A345486 Numbers that are the sum of seven squares in nine or more ways.
69, 70, 78, 79, 81, 82, 85, 87, 88, 90, 91, 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, 128, 129, 130, 131, 132, 133, 134, 135, 136
Offset: 1
Keywords
Examples
70 is a term because 70 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 8^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 + 5^2 = 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 5^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 7^2 = 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 6^2 = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.
Links
- Sean A. Irvine, Table of n, a(n) for n = 1..1000
Programs
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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, 7): tot = sum(pos) keep[tot] += 1 rets = sorted([k for k, v in keep.items() if v >= 9]) for x in range(len(rets)): print(rets[x])
Formula
Conjectures from Chai Wah Wu, Jan 05 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 13.
G.f.: x*(-x^12 + x^11 - x^10 + x^9 - x^8 - x^7 + 2*x^6 - x^5 + x^4 - 7*x^3 + 7*x^2 - 68*x + 69)/(x - 1)^2. (End)