A346803 Numbers that are the sum of nine squares in ten or more ways.

63, 65, 68, 71, 72, 74, 75, 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, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127

Offset: 1

Author

David Consiglio, Jr., Aug 04 2021

Keywords

• nonn

Examples

			65 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 65 is a term.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A025428, A025429, A345497, A345549, A346808.

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, 9):
    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])

Python

def A346803(n): return (63, 65, 68, 71, 72, 74, 75)[n-1] if n<8 else n+69 # Chai Wah Wu, May 09 2024

Formula

From Chai Wah Wu, May 09 2024: (Start)

All integers >= 77 are terms. Proof: since 246 can be written as the sum of 4 positive squares in 10 ways (see A025428) and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 280 can be written as a sum of 9 positive squares in 10 or more ways. Integers from 77 to 279 are terms by inspection.

a(n) = 2*a(n-1) - a(n-2) for n > 9.

G.f.: x*(-x^8 + x^7 - x^6 + x^5 - 2*x^4 + x^2 - 61*x + 63)/(x - 1)^2. (End)