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.

A025285 Numbers that are the sum of 2 nonzero squares in exactly 2 ways.

Original entry on oeis.org

50, 65, 85, 125, 130, 145, 170, 185, 200, 205, 221, 250, 260, 265, 290, 305, 338, 340, 365, 370, 377, 410, 442, 445, 450, 481, 485, 493, 500, 505, 520, 530, 533, 545, 565, 578, 580, 585, 610, 625, 629, 680, 685, 689, 697, 730, 740, 745, 754, 765, 785, 793, 800, 820
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

Order and signs don't count. E.g. 50 = 5^2+5^2 = 7^2+1^2 (= (-5)^2+5^2, but that doesn't count as different).
A131574 is a subsequence. - Zak Seidov, Jan 31 2014
A025426(a(n)) = 2. - Reinhard Zumkeller, Feb 26 2015

Links

Crossrefs

Cf. A025284, A025286, A025287, A025288, A025289, A025290, A025291, A025292, A025293, A025426.
Cf. A007692.

Programs

  • Haskell
    a025285 n = a025285_list !! (n-1)
a025285_list = filter ((== 2) . a025426) [1..]
-- Reinhard Zumkeller, Feb 26 2015
  • Mathematica
    selQ[n_] := Length[ Select[ PowersRepresentations[n, 2, 2], Times @@ # != 0 &]] == 2; Select[Range[1000], selQ] (* Jean-François Alcover, Oct 03 2013 *)
  • PARI
    is(n)=sum(k=sqrtint((n-1)\2)+1,sqrtint(n-1), issquare(n-k^2))==2 \\ Charles R Greathouse IV, May 24 2016
  • PARI
    is(n)=my(v=valuation(n, 2), f=factor(n>>v), t=1); for(i=1, #f[, 1], if(f[i, 1]%4==1, t*=f[i, 2]+1, if(f[i, 2]%2, return(0)))); if(t%2, t-(-1)^v, t)==4 \\ Charles R Greathouse IV, May 24 2016

Formula

a(n) >= A007692(n) with equality only for n <= 16. - Alois P. Heinz, Mar 23 2023

A025295 Numbers that are the sum of 2 nonzero squares in 4 or more ways.

Original entry on oeis.org

1105, 1625, 1885, 2125, 2210, 2405, 2465, 2665, 3145, 3250, 3445, 3485, 3625, 3770, 3965, 4225, 4250, 4420, 4505, 4625, 4745, 4810, 4930, 5125, 5185, 5330, 5365, 5525, 5785, 5945, 6205, 6290, 6305, 6409, 6500, 6565, 6625, 6890, 6970, 7085, 7225, 7250
Offset: 1

Views

Author

David W. Wilson

Keywords

Links

Crossrefs

Cf. A023051, A025287, A025294, A025296, A025332.

Programs

  • Mathematica
    nn = 7250; t = Table[0, {nn}]; lim = Floor[Sqrt[nn - 1]]; Do[num = i^2 + j^2; If[num <= nn, t[[num]]++], {i, lim}, {j, i}]; Flatten[Position[t, ?(# >= 4 &)]] (* _T. D. Noe, Apr 07 2011 *)

A025305 Numbers that are the sum of 2 distinct nonzero squares in exactly 4 ways.

Original entry on oeis.org

1105, 1625, 1885, 2125, 2210, 2405, 2465, 2665, 3145, 3250, 3445, 3485, 3625, 3770, 3965, 4225, 4250, 4420, 4505, 4625, 4745, 4810, 4930, 5125, 5185, 5330, 5365, 5785, 5945, 6205, 6290, 6305, 6409, 6500, 6565, 6625, 6890, 6970, 7085, 7225, 7250, 7345
Offset: 1

Views

Author

David W. Wilson

Keywords

Comments

8450 is the first term in here but not in A025287 [R. Chandler, seqfan list, May 28 2008] [From R. J. Mathar, Nov 30 2008]

Links

Programs

  • Mathematica
    nn = 8450; t = Table[0, {nn}]; lim = Floor[Sqrt[nn - 1]]; Do[num = i^2 + j^2; If[num <= nn, t[[num]]++], {i, lim}, {j, i-1}]; Flatten[Position[t, 4]] (* T. D. Noe, Apr 07 2011 *)

A345865 Numbers that are the sum of two cubes in exactly four ways.

Original entry on oeis.org

6963472309248, 12625136269928, 21131226514944, 26059452841000, 55707778473984, 74213505639000, 95773976104625, 101001090159424, 159380205560856, 169049812119552, 174396242861568, 188013752349696, 208475622728000, 300656502205416, 340878679288056
Offset: 1

Views

Author

David Consiglio, Jr., Jul 05 2021

Keywords

Comments

Differs from A023051 at term 143 because 48988659276962496 = 331954^3 + 231518^3 = 336588^3 + 221424^3 = 342952^3 + 205292^3 = 362753^3 + 107839^3 = 365757^3 + 38787^3.

Examples

			12625136269928 is a term because 12625136269928 = 21869^3 + 12939^3 = 22580^3 + 10362^3 = 23066^3 + 7068^3 = 23237^3 + 4275^3.

Links

Crossrefs

Cf. A023051, A025287, A343969, A344804.

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, 2):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 4])
    for x in range(len(rets)):
        print(rets[x])
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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