A345639 Numbers that are the sum of ten fifth powers in seven or more ways.

555098, 674040, 683166, 707315, 763631, 777852, 778844, 780945, 783224, 893654, 896500, 897668, 920887, 926616, 927819, 928802, 936850, 937631, 944383, 945017, 952897, 953077, 953139, 953350, 953414, 955178, 963131, 975133, 979482, 984133, 985453, 985664

Offset

1,1

Links

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Example

674040 is a term because 674040 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 7^5 + 9^5 + 10^5 + 12^5 + 12^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 = 1^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 = 1^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 = 2^5 + 2^5 + 2^5 + 2^5 + 4^5 + 6^5 + 8^5 + 10^5 + 11^5 + 13^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 9^5 + 14^5 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 7^5 + 8^5 + 12^5 + 13^5.

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, 10):
    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. A345600, A345624, A345638, A345640, A346352.

Sequence in context: A253614 A249687 A267600 * A346352 A196511 A196733

Adjacent sequences: A345636 A345637 A345638 * A345640 A345641 A345642

Keyword

nonn

Author

David Consiglio, Jr., Jun 20 2021

Status

approved