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.

A345791 Numbers that are the sum of eight cubes in exactly nine ways.

Original entry on oeis.org

984, 1080, 1136, 1171, 1192, 1197, 1204, 1223, 1269, 1273, 1280, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1404, 1406, 1415, 1416, 1422, 1425, 1430, 1432, 1438, 1442, 1444, 1445, 1456, 1476, 1481, 1486, 1488, 1494
Offset: 1

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Author

David Consiglio, Jr., Jun 26 2021

Keywords

Comments

Differs from A345539 at term 5 because 1185 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 9^3 = 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 9^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 + 8^3 = 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3 + 7^3.
Likely finite.

Examples

			1080 is a term because 1080 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**3 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 == 9])
        for x in range(len(rets)):
            print(rets[x])