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 1-4 of 4 results.

A025375 Numbers that are the sum of 4 nonzero squares in 10 or more ways.

Original entry on oeis.org

198, 202, 210, 234, 246, 247, 250, 252, 255, 258, 262, 268, 270, 273, 274, 279, 282, 285, 290, 292, 294, 295, 297, 298, 300, 301, 303, 306, 307, 310, 313, 315, 318, 319, 322, 324, 325, 327, 330, 333, 335, 338, 339, 340, 342, 343, 345, 346, 348, 350, 351, 354, 355, 357
Offset: 1

Views

Author

Keywords

Crossrefs

Formula

{n: A025428(n) >= 10}. - R. J. Mathar, Jun 15 2018

A345187 Numbers that are the sum of five third powers in ten or more ways.

Original entry on oeis.org

5860, 6588, 6651, 6859, 6947, 8056, 8289, 8371, 8506, 8569, 8758, 9045, 9080, 9099, 9108, 9227, 9414, 9612, 9801, 9829, 9864, 10009, 10018, 10044, 10277, 10466, 10485, 10522, 10529, 10800, 10963, 10970, 10979, 11008, 11017, 11061, 11089, 11152, 11241, 11385
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3  = 1^3 + 4^3 + 6^3 + 13^3 + 14^3  = 1^3 + 5^3 + 8^3 + 8^3 + 16^3  = 1^3 + 10^3 + 10^3 + 11^3 + 12^3  = 2^3 + 2^3 + 9^3 + 12^3 + 14^3  = 2^3 + 3^3 + 8^3 + 11^3 + 15^3  = 3^3 + 8^3 + 8^3 + 11^3 + 14^3  = 3^3 + 3^3 + 5^3 + 10^3 + 16^3  = 5^3 + 5^3 + 8^3 + 10^3 + 15^3  = 8^3 + 9^3 + 10^3 + 10^3 + 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 >= 10])
    for x in range(len(rets)):
        print(rets[x])

A344802 Numbers that are the sum of five squares in nine or more ways.

Original entry on oeis.org

101, 107, 109, 112, 115, 116, 118, 125, 127, 128, 131, 133, 134, 136, 139, 140, 142, 144, 146, 147, 148, 149, 151, 152, 154, 155, 157, 158, 159, 160, 161, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 178, 179, 180, 181, 182, 183, 184
Offset: 1

Views

Author

Sean A. Irvine, May 29 2021

Keywords

Crossrefs

A345477 Numbers that are the sum of six squares in ten or more ways.

Original entry on oeis.org

81, 84, 86, 89, 92, 93, 95, 100, 101, 102, 104, 105, 107, 108, 110, 111, 113, 114, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			84 = 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 8^2
   = 1^2 + 1^2 + 1^2 + 3^2 + 6^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 4^2 + 4^2 + 7^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 5^2 + 7^2
   = 1^2 + 1^2 + 4^2 + 4^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 5^2 + 5^2 + 5^2
   = 1^2 + 2^2 + 3^2 + 3^2 + 5^2 + 6^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 8^2
   = 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 7^2
   = 2^2 + 4^2 + 4^2 + 4^2 + 4^2 + 4^2
   = 3^2 + 3^2 + 3^2 + 4^2 + 4^2 + 5^2
so 84 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 >= 10])
        for x in range(len(rets)):
            print(rets[x])

Formula

Conjectures from Chai Wah Wu, Jan 05 2024: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 20.
G.f.: x*(-x^19 + x^18 - x^17 + x^16 - x^15 + x^14 - x^13 + x^12 - x^11 + x^10 - 4*x^8 + 3*x^7 + x^6 - 2*x^5 + x^3 - x^2 - 78*x + 81)/(x - 1)^2. (End)
Showing 1-4 of 4 results.