A345618 Numbers that are the sum of eight fifth powers in ten or more ways.

15539667, 22932525, 24393600, 24650406, 24952961, 24953742, 25054306, 25142513, 25201550, 25423794, 26001294, 26851552, 27396567, 27988486, 28609075, 29309819, 29558650, 31052406, 31794336, 32223105, 32527286, 32610600, 32807777, 32890541, 32998317, 33015125

Offset

1,1

Links

Table of n, a(n) for n=1..26.

Example

15539667 is a term because 15539667 = 1^5 + 1^5 + 2^5 + 10^5 + 12^5 + 17^5 + 18^5 + 26^5 = 1^5 + 1^5 + 7^5 + 7^5 + 10^5 + 16^5 + 19^5 + 26^5 = 1^5 + 4^5 + 7^5 + 9^5 + 13^5 + 13^5 + 13^5 + 27^5 = 1^5 + 7^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 27^5 = 2^5 + 2^5 + 3^5 + 8^5 + 9^5 + 16^5 + 23^5 + 24^5 = 3^5 + 5^5 + 10^5 + 19^5 + 19^5 + 20^5 + 20^5 + 21^5 = 3^5 + 10^5 + 12^5 + 12^5 + 18^5 + 18^5 + 20^5 + 24^5 = 4^5 + 11^5 + 13^5 + 13^5 + 15^5 + 15^5 + 22^5 + 24^5 = 5^5 + 6^5 + 13^5 + 15^5 + 15^5 + 19^5 + 20^5 + 24^5 = 6^5 + 9^5 + 11^5 + 11^5 + 15^5 + 21^5 + 22^5 + 22^5.

Prog

(Python)
from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 10])
    for x in range(len(rets)):
        print(rets[x])

Crossrefs

Cf. A345585, A345617, A345627, A345643, A346335.

Sequence in context: A207797 A186050 A121840 * A346335 A004673 A290385

Adjacent sequences: A345615 A345616 A345617 * A345619 A345620 A345621

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved