A345503 Numbers that are the sum of nine squares in six or more ways.

48, 56, 57, 59, 60, 62, 63, 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, 117

Offset

1,1

Comments

Numbers n such that A025433(n) >= 6. - David A. Corneth, Apr 26 2024

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 > 9.

G.f.: x*(-x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - 7*x^2 - 40*x + 48)/(x - 1)^2. (End)

Example

56 is a term because 56 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2.

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, 9):
    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. A025433, A345493, A345502, A345504, A345545, A346805.

Sequence in context: A080854 A255267 A366250 * A259037 A231469 A261546

Adjacent sequences: A345500 A345501 A345502 * A345504 A345505 A345506

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved