A346351 Numbers that are the sum of ten fifth powers in exactly six ways.
392095, 392306, 399839, 406802, 407583, 434676, 491643, 492063, 520261, 521106, 538323, 538534, 540927, 553325, 563526, 582089, 592398, 608190, 611072, 614196, 637833, 639903, 640715, 640895, 640926, 640957, 641106, 643671, 653523, 655327, 656616, 664895
Offset: 1
Keywords
Examples
392095 is a term because 392095 = 2^5 + 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 7^5 + 10^5 + 12^5 = 2^5 + 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 2^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..10000
Programs
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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])
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