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 1-7 of 7 results.

A345152 Numbers that are the sum of four third powers in eight or more ways.

Original entry on oeis.org

21896, 27720, 30429, 31339, 31402, 33579, 34624, 34776, 36162, 36225, 40105, 42120, 42695, 44037, 44163, 44226, 44947, 45162, 45675, 46277, 46683, 46872, 46900, 47600, 48321, 48825, 49042, 50112, 50689, 50806, 50904, 51058, 51408, 51480, 51506, 51597, 51688
Offset: 1

Views

Author

David Consiglio, Jr., Jun 09 2021

Keywords

Examples

			30429 is a term because 30429 = 1^3 + 4^3 + 7^3 + 30^3  = 1^3 + 16^3 + 17^3 + 26^3  = 2^3 + 12^3 + 21^3 + 25^3  = 3^3 + 3^3 + 14^3 + 29^3  = 4^3 + 17^3 + 21^3 + 23^3  = 5^3 + 11^3 + 15^3 + 28^3  = 6^3 + 6^3 + 22^3 + 25^3  = 7^3 + 14^3 + 18^3 + 26^3.
		

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

A345119 Numbers that are the sum of three third powers in nine or more ways.

Original entry on oeis.org

14926248, 16819704, 20168784, 34012224, 44946000, 45580536, 54042624, 58995000, 59768064, 62099136, 66203136, 67956624, 69190848, 69393024, 71319312, 72505152, 78008832, 78716448, 79539832, 80621568, 80996544, 89354448, 90757584, 92853216, 94118760, 95331816
Offset: 1

Views

Author

David Consiglio, Jr., Jun 08 2021

Keywords

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])

A345185 Numbers that are the sum of five third powers in nine or more ways.

Original entry on oeis.org

5860, 6112, 6138, 6462, 6497, 6588, 6651, 6859, 6947, 7001, 7038, 7057, 7064, 7099, 7190, 7316, 7328, 7372, 7433, 7561, 7587, 7703, 7759, 7841, 7902, 8056, 8163, 8289, 8352, 8371, 8443, 8506, 8560, 8569, 8630, 8632, 8758, 8928, 8991, 9017, 9045, 9080, 9099
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 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 >= 9])
    for x in range(len(rets)):
        print(rets[x])

A344926 Numbers that are the sum of four fourth powers in nine or more ways.

Original entry on oeis.org

328118259, 385202034, 395613234, 489597858, 592417938, 625839858, 641398338, 674511618, 677125218, 693239634, 699598578, 722302434, 779889314, 780278643, 780595299, 781388643, 782999714, 791204514, 792005379, 797405714, 797935698, 803898018, 805299699
Offset: 1

Views

Author

David Consiglio, Jr., Jun 02 2021

Keywords

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])

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])

A025374 Numbers that are the sum of 4 nonzero squares in 9 or more ways.

Original entry on oeis.org

162, 178, 198, 202, 207, 210, 220, 223, 225, 226, 231, 234, 242, 243, 246, 247, 250, 252, 253, 255, 258, 262, 265, 266, 267, 268, 270, 271, 273, 274, 278, 279, 282, 283, 285, 286, 287, 290, 291, 292, 294, 295, 297, 298, 300, 301, 303, 306, 307, 309, 310, 313, 314, 315
Offset: 1

Views

Author

Keywords

Crossrefs

Formula

{n: A025428(n) >= 9}. - R. J. Mathar, Jun 15 2018
Showing 1-7 of 7 results.