A345476 Numbers that are the sum of six squares in nine or more ways.

78, 81, 84, 86, 89, 92, 93, 95, 99, 100, 101, 102, 104, 105, 107, 108, 110, 111, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148

Offset

1,1

Links

Sean A. Irvine, Table of n, a(n) for n = 1..1000

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)

a(n) = 2*a(n-1) - a(n-2) for n > 20.

G.f.: x*(-x^19 + x^18 - x^17 + x^16 - x^15 + x^14 - x^13 + x^12 - 3*x^9 + 2*x^8 + x^7 - 2*x^6 + x^4 - x^3 - 75*x + 78)/(x - 1)^2. (End)

Example

81 is a term because 81 = 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 7^2 = 1^2 + 1^2 + 2^2 + 5^2 + 5^2 + 5^2 = 1^2 + 1^2 + 3^2 + 3^2 + 5^2 + 6^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 8^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 7^2 = 1^2 + 4^2 + 4^2 + 4^2 + 4^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 7^2 = 2^2 + 2^2 + 4^2 + 4^2 + 4^2 + 5^2 = 2^2 + 3^2 + 3^2 + 3^2 + 5^2 + 5^2 = 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 6^2.

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, 6):
    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])

Crossrefs

Cf. A344802, A344812, A345477, A345486, A345518.

Sequence in context: A124289 A181467 A344812 * A217006 A053083 A246487

Adjacent sequences: A345473 A345474 A345475 * A345477 A345478 A345479

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved