A345156 Numbers that are the sum of four third powers in exactly ten ways.

21896, 36225, 48825, 51506, 52416, 53200, 58338, 58968, 60480, 66024, 67851, 70434, 70525, 71155, 72819, 76923, 78624, 78912, 85995, 87507, 88641, 90181, 90783, 91728, 93555, 97552, 98280, 98560, 99008, 99225, 99792, 100170, 103040, 104104, 104265, 104958

Offset

1,1

Comments

Differs from A345155 at term 3 because 46872 = 1^3 + 16^3 + 22^3 + 30^3 = 2^3 + 11^3 + 17^3 + 33^3 = 3^3 + 3^3 + 4^3 + 35^3 = 3^3 + 4^3 + 26^3 + 29^3 = 3^3 + 5^3 + 23^3 + 31^3 = 4^3 + 10^3 + 24^3 + 30^3 = 5^3 + 17^3 + 23^3 + 29^3 = 6^3 + 10^3 + 20^3 + 32^3 = 11^3 + 11^3 + 21^3 + 31^3 = 11^3 + 14^3 + 17^3 + 32^3 = 19^3 + 21^3 + 21^3 + 25^3.

Links

David Consiglio, Jr., Table of n, a(n) for n = 1..10000

Example

21896 is a term because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 4):
    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])

Crossrefs

Cf. A025366, A344929, A345122, A345154, A345155, A345188.

Sequence in context: A345152 A345146 A345155 * A031647 A045109 A206126

Adjacent sequences: A345153 A345154 A345155 * A345157 A345158 A345159

Keyword

nonn

Author

David Consiglio, Jr., Jun 09 2021

Status

approved