%I A033890
%S A033890 1,8,55,377,2584,17711,121393,832040,5702887,39088169,267914296,
%T A033890 1836311903,12586269025,86267571272,591286729879,4052739537881,
%U A033890 27777890035288,190392490709135,1304969544928657,8944394323791464
%N A033890 Fibonacci(4n+2).
%C A033890 a(n) = S(n,7)+S(n-1,7) = S(2*n,sqrt(9) = 3), S(n,x) = U(n,x/2) are Chebyshev's 
               polynomials of the 2nd kind. Cf. A049310. S(n,7) = A004187(n+1), 
               S(n,3) = A001906(n+1).
%C A033890 (x,y)=(a(n),a(n+1)) are solutions of (x+y)^2/(1+xy)=9, the other solutions 
               are in A033888. - Floor van Lamoen (fvlamoen(AT)hotmail.com), Dec 
               10 2001
%C A033890 The sequence A033890 provides half of the solutions to the equation 5*x^2 
               + 4 is a square. The other solutions are included in A033888. Lim. 
               n-> Inf. a(n)/a(n-1) = phi^4 = (7 + 3*Sqrt(5))/2 - Gregory V. Richardson 
               (omomom(AT)hotmail.com), Oct 13 2002
%C A033890 a(n) = L(n,-7)*(-1)^n, where L is defined as in A108299; see also A049685 
               for L(n,+7). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Jun 01 2005
%C A033890 General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim n->infinity 
               a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. 
               Examples in OEIS: a(1)=4 gives A002878, primes in it A121534. a(1)=5 
               gives A001834, primes in it A086386. a(1)=6 gives A030221, primes 
               in it not in OEIS {29,139,3191,...}. a(1)=7 gives A002315, primes 
               in it A088165. a(1)=8 gives A033890, primes in it not in OEIS (does 
               there exist any ?). a(1)=9 gives A057080, primes in it not in OEIS 
               {71,34649,16908641,...}. a(1)=10 gives A057081, primes in it not 
               in OEIS {389806471,192097408520951,...}. [From Ctibor O. Zizka (ctibor.zizka(AT)seznam.cz), 
               Sep 02 2008]
%C A033890 Indices of square numbers which are also 12-gonal [From Sture Sjoestedt 
               (sture.sjostedt(AT)spray.se), Jun 01 2009]
%C A033890 a(n) = A167816(4*n+2). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Nov 13 2009]
%H A033890 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A033890 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%H A033890 <a href="Sindx_Ch.html#Cheby">Index entries for sequences related to 
               Chebyshev polynomials.</a>
%F A033890 G.f.: (1+x)/(1-7*x+x^2). a(n)=7*a(n-1)-a(n-2), n>1. a(0)=1, a(1)=8.
%F A033890 a(n) = [ [(7+3*Sqrt(5))^n - [(7-3*Sqrt(5))^n] + 2*[(7+3*Sqrt(5))^(n-1) 
               - [(7-3*Sqrt(5))^(n-1)] ] / (3*(2^n)*Sqrt(5)) - Gregory V. Richardson 
               (omomom(AT)hotmail.com), Oct 13 2002
%F A033890 Let q(n, x)=sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then (-1)^n*q(n, 
               -9)=a(n) - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 10 2002
%F A033890 Define f[x,s] = s x + Sqrt[(s^2-1)x^2+1]; f[0,s]=0. a(n) = f[a(n-1),7/
               2] + f[a(n-2),7/2]. - Marcos Carreira, Dec 27 2006
%F A033890 a(n+1)=8*a(n)-8*a(n-1)+ a(n-2) a(1)=1 , a(2)=8 , a(3)=55 [From Sture 
               Sjostedt (sture.sjostedt(AT)spray.se), May 27 2009]
%t A033890 Table[Fibonacci[4*n+2],{n,0,14}] (Vladimir Orlovsky, Jul 21 2008)
%o A033890 (PARI) a(n)=fibonacci(4*n+2)
%o A033890 (Other) sage: [(lucas_number2(n,7,1)-lucas_number2(n-1,7,1))/5 for n 
               in xrange(1, 21)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Nov 10 2009]
%Y A033890 Sequence in context: A154245 A143420 A075734 this_sequence A010924 A010918 
               A019484
%Y A033890 Adjacent sequences: A033887 A033888 A033889 this_sequence A033891 A033892 
               A033893
%K A033890 nonn,new
%O A033890 0,2
%A A033890 N. J. A. Sloane (njas(AT)research.att.com).