A346335 Numbers that are the sum of eight fifth powers in exactly ten ways.
15539667, 22932525, 24393600, 24650406, 24952961, 24953742, 25142513, 26001294, 27988486, 28609075, 29309819, 31794336, 32223105, 32527286, 32610600, 32807777, 32890541, 32998317, 33015125, 33187858, 33361339, 33550572, 33659175, 33782597, 34029369, 34073650
Offset: 1
Keywords
Examples
15539667 is a term because 15539667 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..8261
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, 8): 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])
Comments