A345830 Numbers that are the sum of seven fourth powers in exactly eight ways.
21252, 21507, 21636, 21876, 25347, 30372, 31412, 31652, 32116, 32356, 33811, 33907, 35637, 35652, 35892, 36261, 37827, 38052, 38596, 38676, 39267, 39347, 39971, 39972, 40212, 40452, 41506, 41731, 41987, 42147, 42227, 42357, 42532, 42771, 42852, 43027, 43282
Offset: 1
Keywords
Examples
21252 is a term because 21252 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 12^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 1^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 1^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 1^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 1^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 7^4 + 8^4 + 9^4 + 9^4.
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**4 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 == 8]) for x in range(len(rets)): print(rets[x])
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