A106331 -id:A106331 - 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

A106330 Numbers k such that k^2 = 24*j^2 + 25.

Original entry on oeis.org

5, 7, 11, 25, 59, 103, 245, 583, 1019, 2425, 5771, 10087, 24005, 57127, 99851, 237625, 565499, 988423, 2352245, 5597863, 9784379, 23284825, 55413131, 96855367, 230496005, 548533447, 958769291, 2281675225, 5429921339, 9490837543, 22586256245, 53750679943

Offset: 1

Pierre CAMI, Apr 29 2005

The ratio k(n) /(2*j(n)) tends to sqrt(6) as n increases.

k(n) = 2*b + 1, for n > 0, where b is a side of the Heronian triangle (5, b, b+1). - Andrés Ventas, Dec 13 2024

Index entries for linear recurrences with constant coefficients, signature (0,0,10,0,0,-1).

Cf. A106331.

Vec(-x*(7*x^5+11*x^4+25*x^3-11*x^2-7*x-5)/(x^6-10*x^3+1) + O(x^100)) \\ Colin Barker, Apr 16 2014

Recurrence: k(1)=5, k(2)=7, k(3)=11, k(4)=25, k(5)=10*k(2)-k(3), k(6)=10*k(3)-k(2) then k(n)=10*k(n-3)-k(n-6).
From Ralf Stephan, Nov 15 2010: (Start)
G.f.: (-7x^5-11x^4-25x^3+11x^2+7x+5)/(x^6-10x^3+1).
a(3n+1) = 5*A001079(n), a(3n+2) = A077409(n), a(3n+3) = A077250(n). (End)

More terms from Ralf Stephan, Nov 15 2010

More terms from Colin Barker, Apr 16 2014

cp's OEIS Frontend

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

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