A346805 Numbers that are the sum of ten squares in six or more ways.

49, 52, 55, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 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

Offset

1,1

Links

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

Example

52 is a term because 52 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3.

Mathematica

Select[Range[120], Count[IntegerPartitions[#, {10}], _?(AllTrue[Sqrt[#], IntegerQ]&)]>5&] (* Harvey P. Dale, Sep 10 2023 *)

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, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 6])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345503, A345554, A346804, A346806.

Sequence in context: A042199 A257443 A020276 * A118073 A043300 A304008

Adjacent sequences: A346802 A346803 A346804 * A346806 A346807 A346808

Keyword

nonn

Author

David Consiglio, Jr., Aug 04 2021

Status

approved