A345503 - OEIS

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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

(list; graph; refs; listen; history; text; internal format)

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])