OFFSET
1,1
LINKS
Sean A. Irvine, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1).
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)
EXAMPLE
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.
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 >= 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
CROSSREFS
KEYWORD
nonn
AUTHOR
David Consiglio, Jr., Aug 04 2021
STATUS
approved