A345643 Numbers that are the sum of seven fifth powers in ten or more ways.
134581976, 189642309, 219063107, 235438301, 252277376, 275782407, 281935070, 290928076, 300919884, 308188849, 309631268, 315635200, 322947868, 327287951, 335530174, 342030094, 358852218, 361946949, 379913293, 384699424, 387538625, 391133568
Offset: 1
Keywords
Examples
189642309 is a term because 189642309 = 1^5 + 1^5 + 2^5 + 19^5 + 30^5 + 36^5 + 40^5 = 1^5 + 2^5 + 6^5 + 7^5 + 18^5 + 20^5 + 45^5 = 1^5 + 6^5 + 21^5 + 27^5 + 29^5 + 36^5 + 39^5 = 2^5 + 9^5 + 19^5 + 23^5 + 33^5 + 33^5 + 40^5 = 3^5 + 4^5 + 21^5 + 28^5 + 29^5 + 34^5 + 40^5 = 6^5 + 7^5 + 11^5 + 29^5 + 33^5 + 36^5 + 37^5 = 7^5 + 12^5 + 17^5 + 20^5 + 29^5 + 32^5 + 42^5 = 8^5 + 11^5 + 21^5 + 21^5 + 22^5 + 34^5 + 42^5 = 13^5 + 14^5 + 14^5 + 19^5 + 21^5 + 38^5 + 40^5 = 20^5 + 21^5 + 24^5 + 24^5 + 24^5 + 38^5 + 38^5.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..5000
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, 7): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v >= 10]) for x in range(len(rets)): print(rets[x])