A345482 Numbers that are the sum of seven squares in five or more ways.

45, 54, 55, 57, 58, 60, 61, 63, 64, 66, 67, 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, 117

Offset

1,1

Links

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

Formula

Conjectures from Chai Wah Wu, Apr 25 2024: (Start)

a(n) = 2*a(n-1) - a(n-2) for n > 13.

G.f.: x*(-x^12 + x^11 - x^10 + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - 8*x^2 - 36*x + 45)/(x - 1)^2. (End)

Example

54 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2
so 54 is a term.

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 5])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A344809, A345481, A345483, A345492, A345523.

Sequence in context: A214572 A295493 A184043 * A295802 A242267 A242226

Adjacent sequences: A345479 A345480 A345481 * A345483 A345484 A345485

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved