A344738 Numbers that are the sum of three fourth powers in exactly eight ways.

5745705602, 8185089458, 11054952818, 14355295682, 21789116258, 22247419922, 26839201298, 29428835618, 31861462178, 37314202562, 38214512882, 41923075922, 46543615202, 51711350418, 54438780578, 56255300738, 59223741122, 62862779042, 63429959138, 71035097042

Offset

1,1

Comments

Differs at term 14 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..100

Example

5745705602 is a term because 5745705602 = 3^4 + 230^4 + 233^4 = 25^4 + 218^4 + 243^4 = 43^4 + 207^4 + 250^4 = 58^4 + 197^4 + 255^4 = 85^4 + 177^4 + 262^4 = 90^4 + 173^4 + 263^4 = 102^4 + 163^4 + 265^4 = 122^4 + 145^4 + 267^4.

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 == 8])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A344730, A344737, A344751, A344925, A345088.

Sequence in context: A365144 A114260 A344737 * A113644 A234059 A370250

Adjacent sequences: A344735 A344736 A344737 * A344739 A344740 A344741

Keyword

nonn

Author

David Consiglio, Jr., May 27 2021

Status

approved