A344751 Numbers that are the sum of three fourth powers in exactly nine ways.

105760443698, 131801075042, 187758243218, 253590205778, 319889609522, 445600096578, 510334859762, 601395185762, 615665999858, 730871934338, 749472385298, 855952663202, 856722174098, 951843993282, 1157106866258, 1186209675378, 1290443616098, 1455023522498

Offset

1,1

Comments

Differs from A344750 at term 1 because 49511121842 = 13^4 + 390^4 + 403^4 = 35^4 + 378^4 + 413^4 = 70^4 + 357^4 + 427^4 = 103^4 + 335^4 + 438^4 = 117^4 + 325^4 + 442^4 = 137^4 + 310^4 + 447^4 = 175^4 + 322^4 + 441^4 = 182^4 + 273^4 + 455^4 = 202^4 + 255^4 + 457^4 = 225^4 + 233^4 + 458^4.

Links

David Consiglio, Jr., Table of n, a(n) for n = 1..23

Example

105760443698 is a term because 105760443698 = 7^4 + 476^4 + 483^4  = 51^4 + 452^4 + 503^4  = 76^4 + 437^4 + 513^4  = 107^4 + 417^4 + 524^4  = 133^4 + 399^4 + 532^4  = 199^4 + 348^4 + 547^4  = 212^4 + 337^4 + 549^4  = 228^4 + 323^4 + 551^4  = 252^4 + 301^4 + 553^4.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 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. A344738, A344750, A344861, A344927, A345120.

Sequence in context: A349321 A168340 A104799 * A221286 A074337 A072143

Adjacent sequences: A344748 A344749 A344750 * A344752 A344753 A344754

Keyword

nonn

Author

David Consiglio, Jr., May 28 2021

Status

approved