A074061 Positive integers k such that 24*k^2 - 23 is a square.

1, 4, 6, 39, 59, 386, 584, 3821, 5781, 37824, 57226, 374419, 566479, 3706366, 5607564, 36689241, 55509161, 363186044, 549484046, 3595171199, 5439331299, 35588525946, 53843828944, 352290088261, 532998958141, 3487312356664

Offset: 0

Author

Michael Somos, Aug 19 2002

Keywords

• nonn
• easy

Comments

Positive values of x (or y) satisfying x^2 - 10xy + y^2 + 23 = 0. - Colin Barker, Feb 09 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• K. S. Brown, Numbers Expressible as (a^2-1)(b^2-1)
• Index entries for two-way infinite sequences
• Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

Programs

Magma

I:=[1,4,6,39]; [n le 4 select I[n] else 10*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Feb 10 2014

Mathematica

CoefficientList[Series[(1 - x) (1 + 5 x + x^2)/(1 - 10 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Feb 10 2014 *)
LinearRecurrence[{0,10,0,-1},{1,4,6,39},30] (* Harvey P. Dale, Jun 06 2015 *)

PARI

{a(n) = if( n<0, a(-1-n), polcoeff( (1 - x) * (1 + 5*x + x^2) / (1 - 10*x^2 + x^4) + x * O(x^n), n))}

Formula

G.f.: (1 - x)*(1 + 5*x + x^2)/(1 - 10*x^2 + x^4).

a(n) = 10*a(n-2) - a(n-4) = a(-1-n).

a(2n-1) = round((1/2)*(1-(1/2)/sqrt(6))*(sqrt(2)+sqrt(3))^(2n)); a(2n)=round(c*(sqrt(2)+sqrt(3))^(2n+1)) with c = 0.191357750597... - Benoit Cloitre, Aug 22 2002