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 1-3 of 3 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

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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

A345149 Numbers that are the sum of four third powers in exactly six ways.

Original entry on oeis.org

6883, 12411, 13923, 14112, 14581, 14896, 14904, 15561, 15876, 16317, 16640, 17208, 17479, 17992, 18739, 18865, 19035, 19080, 19665, 19712, 19763, 19880, 20007, 20384, 20979, 21231, 21420, 21707, 22409, 22617, 23149, 23940, 24355, 25515, 25984, 26208, 26334
Offset: 1

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Author

David Consiglio, Jr., Jun 09 2021

Keywords

Comments

Differs from A345148 at term 3 because 13104 = 1^3 + 10^3 + 16^3 + 18^3 = 1^3 + 11^3 + 14^3 + 19^3 = 2^3 + 9^3 + 15^3 + 19^3 = 4^3 + 6^3 + 14^3 + 20^3 = 4^3 + 9^3 + 10^3 + 21^3 = 5^3 + 7^3 + 11^3 + 21^3 = 8^3 + 9^3 + 14^3 + 19^3.

Examples

			6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3  = 2^3 + 4^3 + 14^3 + 14^3  = 3^3 + 7^3 + 7^3 + 17^3  = 3^3 + 10^3 + 13^3 + 13^3  = 4^3 + 10^3 + 10^3 + 15^3  = 7^3 + 8^3 + 8^3 + 16^3.
		

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, 4):
        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])

A025381 Numbers that are the sum of 4 distinct nonzero squares in exactly 6 ways.

Original entry on oeis.org

174, 234, 254, 258, 266, 273, 279, 285, 291, 309, 327, 333, 339, 359, 362, 365, 371, 380, 391, 395, 396, 403, 413, 443, 444, 449, 451, 476, 487, 492, 521, 572, 596, 604, 643, 668, 692, 696, 772, 796, 936, 1016, 1032, 1064, 1448, 1520, 1584, 1776, 1904, 1968, 2288
Offset: 1

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Keywords

Crossrefs

Formula

{n: A025443(n) = 6}. - R. J. Mathar, Jun 15 2018
Showing 1-3 of 3 results.