A106256 -id:A106256 - cp's OEIS frontend

A273052 Numbers n such that 7*n^2 + 8 is a square.

2, 34, 542, 8638, 137666, 2194018, 34966622, 557271934, 8881384322, 141544877218, 2255836651166, 35951841541438, 572973628011842, 9131626206648034, 145533045678356702, 2319397104647059198, 36964820628674590466, 589117732954146388258, 9388918906637667621662

Offset: 1

Vincenzo Librandi, May 14 2016

Colin Barker, Table of n, a(n) for n = 1..800
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. Numbers n such that k*n^2+(k+1) is a square: A052530 (k=3), this sequence (k=7), A106328 (k=8), A106256 (k=12), A273053 (k=15), A273054 (k=19), A106331 (k=24).

I:=[2,34]; [n le 2 select I[n] else 16*Self(n-1)-Self(n-2): n in [1..30]];

Mathematica
```
LinearRecurrence[{16, -1}, {2, 34}, 30]
```

Vec(x*(2+2*x)/(1-16*x+x^2) + O(x^50)) \\ Colin Barker, May 14 2016

O.g.f.: x*(2 + 2*x)/(1 - 16*x + x^2).
E.g.f.: 2*(1 + (3*sqrt(7)*sinh(3*sqrt(7)*x) - 7*cosh(3*sqrt(7)*x))*exp(8*x)/7). - Ilya Gutkovskiy, May 14 2016
a(n) = 16*a(n-1) - a(n-2).
a(n) = (-(8-3*sqrt(7))^n*(3+sqrt(7))-(-3+sqrt(7))*(8+3*sqrt(7))^n)/sqrt(7). - Colin Barker, May 14 2016

A106257 Numbers k such that k^2 = 12*n^2 + 13.

Original entry on oeis.org

5, 11, 59, 149, 821, 2075, 11435, 28901, 159269, 402539, 2218331, 5606645, 30897365, 78090491, 430344779, 1087660229, 5993929541, 15149152715, 83484668795, 211000477781, 1162791433589, 2938857536219, 16195595401451

Offset: 1

Pierre CAMI, Apr 28 2005

			5^2=12*1^2+13
11^2=12*3^2+13
59^2=12*17^2+13
149^2=12*43^2+13

Index entries for linear recurrences with constant coefficients, signature (0,14,0,-1).

Cf. A106256.

LinearRecurrence[{0,14,0,-1},{5,11,59,149},40] (* Harvey P. Dale, Oct 21 2021 *)

Vec(-x*(x-1)*(5*x^2+16*x+5)/((x^2-4*x+1)*(x^2+4*x+1)) + O(x^100)) \\ Colin Barker, Apr 16 2014

k(1)=5, k(2)=11, k(3)=14*k(1)-k(2), k(4)=14*k(2)-k(1) then k(n)=14*k(n-2)-k(n-4).
G.f.: -x*(x-1)*(5*x^2+16*x+5) / ((x^2-4*x+1)*(x^2+4*x+1)). - Corrected by Colin Barker, Apr 16 2014
a(2n) = (9*A001570(n)+A001570(n+1))/2, a(2n+1) = 5*A001570(n)-6*A007655(n).

Edited by Ralf Stephan, Jun 01 2007

cp's OEIS Frontend

A273052 Numbers n such that 7*n^2 + 8 is a square.

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