A344355 Numbers that are the sum of five fourth powers in exactly four ways.

20995, 21235, 31250, 41474, 43235, 43250, 43315, 43490, 43859, 45139, 46290, 47570, 51939, 53234, 53299, 54994, 56274, 57379, 57410, 57779, 59329, 63970, 67010, 68035, 68290, 71795, 71954, 73730, 73954, 75714, 75794, 77890, 82099, 84499, 86275, 86450, 87730, 92500, 93474, 93859, 94130, 94210, 96194

Offset: 1

Author

David Consiglio, Jr., May 15 2021

Keywords

• nonn

Comments

Differs from A344354 at term 22 because 59779 = 1^4 + 1^4 + 5^4 + 12^4 + 14^4 = 1^4 + 6^4 + 6^4 + 9^4 + 15^4 = 2^4 + 9^4 + 10^4 + 11^4 + 13^4 = 4^4 + 7^4 + 7^4 + 8^4 + 15^4 = 7^4 + 7^4 + 9^4 + 10^4 + 14^4.

Examples

			31250 is a term of this sequence because 31250 = 2^4 + 2^4 + 4^4 + 7^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 13^4 = 4^4 + 6^4 + 7^4 + 9^4 + 12^4 = 5^4 + 5^4 + 10^4 + 10^4 + 10^4.
		

Links

• David Consiglio, Jr., Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A344035, A344244, A344353, A344354, A344359, A344519, A345816.

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1, 50)]
for pos in cwr(power_terms, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 4])
for x in range(len(rets)):
    print(rets[x])