%I #25 Dec 22 2024 17:00:41

%S 5,7,11,25,59,103,245,583,1019,2425,5771,10087,24005,57127,99851,

%T 237625,565499,988423,2352245,5597863,9784379,23284825,55413131,

%U 96855367,230496005,548533447,958769291,2281675225,5429921339,9490837543,22586256245,53750679943

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

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

%C 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

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,10,0,0,-1).

%F 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).

%F From _Ralf Stephan_, Nov 15 2010: (Start)

%F G.f.: (-7x^5-11x^4-25x^3+11x^2+7x+5)/(x^6-10x^3+1).

%F a(3n+1) = 5*A001079(n), a(3n+2) = A077409(n), a(3n+3) = A077250(n). (End)

%o (PARI) 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

%Y Cf. A106331.

%K nonn,easy

%O 1,1

%A _Pierre CAMI_, Apr 29 2005

%E More terms from _Ralf Stephan_, Nov 15 2010

%E More terms from _Colin Barker_, Apr 16 2014