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.

A345523 Numbers that are the sum of seven cubes in five or more ways.

Original entry on oeis.org

627, 768, 838, 845, 857, 864, 874, 881, 894, 900, 920, 937, 950, 955, 962, 969, 976, 981, 983, 990, 1002, 1009, 1011, 1016, 1027, 1046, 1053, 1054, 1060, 1061, 1063, 1072, 1079, 1089, 1096, 1098, 1102, 1105, 1107, 1109, 1115, 1117, 1121, 1124, 1128, 1133
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			768 is a term because 768 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3.

Links

Crossrefs

Cf. A345482, A345514, A345522, A345524, A345535, A345571, A345777.

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, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 5])
    for x in range(len(rets)):
        print(rets[x])

A344809 Numbers that are the sum of six squares in five or more ways.

Original entry on oeis.org

54, 57, 60, 62, 63, 65, 66, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 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

			57 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2
so 57 is a term.

Links

Crossrefs

Cf. A344798, A344808, A344810, A345482, A345514.

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, 6):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 5])
    for x in range(len(rets)):
        print(rets[x])

A345481 Numbers that are the sum of seven squares in four or more ways.

Original entry on oeis.org

37, 40, 42, 45, 46, 48, 49, 50, 52, 53, 54, 55, 57, 58, 60, 61, 62, 63, 64, 65, 66, 67, 69, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			40 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2
so 40 is a term.

Links

Crossrefs

Cf. A344808, A345480, A345482, A345491, A345522.

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 >= 4])
    for x in range(len(rets)):
        print(rets[x])

A345483 Numbers that are the sum of seven squares in six or more ways.

Original entry on oeis.org

55, 58, 61, 63, 64, 66, 67, 69, 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 + 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. A344810, A345482, A345484, A345493, A345524.

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 >= 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 > 9.
G.f.: x*(-x^8 + x^7 - x^6 + x^5 - x^4 - x^3 - 52*x + 55)/(x - 1)^2. (End)

A345492 Numbers that are the sum of eight squares in five or more ways.

Original entry on oeis.org

46, 47, 49, 53, 54, 55, 56, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			47 is a term because 47 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2.

Links

Crossrefs

Cf. A345482, A345491, A345493, A345502, A345535.

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 >= 5])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 results.

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