A345638 Numbers that are the sum of ten fifth powers in six or more ways.

392095, 392306, 399839, 406802, 407583, 434676, 491643, 492063, 520261, 521106, 538323, 538534, 540927, 553325, 555098, 563526, 582089, 592398, 608190, 611072, 614196, 637833, 639903, 640715, 640895, 640926, 640957, 641106, 643671, 653523, 655327, 656616

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			392306 is a term because 392306 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 1^5 + 2^5 + 3^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 2^5 + 3^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A345599, A345623, A345637, A345639, A346351.

Programs

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 6])
    for x in range(len(rets)):
        print(rets[x])