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.

A025371 Numbers that are the sum of 4 nonzero squares in 6 or more ways.

Original entry on oeis.org

90, 124, 130, 133, 135, 138, 147, 148, 150, 154, 156, 157, 159, 162, 163, 165, 166, 170, 171, 172, 174, 175, 177, 178, 180, 182, 183, 186, 187, 188, 189, 190, 193, 195, 196, 198, 199, 201, 202, 203, 205, 207, 210, 213, 214, 215, 217, 218, 219, 220, 222, 223, 225, 226
Offset: 1

Views

Author

Keywords

Crossrefs

Programs

  • Python
    limit = 226
    from functools import lru_cache
    sq = [k**2 for k in range(1, int(limit**.5)+2) if k**2 + 3 <= limit]
    sqs = set(sq)
    @lru_cache(maxsize=None)
    def findsums(n, m):
      if m == 1: return {(n, )} if n in sqs else set()
      return set(tuple(sorted(t+(s,))) for s in sqs for t in findsums(n-s, m-1))
    print([n for n in range(4, limit+1) if len(findsums(n, 4)) >= 6]) # Michael S. Branicky, Apr 20 2021

Formula

{n: A025428(n) >= 6}. Union of A025372 and A025362. - R. J. Mathar, Jun 15 2018

A345174 Numbers that are the sum of five third powers in six or more ways.

Original entry on oeis.org

2430, 2979, 3214, 3249, 3312, 3492, 3520, 3737, 3753, 3788, 3816, 3842, 3942, 3968, 4121, 4185, 4213, 4267, 4355, 4392, 4411, 4418, 4446, 4453, 4456, 4465, 4472, 4482, 4509, 4544, 4563, 4600, 4626, 4663, 4670, 4723, 4753, 4896, 4905, 4915, 4924, 4938, 4941
Offset: 1

Views

Author

David Consiglio, Jr., Jun 10 2021

Keywords

Examples

			2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3  = 1^3 + 3^3 + 4^3 + 7^3 + 11^3  = 2^3 + 2^3 + 6^3 + 6^3 + 11^3  = 2^3 + 3^3 + 3^3 + 9^3 + 10^3  = 3^3 + 5^3 + 8^3 + 8^3 + 8^3  = 3^3 + 4^3 + 7^3 + 8^3 + 9^3.
		

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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

A344798 Numbers that are the sum of five squares in five or more ways.

Original entry on oeis.org

53, 56, 59, 61, 64, 67, 68, 74, 77, 79, 80, 83, 85, 86, 88, 91, 92, 93, 94, 95, 96, 98, 99, 100, 101, 102, 103, 104, 106, 107, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 130, 131, 133, 134, 135, 136, 137
Offset: 1

Views

Author

Sean A. Irvine, May 28 2021

Keywords

Crossrefs

A344800 Numbers that are the sum of five squares in seven or more ways.

Original entry on oeis.org

77, 83, 85, 88, 91, 94, 99, 101, 104, 106, 107, 109, 112, 115, 116, 118, 119, 120, 122, 123, 124, 125, 126, 127, 128, 130, 131, 133, 134, 136, 137, 138, 139, 140, 141, 142, 143, 144, 146, 147, 148, 149, 150, 151, 152, 154, 155, 156, 157, 158, 159, 160, 161
Offset: 1

Views

Author

Sean A. Irvine, May 28 2021

Keywords

Crossrefs

A344810 Numbers that are the sum of six squares in six or more ways.

Original entry on oeis.org

54, 57, 60, 62, 65, 68, 69, 71, 72, 75, 76, 77, 78, 80, 81, 83, 84, 86, 87, 88, 89, 90, 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

			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.
		

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 >= 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 > 24.
G.f.: x*(-x^23 + x^22 - x^18 + x^17 - x^16 + x^15 - x^14 + x^13 - 2*x^10 + 2*x^9 - x^8 + x^7 - 2*x^6 + x^4 - x^3 - 51*x + 54)/(x - 1)^2. (End)
Showing 1-5 of 5 results.