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.

A345538 Numbers that are the sum of eight cubes in eight or more ways.

Original entry on oeis.org

970, 977, 984, 1054, 1073, 1075, 1080, 1090, 1099, 1106, 1110, 1125, 1129, 1136, 1148, 1160, 1166, 1171, 1178, 1181, 1185, 1186, 1188, 1192, 1197, 1204, 1206, 1211, 1217, 1218, 1223, 1225, 1230, 1232, 1234, 1236, 1237, 1242, 1243, 1249, 1262, 1263, 1269, 1273
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			977 is a term because 977 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 5^3 + 8^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 7^3 = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 8^3 = 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 6^3.

Links

Crossrefs

Cf. A345495, A345526, A345537, A345539, A345547, A345583, A345790.

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

A345485 Numbers that are the sum of seven squares in eight or more ways.

Original entry on oeis.org

61, 66, 69, 70, 72, 73, 76, 77, 78, 79, 81, 82, 84, 85, 86, 87, 88, 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, 124, 125, 126, 127, 128, 129
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			66 is a term because 66 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 6^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 6^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2.

Links

Crossrefs

Cf. A344812, A345484, A345486, A345495, A345526.

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

A345494 Numbers that are the sum of eight squares in seven or more ways.

Original entry on oeis.org

56, 59, 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, 121
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. A345484, A345493, A345495, A345504, A345537.

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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

A345496 Numbers that are the sum of eight squares in nine or more ways.

Original entry on oeis.org

62, 64, 67, 70, 71, 73, 74, 76, 77, 78, 79, 80, 81, 82, 83, 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, 124, 125, 126, 127
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			64 is a term because 64 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2 = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2 = 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2 = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2 = 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2.

Links

Crossrefs

Cf. A345486, A345495, A345497, A345539.

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

A345505 Numbers that are the sum of nine squares in eight or more ways.

Original entry on oeis.org

57, 60, 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, 122
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. A345495, A345504, A345547, A346807.

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 >= 8])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-5 of 5 results.

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