%I #26 Apr 23 2023 11:47:13
%S 3,54,969,17388,312015,5598882,100467861,1802822616,32350339227,
%T 580503283470,10416708763233,186920254454724,3354147871421799,
%U 60187741431137658,1080025197889056045,19380265820571871152
%N a(0) = c, a(1) = p*c^3; a(n+2) = p*c^2*a(n+1) - a(n), for p = 2, c = 3.
%C Integer values of Fibonacci numbers * 3/8 (see 2nd formula). - _Vladimir Joseph Stephan Orlovsky_, Oct 25 2009
%H Harry J. Smith, <a href="/A065102/b065102.txt">Table of n, a(n) for n=0,...,100</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H J.-P. Ehrmann et al., <a href="http://forumgeom.fau.edu/POLYA/ProblemCenter/POLYA002.html">Problem POLYA002</a>, Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).
%F G.f.: 3/(1 - 18*x + x^2). - _Floor van Lamoen_, Feb 07 2002
%F a(n) = 3*A049660(n+1). - _R. J. Mathar_, Sep 27 2014
%t a[0] = c; a[1] = p*c^3; a[n_] := a[n] = p*c^2*a[n - 1] - a[n - 2]; p = 2; c = 3; Table[ a[n], {n, 0, 20} ]
%t Clear[f,lst,n,a] f[n_]:=Fibonacci[n]; lst={};Do[a=f[n]*(3/8);If[IntegerQ[a],AppendTo[lst,a]],{n,0,5!}];lst (* _Vladimir Joseph Stephan Orlovsky_, Oct 25 2009 *)
%t nxt[{a_,b_}]:={b,18b-a}; NestList[nxt,{3,54},20][[;;,1]] (* or *) LinearRecurrence[{18,-1},{3,54},20] (* _Harvey P. Dale_, Apr 23 2023 *)
%o (PARI): polya002(2,3,17). For definition of function polya002 see A052530.
%o (PARI) { p=2; c=3; k=p*c^2; for (n=0, 100, if (n>1, a=k*a1 - a2; a2=a1; a1=a, if (n, a=a1=k*c, a=a2=c)); write("b065102.txt", n, " ", a) ) } \\ _Harry J. Smith_, Oct 07 2009
%K easy,nonn
%O 0,1
%A _N. J. A. Sloane_, Nov 12 2001