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.

3, 54, 969, 17388, 312015, 5598882, 100467861, 1802822616, 32350339227, 580503283470, 10416708763233, 186920254454724, 3354147871421799, 60187741431137658, 1080025197889056045, 19380265820571871152

Offset

0,1

Comments

Integer values of Fibonacci numbers * 3/8 (see 2nd formula). - Vladimir Joseph Stephan Orlovsky, Oct 25 2009

Links

Harry J. Smith, Table of n, a(n) for n=0,...,100

Tanya Khovanova, Recursive Sequences

J.-P. Ehrmann et al., Problem POLYA002, Integer pairs (x,y) for which (x^2+y^2)/(1+pxy) is an integer.

Index entries for linear recurrences with constant coefficients, signature (18,-1).

Formula

G.f.: 3/(1 - 18*x + x^2). - Floor van Lamoen, Feb 07 2002

a(n) = 3*A049660(n+1). - R. J. Mathar, Sep 27 2014

Mathematica

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} ]
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 *)
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 *)

Prog

(PARI): polya002(2, 3, 17). For definition of function polya002 see A052530.
(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

Crossrefs

Sequence in context: A157541 A374522 A182159 * A223926 A003776 A222203

Adjacent sequences: A065099 A065100 A065101 * A065103 A065104 A065105

Keyword

easy,nonn

Author

N. J. A. Sloane, Nov 12 2001

Status

approved