0,2

In any generalized Fibonacci sequence {f(i)}, sum_{i=0..4n+1} f(i) = a(n) f(2n+2). - Lekraj Beedassy (blekraj(AT)yahoo.com), 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 (benoit7848c(AT)orange.fr), 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 (kolotoko(AT)wanadoo.fr), 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 (ctibor.zizka(AT)seznam.cz), 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 (qntmpkt(AT)yahoo.com), Nov 26 2008]

a(n+1) is the Hankel transform of A001700(n)+A001700(n+1). [From Paul Barry (pbarry(AT)wit.ie), Apr 21 2009]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

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 (qntmpkt(AT)yahoo.com), 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.

Index entries for sequences related to linear recurrences with constant coefficients

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.

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

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 (benoit7848c(AT)orange.fr), Nov 10 2002

a(n) = A005248(n+1) - A005248(n) = sum(A005248:0, n) - 1. - Lekraj Beedassy (blekraj(AT)yahoo.com), 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 (kolotoko(AT)wanadoo.fr), Mar 01 2004

a(n)=(-1)^n*sum(k=0, n, (-5)^k*binomial(n+k, n-k)) - Benoit Cloitre (benoit7848c(AT)orange.fr), May 09 2004

Both bisection and binomial transform of A000204. a(n)=Fib(2n)+Fib(2n+2). - Paul Barry (pbarry(AT)wit.ie), May 27 2004

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)=(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 (ppl(AT)spl.at), Nov 21 2008]

a(n)=the numerators of sinh((2*n-1)*psi) where the denominators are 2. Psi=ln((1+sqrt5)/2). Offset 1. a(3)=11. [From Al Hakanson (hawkuu(AT)gmail.com), Mar 25 2009]

A002878:=(1+z)/(1-3*z+z**2); [Conjectured (correctly) by S. Plouffe in his 1992 dissertation.]

f[n_] := FullSimplify[GoldenRatio^n - GoldenRatio^-n]; Table[f[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] (* or *)

(Other) sage: [(lucas_number2(n, 3, 1)-lucas_number2(n-1, 3, 1)) for n in xrange(1, 28)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 2009]

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: A027968 A027970 A027972 this_sequence A098149 A124861 A110579

Adjacent sequences: A002875 A002876 A002877 this_sequence A002879 A002880 A002881

nonn,easy,new

N. J. A. Sloane (njas(AT)research.att.com).

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