A344945 Numbers that are the sum of five fourth powers in exactly eight ways.

534130, 654754, 663155, 729219, 737459, 742770, 758354, 810034, 813459, 816579, 831250, 906034, 930499, 954930, 1009954, 1055619, 1083955, 1099459, 1101859, 1103554, 1106019, 1157634, 1167794, 1180003, 1215394, 1217539, 1246354, 1253074, 1255539, 1278690

Offset

1,1

Comments

Differs from A344944 at term 2 because 619090 = 1^4 + 2^4 + 18^4 + 22^4 + 23^4 = 1^4 + 3^4 + 4^4 + 8^4 + 28^4 = 1^4 + 11^4 + 14^4 + 22^4 + 24^4 = 2^4 + 2^4 + 8^4 + 17^4 + 27^4 = 2^4 + 13^4 + 13^4 + 18^4 + 26^4 = 3^4 + 6^4 + 12^4 + 16^4 + 27^4 = 4^4 + 12^4 + 14^4 + 23^4 + 23^4 = 9^4 + 12^4 + 16^4 + 21^4 + 24^4 = 14^4 + 16^4 + 18^4 + 19^4 + 23^4.

Links

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

Example

534130 is a term 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.

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

Crossrefs

Cf. A341892, A344925, A344943, A344944, A345184, A345820.

Sequence in context: A061730 A235076 A344944 * A269322 A227497 A183650

Adjacent sequences: A344942 A344943 A344944 * A344946 A344947 A344948

Keyword

nonn

Author

David Consiglio, Jr., Jun 03 2021

Status

approved