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.

A345511 Numbers that are the sum of six cubes in two or more ways.

Original entry on oeis.org

158, 165, 184, 221, 228, 235, 247, 254, 256, 261, 268, 273, 275, 280, 282, 284, 287, 291, 294, 306, 310, 313, 317, 324, 331, 332, 343, 345, 347, 350, 352, 362, 369, 371, 373, 376, 378, 380, 385, 387, 388, 392, 395, 399, 404, 406, 408, 411, 418, 425, 430, 432
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

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, 6):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 2])
        for x in range(len(rets)):
            print(rets[x])

A344795 Numbers that are the sum of five squares in two or more ways.

Original entry on oeis.org

20, 29, 32, 35, 37, 38, 40, 41, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 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
Offset: 1

Views

Author

Sean A. Irvine, May 28 2021

Keywords

Crossrefs

A344807 Numbers that are the sum of six squares in three or more ways.

Original entry on oeis.org

30, 33, 36, 38, 39, 41, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 59, 60, 61, 62, 63, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			33 = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2
   = 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2
so 33 is a term.
		

Crossrefs

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 >= 3])
        for x in range(len(rets)):
            print(rets[x])

A345479 Numbers that are the sum of seven squares in two or more ways.

Original entry on oeis.org

22, 25, 28, 30, 31, 33, 34, 37, 38, 39, 40, 42, 43, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 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
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			25 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2
so 25 is a term.
		

Crossrefs

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 >= 2])
        for x in range(len(rets)):
            print(rets[x])

A344805 Numbers that are the sum of six squares in one or more ways.

Original entry on oeis.org

6, 9, 12, 14, 15, 17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 1

Views

Author

David Consiglio, Jr., Jun 19 2021

Keywords

Examples

			9 is a term because 9 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2.
		

Crossrefs

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 >= 1])
        for x in range(len(rets)):
            print(rets[x])

Formula

From Chai Wah Wu, Jun 12 2025: (Start)
All integers >= 20 are terms. See A345508 for a similar proof.
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 - 3*x + 6)/(x - 1)^2. (End)
Showing 1-5 of 5 results.