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A033890
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Fibonacci(4n+2).
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32
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1, 8, 55, 377, 2584, 17711, 121393, 832040, 5702887, 39088169, 267914296, 1836311903, 12586269025, 86267571272, 591286729879, 4052739537881, 27777890035288, 190392490709135, 1304969544928657, 8944394323791464, 61305790721611591, 420196140727489673
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| 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).
(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
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
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
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]
Indices of square numbers which are also 12-gonal [From Sture Sjoestedt (sture.sjostedt(AT)spray.se), Jun 01 2009]
a(n) = A167816(4*n+2). [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Nov 13 2009]
For positive n, a(n) equals the permanent of the (2n)X(2n) tridiagonal matrix with 3's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). [From John M. Campbell, Jul 08 2011]
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LINKS
| Index entries for sequences related to linear recurrences with constant coefficients, signature (7,-1).
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
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FORMULA
| 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.
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
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
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
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]
a(n)=b such that (-1)^n*Integral_{0..Pi/2} (cos((2*n+1)*x))/(3/2-sin(x)) dx = c + b*ln(3) [From Francesco Daddi (francesco.daddi(AT)libero.it), Aug 01 2011]
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MATHEMATICA
| Table[Fibonacci[4*n+2], {n, 0, 14}] (Vladimir Orlovsky, Jul 21 2008)
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PROG
| (PARI) a(n)=fibonacci(4*n+2)
(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]
(MAGMA) [Fibonacci(4*n +2): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011
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CROSSREFS
| Sequence in context: A154245 A143420 A075734 * A010924 A010918 A019484
Adjacent sequences: A033887 A033888 A033889 * A033891 A033892 A033893
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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