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.

A344927 Numbers that are the sum of four fourth powers in exactly nine ways.

Original entry on oeis.org

328118259, 385202034, 395613234, 489597858, 625839858, 641398338, 674511618, 693239634, 699598578, 722302434, 779889314, 780278643, 782999714, 791204514, 792005379, 797405714, 797935698, 805299699, 815120658, 822938754, 851527314, 857962914, 870861618
Offset: 1

Views

Author

David Consiglio, Jr., Jun 02 2021

Keywords

Comments

Differs from A344926 at term 5 because 328118259 is a term because 328118259 = 2^4 + 77^4 + 109^4 + 111^4 = 8^4 + 79^4 + 93^4 + 121^4 = 18^4 + 79^4 + 97^4 + 119^4 = 21^4 + 77^4 + 98^4 + 119^4 = 27^4 + 77^4 + 94^4 + 121^4 = 34^4 + 77^4 + 89^4 + 123^4 = 46^4 + 57^4 + 103^4 + 119^4 = 49^4 + 77^4 + 77^4 + 126^4 = 61^4 + 66^4 + 77^4 + 127^4.

Examples

			328118259 is a term because 328118259 = 2^4 + 77^4 + 109^4 + 111^4  = 8^4 + 79^4 + 93^4 + 121^4  = 18^4 + 79^4 + 97^4 + 119^4  = 21^4 + 77^4 + 98^4 + 119^4  = 27^4 + 77^4 + 94^4 + 121^4  = 34^4 + 77^4 + 89^4 + 123^4  = 46^4 + 57^4 + 103^4 + 119^4  = 49^4 + 77^4 + 77^4 + 126^4  = 61^4 + 66^4 + 77^4 + 127^4.
		

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

A341898 Numbers that are the sum of five fourth powers in exactly ten ways.

Original entry on oeis.org

954979, 1205539, 1574850, 1713859, 1863859, 1877394, 1882579, 2071939, 2109730, 2225859, 2288179, 2419954, 2492434, 2495939, 2605314, 2711394, 2784499, 2835939, 2847394, 2880994, 2924674, 3007474, 3061939, 3071379, 3074179, 3117235, 3127219, 3174834, 3190899
Offset: 1

Views

Author

David Consiglio, Jr., Jun 04 2021

Keywords

Comments

Differs at term 5 because
1801459 = 1^4 + 4^4 + 5^4 + 28^4 + 33^4
= 1^4 + 4^4 + 12^4 + 23^4 + 35^4
= 1^4 + 7^4 + 16^4 + 30^4 + 31^4
= 1^4 + 16^4 + 18^4 + 19^4 + 35^4
= 3^4 + 6^4 + 18^4 + 21^4 + 35^4
= 5^4 + 7^4 + 19^4 + 24^4 + 34^4
= 5^4 + 9^4 + 14^4 + 29^4 + 32^4
= 7^4 + 9^4 + 16^4 + 25^4 + 34^4
= 7^4 + 14^4 + 16^4 + 21^4 + 35^4
= 8^4 + 9^4 + 20^4 + 29^4 + 31^4
= 10^4 + 19^4 + 19^4 + 21^4 + 34^4.

Examples

			954979 =  1^4 +  2^4 + 11^4 + 19^4 + 30^4
       =  1^4 +  7^4 + 18^4 + 25^4 + 26^4
       =  3^4 +  8^4 + 17^4 + 20^4 + 29^4
       =  4^4 +  8^4 + 13^4 + 25^4 + 27^4
       =  4^4 +  9^4 + 10^4 + 11^4 + 31^4
       =  6^4 +  6^4 + 15^4 + 21^4 + 29^4
       =  7^4 + 10^4 + 18^4 + 19^4 + 29^4
       = 11^4 + 11^4 + 20^4 + 22^4 + 27^4
       = 16^4 + 17^4 + 17^4 + 24^4 + 25^4
       = 18^4 + 19^4 + 20^4 + 23^4 + 23^4
so 954979 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**4 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])

A344928 Numbers that are the sum of four fourth powers in ten or more ways.

Original entry on oeis.org

592417938, 677125218, 780595299, 781388643, 803898018, 806692194, 937239954, 940415058, 980421939, 1164012003, 1269819378, 1355899923, 1403089314, 1488645939, 1539221154, 1599073938, 1635878754, 1657885698, 1666044963, 1701067683, 1734489603, 1758151458
Offset: 1

Views

Author

David Consiglio, Jr., Jun 02 2021

Keywords

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

A345156 Numbers that are the sum of four third powers in exactly ten ways.

Original entry on oeis.org

21896, 36225, 48825, 51506, 52416, 53200, 58338, 58968, 60480, 66024, 67851, 70434, 70525, 71155, 72819, 76923, 78624, 78912, 85995, 87507, 88641, 90181, 90783, 91728, 93555, 97552, 98280, 98560, 99008, 99225, 99792, 100170, 103040, 104104, 104265, 104958
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345155 at term 3 because 46872 = 1^3 + 16^3 + 22^3 + 30^3 = 2^3 + 11^3 + 17^3 + 33^3 = 3^3 + 3^3 + 4^3 + 35^3 = 3^3 + 4^3 + 26^3 + 29^3 = 3^3 + 5^3 + 23^3 + 31^3 = 4^3 + 10^3 + 24^3 + 30^3 = 5^3 + 17^3 + 23^3 + 29^3 = 6^3 + 10^3 + 20^3 + 32^3 = 11^3 + 11^3 + 21^3 + 31^3 = 11^3 + 14^3 + 17^3 + 32^3 = 19^3 + 21^3 + 21^3 + 25^3.

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])

A344861 Numbers that are the sum of three fourth powers in exactly ten ways.

Original entry on oeis.org

49511121842, 364765611938, 703409488418, 792177949472, 2667500248322, 3602781562562, 3999861055442, 4010400869202, 5698033074818, 5836249791008, 6330685395762, 7250378688098, 7695882509378, 8746828790882, 10383571090802, 11254551814688, 12160605587858
Offset: 1

Views

Author

David Consiglio, Jr., May 31 2021

Keywords

Comments

Differs from A344862 at term 2 because 281539574498 = 7^4 + 609^4 + 616^4 = 41^4 + 591^4 + 632^4 = 81^4 + 568^4 + 649^4 = 99^4 + 557^4 + 656^4 = 121^4 + 543^4 + 664^4 = 168^4 + 511^4 + 679^4 = 224^4 + 469^4 + 693^4 = 239^4 + 457^4 + 696^4 = 256^4 + 443^4 + 699^4 = 269^4 + 432^4 + 701^4 = 293^4 + 411^4 + 704^4 = 336^4 + 371^4 + 707^4.

Examples

			49511121842 is a term because 49511121842 = 13^4 + 390^4 + 403^4  = 35^4 + 378^4 + 413^4  = 70^4 + 357^4 + 427^4  = 103^4 + 335^4 + 438^4  = 117^4 + 325^4 + 442^4  = 137^4 + 310^4 + 447^4  = 175^4 + 322^4 + 441^4  = 182^4 + 273^4 + 455^4  = 202^4 + 255^4 + 457^4  = 225^4 + 233^4 + 458^4.
		

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, 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 01 2021
Showing 1-5 of 5 results.