A344943 Numbers that are the sum of five fourth powers in exactly seven ways.

197779, 211059, 217154, 236675, 431155, 444019, 480739, 503539, 530659, 548994, 564979, 568450, 571539, 602450, 602770, 621859, 625635, 625939, 626194, 650659, 651954, 653059, 654130, 666739, 687314, 692754, 692899, 698019, 708499, 716739, 728914, 730914

Offset

1,1

Comments

Differs from A344942 at term 10 because 534130 = 1^4 + 3^4 + 16^4 + 22^4 + 22^4 = 2^4 + 2^4 + 4^4 + 7^4 + 27^4 = 2^4 + 3^4 + 6^4 + 6^4 + 27^4 = 2^4 + 6^4 + 9^4 + 21^4 + 24^4 = 4^4 + 16^4 + 17^4 + 18^4 + 23^4 = 6^4 + 8^4 + 11^4 + 22^4 + 23^4 = 7^4 + 8^4 + 16^4 + 19^4 + 24^4 = 13^4 + 14^4 + 14^4 + 21^4 + 22^4.

Links

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

Example

197779 is a term because 197779 = 1^4 + 5^4 + 6^4 + 16^4 + 19^4  = 1^4 + 7^4 + 11^4 + 12^4 + 20^4  = 1^4 + 10^4 + 12^4 + 17^4 + 17^4  = 2^4 + 4^4 + 5^4 + 7^4 + 21^4  = 3^4 + 5^4 + 6^4 + 6^4 + 21^4  = 4^4 + 7^4 + 9^4 + 13^4 + 20^4  = 11^4 + 13^4 + 14^4 + 15^4 + 16^4.

Prog

(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, 1000)]
for pos in cwr(power_terms, 5):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 7])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A344923, A344941, A344942, A344945, A345181, A345819.

Sequence in context: A185519 A345659 A344942 * A213817 A215996 A205234

Adjacent sequences: A344940 A344941 A344942 * A344944 A344945 A344946

Keyword

nonn

Author

David Consiglio, Jr., Jun 03 2021

Status

approved