0,2

Index entries for sequences related to linear recurrences with constant coefficients

Tanya Khovanova, Recursive Sequences

a(n) ~ 1/4*(sqrt(5) + 2)^(2*n+1) - Joe Keane (jgk(AT)jgk.org), May 15 2002

For all members x of the sequence, 20*x^2 + 5 is a square. Lim. n -> Inf. a(n)/a(n-1) = 9 + 2*Sqrt(20) = 9 + 4*Sqrt(5). The 20 can be seen to derive from the equation "20*x^2 + 5 is a square". - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 12 2002

a(n) = [ [(9 + 4*Sqrt(5))^N - (9 - 4*Sqrt(5))^N] + [(9 + 4*Sqrt(5))^(N-1) - (9 - 4*Sqrt(5))^(N-1)] / (8*Sqrt(5)) - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 12 2002

G.f.: (1+x)/(1-18x+x^2). a(n)=A049660(n)+A049660(n+1). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 04 2008]

a(n)=18*a(n-1)-a(n-2) for n>1; a(0)=1, a(1)=19. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 17 2008]

q=10; s=0; lst={}; Do[s+=n; If[Sqrt[q*s+1]==Floor[Sqrt[q*s+1]], AppendTo[lst, Sqrt[q*s+1]]], {n, 0, 9!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Apr 02 2009]

(Other) sage: [(lucas_number2(n, 18, 1)-lucas_number2(n-1, 18, 1))/16 for n in xrange(1, 13)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 2009]

Bisection of A001077 divided by 2

Sequence in context: A121324 A093973 A142549 this_sequence A144745 A049664 A078368

Adjacent sequences: A049626 A049627 A049628 this_sequence A049630 A049631 A049632

nonn,new

Clark Kimberling (ck6(AT)evansville.edu)