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A065102 a(0) = c, a(1) = p*c^3; a(n+2) = p*c^2*a(n+1) - a(n), for p = 2, c = 3.

%I A065102 #29 Mar 18 2025 10:56:37
%S A065102 3,54,969,17388,312015,5598882,100467861,1802822616,32350339227,
%T A065102 580503283470,10416708763233,186920254454724,3354147871421799,
%U A065102 60187741431137658,1080025197889056045,19380265820571871152
%N A065102 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 A065102 Integer values of Fibonacci numbers * 3/8 (see 2nd formula). - _Vladimir Joseph Stephan Orlovsky_, Oct 25 2009
%H A065102 Harry J. Smith, <a href="/A065102/b065102.txt">Table of n, a(n) for n=0,...,100</a>
%H A065102 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H A065102 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 A065102 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (18,-1).
%F A065102 G.f.: 3/(1 - 18*x + x^2). - _Floor van Lamoen_, Feb 07 2002
%F A065102 a(n) = 3*A049660(n+1). - _R. J. Mathar_, Sep 27 2014
%t A065102 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 A065102 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 A065102 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 A065102 (PARI) polya002(2,3,17) \\ For definition of function polya002 see A052530.
%o A065102 (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 A065102 easy,nonn
%O A065102 0,1
%A A065102 _N. J. A. Sloane_, Nov 12 2001