cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A346354 Numbers that are the sum of ten fifth powers in exactly nine ways.

Original entry on oeis.org

1192180, 1226654, 1242437, 1431399, 1431430, 1431672, 1431883, 1432453, 1432664, 1434765, 1439174, 1441695, 1442718, 1447602, 1448447, 1455346, 1455377, 1464166, 1474431, 1474462, 1475485, 1491978, 1497619, 1531429, 1539173, 1614736, 1671199, 1671410, 1672937
Offset: 1

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Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345641 at term 6 because 1431641 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 1^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 1^5 + 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 1^5 + 1^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 3^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.

Examples

			1192180 is a term because 1192180 = 5^5 + 5^5 + 5^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 16^5 = 2^5 + 5^5 + 5^5 + 5^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 16^5 = 3^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 6^5 + 8^5 + 13^5 + 15^5 = 3^5 + 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 7^5 + 7^5 + 13^5 + 15^5 = 2^5 + 2^5 + 2^5 + 3^5 + 8^5 + 8^5 + 9^5 + 9^5 + 12^5 + 15^5 = 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 12^5 + 13^5 + 14^5 = 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 12^5 + 13^5 + 13^5 = 1^5 + 2^5 + 2^5 + 2^5 + 4^5 + 11^5 + 11^5 + 12^5 + 12^5 + 13^5 = 6^5 + 9^5 + 9^5 + 10^5 + 11^5 + 11^5 + 11^5 + 11^5 + 11^5 + 11^5.
		

Crossrefs

Programs

  • 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 == 9])
        for x in range(len(rets)):
            print(rets[x])