A345496 Numbers that are the sum of eight squares in nine or more ways.

62, 64, 67, 70, 71, 73, 74, 76, 77, 78, 79, 80, 81, 82, 83, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127

Offset

1,1

Links

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

Example

64 is a term because 64 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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. A345486, A345495, A345497, A345539.

Sequence in context: A188631 A076764 A367090 * A214251 A031058 A084801

Adjacent sequences: A345493 A345494 A345495 * A345497 A345498 A345499

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved