A345617 Numbers that are the sum of eight fifth powers in nine or more ways.

8742208, 15539667, 18913169, 19987308, 20135313, 21505583, 21512966, 21563089, 21727552, 22237510, 22256608, 22438990, 22545600, 22686818, 22932525, 23106589, 23122550, 23189782, 23221517, 23287858, 23346048, 23477344, 23798742, 23847285, 23931325, 24138358

Offset

1,1

Links

Table of n, a(n) for n=1..26.

Example

15539667 is a term because 15539667 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 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 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5.

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, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 9])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345584, A345616, A345618, A345626, A345631, A346334.

Sequence in context: A151936 A186824 A179577 * A346334 A203669 A233476

Adjacent sequences: A345614 A345615 A345616 * A345618 A345619 A345620

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved