A344810 Numbers that are the sum of six squares in six or more ways.

54, 57, 60, 62, 65, 68, 69, 71, 72, 75, 76, 77, 78, 80, 81, 83, 84, 86, 87, 88, 89, 90, 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

Formula

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

a(n) = 2*a(n-1) - a(n-2) for n > 24.

G.f.: x*(-x^23 + x^22 - x^18 + x^17 - x^16 + x^15 - x^14 + x^13 - 2*x^10 + 2*x^9 - x^8 + x^7 - 2*x^6 + x^4 - x^3 - 51*x + 54)/(x - 1)^2. (End)

Example

57 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2
so 57 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, 6):
    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. A344799, A344809, A344811, A345483, A345515.

Sequence in context: A326181 A300447 A344809 * A295696 A095561 A095551

Adjacent sequences: A344807 A344808 A344809 * A344811 A344812 A344813

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved