A345575 Numbers that are the sum of seven fourth powers in nine or more ways.

19491, 21267, 21332, 23652, 31251, 35427, 36052, 37812, 38067, 39891, 40356, 41732, 41747, 43267, 43876, 43891, 43956, 44131, 44196, 44532, 44547, 44612, 45156, 45171, 45411, 45651, 45652, 45827, 45891, 45892, 45907, 46276, 46451, 46516, 47427, 47667, 47971

Offset

1,1

Links

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Example

21267 is a term because 21267 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 2^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.

Prog

(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 >= 9])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345527, A345566, A345574, A345576, A345584, A345631, A345831.

Sequence in context: A253858 A336982 A345574 * A345831 A322101 A206274

Adjacent sequences: A345572 A345573 A345574 * A345576 A345577 A345578

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved