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

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.

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.

Tanya Khovanova, Recursive Sequences

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).

A002878:=(1+z)/(1-3*z+z**2); [Conjectured 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 *)

Cf. A000204, A005248. a(n)= A060923(n, 0).

Adjacent sequences: A002875 A002876 A002877 this_sequence A002879 A002880 A002881

Sequence in context: A027968 A027970 A027972 this_sequence A098149 A124861 A110579

nonn,easy

njas

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