A345860 Numbers that are the sum of ten fourth powers in exactly eight ways.
6675, 6740, 6755, 6805, 6995, 7015, 7030, 7045, 7095, 7270, 7300, 7365, 7429, 7494, 7525, 7540, 7590, 7605, 7750, 7780, 7845, 7955, 8005, 8085, 8150, 8195, 8215, 8310, 8450, 8470, 8500, 8630, 8644, 8709, 8710, 8790, 8885, 8949, 9124, 9189, 9190, 9250, 9255
Offset: 1
Keywords
Examples
6740 is a term because 6740 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 9^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 7^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4 = 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 6^4 + 8^4.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..8900
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, 10): 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])
Comments