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Numbers j such that 24*(j^2) + 25 = k^2.
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%I #24 Sep 17 2019 10:31:20

%S 0,1,2,5,12,21,50,119,208,495,1178,2059,4900,11661,20382,48505,115432,

%T 201761,480150,1142659,1997228,4752995,11311158,19770519,47049800,

%U 111968921,195707962,465745005,1108378052,1937309101,4610400250

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

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

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,10,0,0,-1). [_R. J. Mathar_, May 22 2010]

%F j(1)=0, j(2)=1, j(3)=2, j(4)=5, j(5)=10*j(2)+j(3), j(6)=10*j(3)+j(2), j(7)=10*j(4)+j(1) then j(n)=10*j(n-3)-j(n-6).

%F a(n) = +10*a(n-3) -a(n-6). G.f.: x^2*(1+2*x+5*x^2+2*x^3+x^4)/(1-10*x^3+x^6). [_R. J. Mathar_, May 22 2010]

%F a(3*n+1) = 5*A004189(n), a(3*n+2) = A077251(n), a(3*n+3) = A077249(n). - _Ralf Stephan_, Nov 15 2010

%Y Cf. A106330.

%K nonn,easy

%O 1,3

%A _Pierre CAMI_, Apr 29 2005

%E More terms from _Jon E. Schoenfield_, May 16 2010