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.

A345537 Numbers that are the sum of eight cubes in seven or more ways.

Original entry on oeis.org

902, 908, 921, 938, 958, 963, 970, 977, 982, 984, 991, 996, 1003, 1008, 1010, 1017, 1019, 1028, 1029, 1033, 1047, 1054, 1055, 1058, 1061, 1062, 1070, 1073, 1075, 1080, 1087, 1090, 1091, 1094, 1096, 1097, 1099, 1104, 1106, 1108, 1110, 1111, 1113, 1115, 1116
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			908 is a term because 908 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3.

Links

Crossrefs

Cf. A345494, A345525, A345536, A345538, A345546, A345582, A345789.

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, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 7])
    for x in range(len(rets)):
        print(rets[x])

A345484 Numbers that are the sum of seven squares in seven or more ways.

Original entry on oeis.org

55, 58, 61, 63, 64, 66, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			58 is a term because 58 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2.

Links

Crossrefs

Cf. A344811, A345483, A345485, A345494, A345525.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 for x in range(1, 1000)]
for pos in cwr(power_terms, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 7])
    for x in range(len(rets)):
        print(rets[x])

Formula

Conjectures from Chai Wah Wu, Apr 25 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 21.
G.f.: x*(-x^20 + x^19 - x^9 + x^8 - 2*x^7 + x^6 + x^5 - x^4 - x^3 - 52*x + 55)/(x - 1)^2. (End)

A345493 Numbers that are the sum of eight squares in six or more ways.

Original entry on oeis.org

56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			58 is a term because 58 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 5^2.

Links

Crossrefs

Cf. A345483, A345492, A345494, A345503, A345536.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 for x in range(1, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 6])
    for x in range(len(rets)):
        print(rets[x])

Formula

Conjectures from Chai Wah Wu, Apr 25 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 11.
G.f.: x*(-x^10 + x^9 - x^8 + x^7 - x^6 + x^5 - x^4 + x^3 - x^2 - 54*x + 56)/(x - 1)^2. (End)

A345495 Numbers that are the sum of eight squares in eight or more ways.

Original entry on oeis.org

56, 59, 62, 64, 65, 67, 68, 70, 71, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			59 is a term because 59 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345485, A345494, A345496, A345505, A345538.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 for x in range(1, 1000)]
for pos in cwr(power_terms, 8):
    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])

A345504 Numbers that are the sum of nine squares in seven or more ways.

Original entry on oeis.org

57, 60, 62, 63, 65, 66, 68, 69, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			60 is a term because 60 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2 = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2.

Links

Crossrefs

Cf. A345494, A345503, A345505, A345546, A346806.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 for x in range(1, 1000)]
for pos in cwr(power_terms, 9):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 7])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 results.

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