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.

Showing 1-5 of 5 results.

A345154 Numbers that are the sum of four third powers in exactly nine ways.

Original entry on oeis.org

42120, 46683, 50806, 50904, 51408, 51480, 51688, 52208, 53865, 54971, 56385, 57113, 60515, 60984, 62433, 65303, 66276, 66339, 66430, 67158, 69048, 69832, 69930, 71162, 72072, 72520, 72576, 72800, 73017, 77714, 77903, 79345, 79667, 79849, 80066, 80073, 81207
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345146 at term 1 because 21896 = 1^3 + 11^3 + 19^3 + 22^3 = 2^3 + 2^3 + 12^3 + 26^3 = 2^3 + 3^3 + 19^3 + 23^3 = 2^3 + 5^3 + 15^3 + 25^3 = 3^3 + 10^3 + 16^3 + 24^3 = 3^3 + 17^3 + 19^3 + 19^3 = 4^3 + 6^3 + 20^3 + 22^3 = 5^3 + 8^3 + 14^3 + 25^3 = 7^3 + 11^3 + 17^3 + 23^3 = 8^3 + 9^3 + 19^3 + 22^3.

Examples

			42120 is a term because 42120 = 1^3 + 19^3 + 22^3 + 27^3  = 2^3 + 3^3 + 13^3 + 33^3  = 2^3 + 6^3 + 17^3 + 32^3  = 3^3 + 3^3 + 20^3 + 31^3  = 3^3 + 17^3 + 20^3 + 29^3  = 3^3 + 13^3 + 14^3 + 32^3  = 6^3 + 15^3 + 16^3 + 31^3  = 7^3 + 17^3 + 23^3 + 27^3  = 11^3 + 13^3 + 21^3 + 29^3.
		

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, 4):
        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])

A345155 Numbers that are the sum of four third powers in ten or more ways.

Original entry on oeis.org

21896, 36225, 46872, 48321, 48825, 51506, 52416, 53200, 55575, 58338, 58968, 59059, 60480, 62244, 66024, 67536, 67851, 70434, 70525, 71155, 72819, 73808, 76384, 76923, 77896, 78624, 78912, 81081, 81991, 85995, 87507, 88641, 90181, 90783, 91448, 91728, 92008
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Examples

			21896 is a term because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.
		

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, 4):
        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])

A344929 Numbers that are the sum of four fourth powers in exactly ten ways.

Original entry on oeis.org

592417938, 806692194, 940415058, 980421939, 1269819378, 1355899923, 1488645939, 1599073938, 1635878754, 1657885698, 1666044963, 1758151458, 1797373314, 1813434483, 1991146899, 2064726483, 2198975058, 2246905683, 2266525314, 2302589298, 2302698258, 2502041283
Offset: 1

Views

Author

David Consiglio, Jr., Jun 02 2021

Keywords

Comments

Differs from A344928 at term 2 because 677125218 = 2^4 + 109^4 + 111^4 + 140^4 = 21^4 + 98^4 + 119^4 + 140^4 = 27^4 + 94^4 + 121^4 + 140^4 = 28^4 + 35^4 + 42^4 + 161^4 = 34^4 + 89^4 + 123^4 + 140^4 = 36^4 + 98^4 + 109^4 + 145^4 = 44^4 + 75^4 + 128^4 + 139^4 = 49^4 + 77^4 + 126^4 + 140^4 = 61^4 + 66^4 + 127^4 + 140^4 = 70^4 + 119^4 + 119^4 + 126^4 = 75^4 + 76^4 + 98^4 + 151^4.

Examples

			592417938 is a term because 592417938 = 6^4 + 59^4 + 65^4 + 154^4  = 7^4 + 11^4 + 20^4 + 156^4  = 10^4 + 17^4 + 17^4 + 156^4  = 12^4 + 112^4 + 115^4 + 127^4  = 15^4 + 86^4 + 107^4 + 142^4  = 21^4 + 49^4 + 70^4 + 154^4  = 25^4 + 107^4 + 112^4 + 132^4  = 26^4 + 45^4 + 71^4 + 154^4  = 28^4 + 105^4 + 112^4 + 133^4  = 63^4 + 77^4 + 112^4 + 140^4.
		

