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A002878
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Bisection of Lucas sequence: a(n) = L(2*n+1).
(Formerly M3420 N1384)
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64
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1, 4, 11, 29, 76, 199, 521, 1364, 3571, 9349, 24476, 64079, 167761, 439204, 1149851, 3010349, 7881196, 20633239, 54018521, 141422324, 370248451, 969323029, 2537720636, 6643838879, 17393796001, 45537549124, 119218851371, 312119004989, 817138163596, 2139295485799
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| In any generalized Fibonacci sequence {f(i)}, sum_{i=0..4n+1} f(i) = a(n)*f(2n+2). - Lekraj Beedassy, Dec 31 2002
The continued fraction expansion for F((2n+1)*(k+1))/F((2n+1)*k) k>=1 is [a(n),a(n),...,a(n)] where there are exactly k elements (F(n) denotes the n-th Fibonacci number). E.g. continued fraction for F(12)/F(9) is [4, 4,4]. - Benoit Cloitre, Apr 10 2003
See A135064 for a possible connection with Galois groups of quintics.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(5)). - Thomas Baruchel Sep 15 2003
All positive integer solutions of Pell equation a(n)^2 - 5*b(n)^2 = -4 together with b(n)=A001519(n), n>=0.
a(n) = L(n,-3)*(-1)^n, where L is defined as in A108299; see also A001519 for L(n,+3).
Inverse binomial transform of A030191 . - Philippe DELEHAM, Oct 04 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, Sep 02 2008]
Let r = (2n+1), then a(n), n>0 = PRODUCT_{k=1,[(r-1)/2] (1 + Sin^2 k*Pi/r); e.g., a(3) = 29 = (3.4450418679...)*(4.801937735...)*(1.753020396...). [From Gary W. Adamson, Nov 26 2008]
a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). [From Paul Barry, Apr 21 2009]
a(n) is equal to the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(5)'s along the main diagonal, i's along the superdiagonal and the subdiagonal (i is the imaginary unit), and 0's everywhere else. [From John M. Campbell, June 9 2011]
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REFERENCES
| N. D. Cahill, J. D. D'Errico and J. P. Spencer, "Complex Factorizations of the Fibonacci and Lucas Numbers"; Fibonacci Quarterly, 1(41):13-19, 2003. [From Gary W. Adamson, Nov 26 2008]
A. Gougenheim, About the linear sequence of integers such that each term is the sum of the two preceding, Fib. Quart., 9 (1971), 277-295, 298.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
Eric Weisstein's World of Mathematics, Fibonacci Polynomial
Index entries for sequences related to Chebyshev polynomials.
Index entries for sequences related to linear recurrences with constant coefficients, signature (3,-1).
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FORMULA
| a(n+1) = 3*a(n)-a(n-1).
G.f.: (1+x)/(1-3*x+x^2).
a(n)= S(2*n, sqrt(5)) = S(n, 3)+S(n-1, 3); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 3)= A001906(n+1) (even indexed Fibonacci numbers).
a(n) ~ phi^(2*n+1) - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = phi^(2*n+1)-phi^(-2*n-1), n >=0. - Paolo P. Lava, Jan 03 2011
Let q(n, x)=sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then (-1)^n*q(n, -1)=a(n) - Benoit Cloitre, Nov 10 2002
a(n) = A005248(n+1) - A005248(n) = sum(A005248:0, n) - 1. - Lekraj Beedassy, Dec 31 2002
a(n) = 2^(-n)*A082762(n) = 4^(-n)*Sum_{k>=0} binomial(2*n+1, 2*k)*5^k; see A091042 . - DELEHAM Philippe, Mar 01 2004
a(n)=(-1)^n*sum(k=0, n, (-5)^k*binomial(n+k, n-k)) - Benoit Cloitre, May 09 2004
Both bisection and binomial transform of A000204. a(n)=Fib(2n)+Fib(2n+2). - Paul Barry, May 27 2004
a(n)=(1/2)*[(3/2)+(1/2)*sqrt(5)]^n+(1/2)*[(3/2)+(1/2)*sqrt(5)]^n*sqrt(5)-(1/2)*[(3/2)-(1/2)*sqrt(5)]^n *sqrt(5)+(1/2)*[(3/2)-(1/2)*sqrt(5)]^n, with n>=0 [From Paolo P. Lava, Nov 21 2008]
a(n)=the numerators of sinh((2*n-1)*psi) where the denominators are 2. Psi=log((1+sqrt5)/2). Offset 1. a(3)=11. [From Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009]
a(n) = A001906(n) + A001906(n+1). [Reinhard Zumkeller, Jan 11 2012]
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MAPLE
| A002878:=(1+z)/(1-3*z+z**2); [Conjectured (correctly) by S. Plouffe in his 1992 dissertation.]
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MATHEMATICA
| a[n_] := FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[a[n], {n, 1, 55, 2}] (* or *)
a[1] = 1; a[2] = 4; a[n_] := a[n] = 3a[n - 1] - a[n - 2]; Array[a, 28]
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PROG
| (MAGMA) [ Lucas(2*n +1): n in [0..120]]; // Vincenzo Librandi, Apr 16 2011
(PARI) a(n)=fibonacci(2*n)+fibonacci(2*n+2) \\ Charles R Greathouse IV, Jun 16 2011
(Haskell)
a002878 n = a002878_list !! n
a002878_list = zipWith (+) (tail a001906_list) a001906_list
-- Reinhard Zumkeller, Jan 11 2012
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CROSSREFS
| Cf. A000204. a(n)= A060923(n, 0).
Cf. A005248 [L(2n) = bisection (even n) of Lucas sequence].
Cf. A001906 [F(2n) = bisection (even n) of Fibonacci sequence].
Sequence in context: A027970 A027972 A098149 * A124861 A110579 A084378
Adjacent sequences: A002875 A002876 A002877 * A002879 A002880 A002881
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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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EXTENSIONS
| More terms from James A. Sellers (sellersj(AT)math.psu.edu), May 29 2000
Chebyshev and Pell comments from W. Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Aug 31 2004
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