A345487 Numbers that are the sum of seven squares in ten or more ways.

70, 79, 82, 85, 87, 88, 90, 91, 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, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			79 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 4^2 + 6^2
   = 2^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
so 79 is a term.
		

Links

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

Crossrefs

Cf. A345477, A345486, A345497, A345506.

Programs

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 10])
    for x in range(len(rets)):
        print(rets[x])

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)

a(n) = 2*a(n-1) - a(n-2) for n > 10.

G.f.: x*(-x^9 + x^8 - x^7 + x^6 - x^5 - x^4 - 6*x^2 - 61*x + 70)/(x - 1)^2. (End)