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A344738
Numbers that are the sum of three fourth powers in exactly eight ways.
6
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
KEYWORD
nonn
AUTHOR
STATUS
approved