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

%I #77 Jan 18 2023 02:21:59

%S 1,4,6,39,59,386,584,3821,5781,37824,57226,374419,566479,3706366,

%T 5607564,36689241,55509161,363186044,549484046,3595171199,5439331299,

%U 35588525946,53843828944,352290088261,532998958141,3487312356664

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

%C Positive values of x (or y) satisfying x^2 - 10xy + y^2 + 23 = 0. - _Colin Barker_, Feb 09 2014

%H Vincenzo Librandi, <a href="/A074061/b074061.txt">Table of n, a(n) for n = 0..200</a>

%H K. S. Brown, <a href="http://www.mathpages.com/home/kmath275.htm">Numbers Expressible as (a^2-1)(b^2-1)</a>

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,10,0,-1).

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

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

%F 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

%t CoefficientList[Series[(1 - x) (1 + 5 x + x^2)/(1 - 10 x^2 + x^4), {x, 0, 30}], x] (* _Vincenzo Librandi_, Feb 10 2014 *)

%t LinearRecurrence[{0,10,0,-1},{1,4,6,39},30] (* _Harvey P. Dale_, Jun 06 2015 *)

%o (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))}

%o (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

%K nonn,easy

%O 0,2

%A _Michael Somos_, Aug 19 2002