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A346339
Numbers that are the sum of nine fifth powers in exactly four ways.
7
55542, 120350, 143507, 167241, 182549, 192233, 202890, 326685, 327986, 328247, 329028, 329809, 333257, 351722, 358474, 358968, 359210, 359538, 359813, 365404, 367071, 367313, 374034, 374846, 375627, 376619, 377158, 379259, 381157, 383910, 384765, 390396
OFFSET
1,1
COMMENTS
Differs from A345621 at term 37 because 392063 = 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.
LINKS
EXAMPLE
55542 is a term because 55542 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.
PROG
(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 9):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 4])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved