A345153 Numbers that are the sum of four third powers in exactly eight ways.

27720, 30429, 31339, 31402, 33579, 34624, 34776, 36162, 40105, 42695, 44037, 44163, 44226, 44947, 45162, 45675, 46277, 46900, 47600, 49042, 50112, 50689, 51058, 51597, 51805, 52227, 52264, 52507, 53144, 54271, 54873, 55692, 55790, 56240, 58032, 58221, 58312

Offset

1,1

Comments

Differs from A345152 at term 1 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.

Links

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

Example

30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3  = 1^3 + 16^3 + 17^3 + 26^3  = 2^3 + 12^3 + 21^3 + 25^3  = 3^3 + 3^3 + 14^3 + 29^3  = 4^3 + 17^3 + 21^3 + 23^3  = 5^3 + 11^3 + 15^3 + 28^3  = 6^3 + 6^3 + 22^3 + 25^3  = 7^3 + 14^3 + 18^3 + 26^3.

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 == 8])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A025364, A344925, A345088, A345151, A345152, A345154, A345184.

Sequence in context: A023198 A230608 A204831 * A190111 A068404 A279091

Adjacent sequences: A345150 A345151 A345152 * A345154 A345155 A345156

Keyword

nonn

Author

David Consiglio, Jr., Jun 09 2021

Status

approved