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 1-4 of 4 results.

A345120 Numbers that are the sum of three third powers in exactly nine ways.

Original entry on oeis.org

14926248, 16819704, 20168784, 44946000, 45580536, 54042624, 59768064, 62099136, 66203136, 67956624, 69393024, 78008832, 78716448, 79539832, 80621568, 80996544, 89354448, 90757584, 99616392, 100088568, 101352168, 101943360, 112216896, 112720896, 114306984
Offset: 1

Views

Author

David Consiglio, Jr., Jun 08 2021

Keywords

Comments

Differs from A345119 at term 4 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.

Examples

			14926248 is a term because 14926248 = 2^3 + 33^3 + 245^3  = 11^3 + 185^3 + 203^3  = 14^3 + 32^3 + 245^3  = 50^3 + 113^3 + 236^3  = 71^3 + 89^3 + 239^3  = 74^3 + 189^3 + 196^3  = 89^3 + 185^3 + 197^3  = 98^3 + 148^3 + 219^3  = 105^3 + 149^3 + 217^3.
		

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

A345121 Numbers that are the sum of three third powers in ten or more ways.

Original entry on oeis.org

34012224, 58995000, 69190848, 71319312, 72505152, 92853216, 94118760, 95331816, 119095488, 119409984, 139755888, 147545280, 150506000, 150547032, 157464000, 159874560, 161023680, 161350272, 164186352, 171904032, 175986000, 176175000, 182393856, 184909824
Offset: 1

Views

Author

David Consiglio, Jr., Jun 08 2021

Keywords

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

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 1-4 of 4 results.