A345574 Numbers that are the sum of seven fourth powers in eight or more ways.

19491, 21252, 21267, 21332, 21507, 21636, 21876, 23652, 25347, 30372, 31251, 31412, 31652, 32116, 32356, 33811, 33907, 35427, 35637, 35652, 35892, 36052, 36261, 37812, 37827, 38052, 38067, 38596, 38676, 39267, 39347, 39891, 39971, 39972, 40212, 40356, 40452

Offset

1,1

Links

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

Example

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.

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

Crossrefs

Cf. A345526, A345565, A345573, A345575, A345583, A345630, A345830.

Sequence in context: A254493 A253858 A336982 * A345575 A345831 A322101

Adjacent sequences: A345571 A345572 A345573 * A345575 A345576 A345577

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved