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.

A346365 Numbers that are the sum of six fifth powers in exactly ten ways.

Original entry on oeis.org

55302546200, 89999127392, 96110537743, 104484239200, 120492759200, 121258798144, 127794946400, 133364991375, 135030535200, 136156575744, 151305014432, 155434423925, 174388570400, 177099008000, 179272687000, 182844944832, 184948721056, 187873845500
Offset: 1

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Author

David Consiglio, Jr., Jul 18 2021

Keywords

Comments

This sequence differs from A344196:
180336745600 = 48^5 + 54^5 + 66^5 + 66^5 + 112^5 + 174^5
= 9^5 + 21^5 + 93^5 + 112^5 + 117^5 + 168^5
= 11^5 + 44^5 + 73^5 + 92^5 + 133^5 + 167^5
= 15^5 + 81^5 + 94^5 + 95^5 + 129^5 + 166^5
= 1^5 + 49^5 + 62^5 + 107^5 + 138^5 + 163^5
= 35^5 + 69^5 + 75^5 + 98^5 + 141^5 + 162^5
= 18^5 + 81^5 + 105^5 + 112^5 + 135^5 + 159^5
= 14^5 + 50^5 + 62^5 + 86^5 + 150^5 + 158^5
= 2^5 + 52^5 + 54^5 + 108^5 + 146^5 + 158^5
= 14^5 + 22^5 + 66^5 + 118^5 + 142^5 + 158^5
= 4^5 + 50^5 + 58^5 + 102^5 + 150^5 + 156^5,
so 180336745600 is in A344196, but is not in this sequence.

Examples

			55302546200 = 34^5 + 38^5 + 50^5 + 57^5 + 95^5 + 136^5
            = 23^5 + 49^5 + 61^5 + 69^5 + 107^5 + 131^5
            = 24^5 + 37^5 + 63^5 + 81^5 + 104^5 + 131^5
            = 21^5 + 35^5 + 60^5 + 94^5 + 100^5 + 130^5
            = 57^5 + 60^5 + 71^5 + 75^5 + 109^5 + 128^5
            = 19^5 + 37^5 + 56^5 + 96^5 + 104^5 + 128^5
            = 35^5 + 41^5 + 53^5 + 69^5 + 115^5 + 127^5
            = 16^5 + 49^5 + 53^5 + 83^5 + 112^5 + 127^5
            = 35^5 + 37^5 + 40^5 + 88^5 + 119^5 + 121^5
            = 11^5 + 24^5 + 71^5 + 104^5 + 109^5 + 121^5
so 55302546200 is a term.
		

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, 6):
        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])