Crossrefs

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
    from collections import defaultdict
    keep = defaultdict(lambda: 0)
    power_terms = [x**4 for x in range(1, 1000)]
    for pos in cwr(power_terms, 4):
        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])

Extensions

More terms from Sean A. Irvine, Jun 03 2021

A345188 Numbers that are the sum of five third powers in exactly ten ways.

Original entry on oeis.org

5860, 6588, 6651, 6859, 6947, 8056, 8289, 8569, 8758, 9045, 9099, 9227, 9414, 9612, 9829, 10009, 10277, 10485, 10522, 10529, 10800, 10963, 10970, 11008, 11061, 11089, 11241, 11385, 11458, 11656, 11719, 11782, 11817, 11845, 11934, 11990, 12016, 12060, 12088
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Comments

Differs from A345187 at term 8 because 8371 = 1^3 + 1^3 + 11^3 + 11^3 + 16^3 = 1^3 + 4^3 + 5^3 + 12^3 + 17^3 = 1^3 + 8^3 + 9^3 + 11^3 + 16^3 = 3^3 + 3^3 + 4^3 + 15^3 + 15^3 = 3^3 + 3^3 + 8^3 + 8^3 + 18^3 = 3^3 + 3^3 + 3^3 + 5^3 + 19^3 = 3^3 + 7^3 + 9^3 + 9^3 + 17^3 = 4^3 + 6^3 + 6^3 + 11^3 + 17^3 = 5^3 + 9^3 + 10^3 + 11^3 + 15^3 = 6^3 + 6^3 + 12^3 + 13^3 + 13^3 = 8^3 + 8^3 + 9^3 + 9^3 + 16^3.

Examples

			6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3  = 1^3 + 4^3 + 6^3 + 13^3 + 14^3  = 1^3 + 5^3 + 8^3 + 8^3 + 16^3  = 1^3 + 10^3 + 10^3 + 11^3 + 12^3  = 2^3 + 2^3 + 9^3 + 12^3 + 14^3  = 2^3 + 3^3 + 8^3 + 11^3 + 15^3  = 3^3 + 8^3 + 8^3 + 11^3 + 14^3  = 3^3 + 3^3 + 5^3 + 10^3 + 16^3  = 5^3 + 5^3 + 8^3 + 10^3 + 15^3  = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.
		

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, 5):
        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])

A345122 Numbers that are the sum of three third powers in exactly ten ways.

Original entry on oeis.org

34012224, 58995000, 71319312, 72505152, 92853216, 94118760, 95331816, 139755888, 147545280, 150506000, 157464000, 159874560, 161023680, 164186352, 171904032, 182393856, 184909824, 188224128, 189771336, 191260224, 199108125, 201342240, 202440384, 217054720
Offset: 1

Views

Author

David Consiglio, Jr., Jun 08 2021

Keywords

Comments

Differs from A345121 at term 3 because 69190848 = 23^3 + 107^3 + 407^3 = 23^3 + 191^3 + 395^3 = 33^3 + 271^3 + 365^3 = 35^3 + 299^3 + 347^3 = 50^3 + 137^3 + 404^3 = 89^3 + 308^3 + 338^3 = 95^3 + 178^3 + 396^3 = 107^3 + 179^3 + 395^3 = 121^3 + 149^3 + 399^3 = 152^3 + 254^3 + 365^3 = 206^3 + 215^3 + 368^3.

Examples

			34012224 is a term because 34012224 = 35^3 + 215^3 + 287^3  = 38^3 + 152^3 + 311^3  = 40^3 + 113^3 + 318^3  = 44^3 + 245^3 + 266^3  = 71^3 + 113^3 + 317^3  = 99^3 + 191^3 + 295^3  = 101^3 + 226^3 + 276^3  = 117^3 + 185^3 + 295^3  = 161^3 + 215^3 + 269^3  = 172^3 + 213^3 + 266^3.
		

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, 3):
        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])

Extensions

More terms from Sean A. Irvine, Jun 08 2021
Showing 1-5 of 5 results.