A344359 Numbers that are the sum of five fourth powers in exactly five ways.

59779, 67859, 93394, 108274, 112850, 136915, 142354, 151475, 161459, 168979, 181219, 183539, 183604, 185299, 187699, 189394, 193379, 195394, 199090, 199474, 200979, 201874, 202979, 203299, 205859, 211330, 212419, 213730, 217810, 217890, 221779, 223090, 223155, 223714, 226514, 227779, 231235

Offset

1,1

Comments

Differs from A344358 at term 8 because 151300 = 3^4 + 3^4 + 3^4 + 12^4 + 19^4 = 3^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 13^4 + 13^4 + 13^4 + 16^4 = 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

Links

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

Example

93394 is a term of this sequence because 93394 = 1^4 + 4^4 + 8^4 + 14^4 + 15^4 = 1^4 + 6^4 + 12^4 + 12^4 + 15^4 = 1^4 + 9^4 + 10^4 + 14^4 + 14^4 = 5^4 + 6^4 + 11^4 + 14^4 + 14^4 = 5^4 + 7^4 + 8^4 + 12^4 + 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, 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 == 5])
for x in range(len(rets)):
    print(rets[x])

Crossrefs

Cf. A343988, A344355, A344357, A344358, A344941, A345817, A346257.

Sequence in context: A031685 A238077 A344358 * A043635 A106774 A189985

Adjacent sequences: A344356 A344357 A344358 * A344360 A344361 A344362

Keyword

nonn

Author

David Consiglio, Jr., May 15 2021

Status

approved