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A344355 Numbers that are the sum of five fourth powers in exactly four ways. 8
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
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.
LINKS
David Consiglio, Jr., Table of n, a(n) for n = 1..20000
EXAMPLE
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.
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 == 4])
for x in range(len(rets)):
print(rets[x])
CROSSREFS
Sequence in context: A233649 A262668 A344354 * A231314 A157083 A183639
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified April 23 15:20 EDT 2024. Contains 371916 sequences. (Running on oeis4.)