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 1-2 of 2 results.

A224419 Numbers n such that triangular(n) + triangular(2*n) is a square.

Original entry on oeis.org

0, 1, 25, 216, 1849, 36481, 311904, 2666689, 52606009, 449765784, 3845364121, 75857828929, 648561949056, 5545012396225, 109386936710041, 935225880773400, 7995904029992761, 157735886878050625, 1348595071513294176, 11530088066237165569, 227455039491212291641, 1944673157896289428824, 16626378995609962758169, 327990009210441246496129
Offset: 1

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Author

Alex Ratushnyak, Apr 18 2013

Keywords

Comments

8 of the first 10 terms are of the form x^y. The two exceptions are a(7) = 311904 = 2^5 * 3^3 * 19^2 and a(10) = 449765784 = 2^3 * 3^5 * 13^2 * 37^2.
The corresponding squares are given by A075873(2*n-1)^2. E.g., triangular(a(10)) + triangular(2*a(10)) = 711142146^2 = A075873(19)^2.
Locations of squares in A147875, equivalent to solving the Diophantine equation n*(5*n+3)=2*s^2. - R. J. Mathar, Apr 19 2013

Crossrefs

Cf. A220186 (numbers n such that triangular(2*n) - triangular(n) is a square).

Programs

  • Mathematica
    LinearRecurrence[{1,0,1442,-1442,0,-1,1},{0,1,25,216,1849,36481,311904},30]  (* Harvey P. Dale, Jan 23 2015 *)
  • Python
    import math
    for i in range(1<<30):
            s = i*(i+1)/2 + i*(2*i+1)
            t = int(math.sqrt(s))
            if s == t*t:  print(i)

Formula

a(n) = (A228209(2*n-1) - 3) / 10. - Max Alekseyev, Sep 04 2013
G.f.: x^2*(x+1)*(x^4 + 23*x^3 + 168*x^2 + 23*x + 1) / (x^6 - 1442*x^3 + 1) / (1-x). - Max Alekseyev, Sep 04 2013

Extensions

Terms a(11) onward from Max Alekseyev, Sep 04 2013

A343034 Positive numbers m such that m^2 with last digit z deleted is still a perfect square k^2, and z divides m-k.

Original entry on oeis.org

1, 13, 19, 487, 721, 18493, 27379, 702247, 1039681, 26666893, 39480499, 1012639687, 1499219281, 38453641213, 56930852179, 1460225726407, 2161873163521, 55450123962253, 82094249361619, 2105644484839207, 3117419602578001, 79959040299927613, 118379850648602419, 3036337886912410087
Offset: 1

Views

Author

Bernard Schott, Apr 03 2021

Keywords

Comments

This sequence is the answer to the problem A1872 proposed on French mathematical site Diophante (see link).
Equivalent to the two Diophantine equations: m^2 = 10*k^2 + z and m-k = q*z for some q >= 1.
There exist solutions iff z = 1 or z = 9.
When (m, k, z) is a solution, then (19m+60k, 6m+19k, z) is another solution.
There is only one solution such that z = m-k: (13, 4, 9), see 1st example.
There exist two distinct families of solutions corresponding to z = 1 and z = 9, odd indices correspond to z = 1 and even indices to z = 9.
-> For z = 1, all solutions m of Pell equation m^2 - 10*k^2 = 1 are terms because z = 1 divides every m-k.
First few solutions (m, k) are (1, 0), (19, 6), (721, 228), (27379, 86568), ... with m = A078986(q) and corresponding k = 6*A078987(q).
-> For z = 9, solutions m must satisfy m^2 - 10*k^2 = 9 with 9 divides m-k. Among the 3 fundamental solutions (3, 0), (7, 2), (13, 4) of Pell equation m^2 -10*k^2 = 9, only (13, 4) gives solutions where 9 divides m-k.
First few solutions (m, k) are (13, 4), (487, 154), (18493, 5848), ... with m = A228209(3q).

Examples

			For m = 13, 13^2 = 169, 4^2 = 16, 13^2 - 10*4^2 = 9 and 9 = 13-4 divides 13-4.
For m = 19, 19^2 = 361, 6^2 = 36, 19^2 - 10*6^2 = 1 and 1 divides 19-6 = 13.
For m = 487, 487^2 = 237169, 154^2 = 23716, 487^2 - 10*154^2 = 9 and 9 divides 487-154 = 333 = 9*37.
		

Crossrefs

Subsequence of A031149.
A078986 is a subsequence.

Programs

  • Mathematica
    LinearRecurrence[{0, 38, 0, -1}, {1, 13, 19, 487}, 24] (* Amiram Eldar, Apr 03 2021 *)

Formula

a(2n+1) = A078986(n) for n >= 0.
a(2n) = A228209(3n) for n >= 1.
a(n+4) = 38*a(n+2) - a(n), a(1) = 1, a(2) = 13, a(3)= 19, a(4) = 487.
G.f.: x*(1 + 13*x - 19*x^2 - 7*x^3)/(1 - 38*x^2 + x^4). - Stefano Spezia, Apr 03 2021
Showing 1-2 of 2 results.