A345640 Numbers that are the sum of ten fifth powers in eight or more ways.

944383, 953139, 953414, 985453, 1118585, 1151438, 1185375, 1192180, 1198879, 1206546, 1209912, 1216569, 1217172, 1218912, 1223321, 1225398, 1226654, 1234631, 1241834, 1242437, 1251195, 1251406, 1252123, 1259685, 1265563, 1265594, 1267937, 1275375, 1281736

Offset

1,1

Links

Sean A. Irvine, Table of n, a(n) for n = 1..1000

Example

953139 is a term because 953139 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 10^5 + 10^5 + 10^5 + 12^5 + 13^5 = 1^5 + 2^5 + 2^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 12^5 + 14^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 12^5 + 12^5 + 13^5 = 2^5 + 2^5 + 3^5 + 3^5 + 7^5 + 7^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 11^5 + 11^5 + 12^5 + 13^5 = 2^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 10^5 + 10^5 + 13^5 + 13^5 = 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 + 15^5.

Prog

(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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345601, A345625, A345639, A345641, A346353.

Sequence in context: A198780 A104940 A203893 * A346353 A203823 A251911

Adjacent sequences: A345637 A345638 A345639 * A345641 A345642 A345643

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved