + 179^3 + 395^3 = 121^3 + 149^3 + 399^3 = 152^3 + 254^3 + 365^3 = 206^3 + 215^3 + 368^3.

Links

Sean A. Irvine, Table of n, a(n) for n = 1..2238

Example

34012224 is a term 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.

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 == 10])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A025330, A344861, A345120, A345121, A345156.

Sequence in context: A205943 A232636 A345121 * A375343 A363295 A172687

Adjacent sequences: A345119 A345120 A345121 * A345123 A345124 A345125

Keyword

nonn

Author

David Consiglio, Jr., Jun 08 2021

Extensions

More terms from Sean A. Irvine, Jun 08 2021

Status

approved