0,11

For n>3, number of consecutive "11's" after the (n+3) "1's" in the continued fraction for sqrt(L(n+2)/L(n)) where L(n) is the n-th Lucas number A000002 (see example). E.g. the continued fraction for sqrt(L(11)/L(9)) is [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 11, 58, 2, 4, 1, ....] with 12 consecutive ones followed by floor(11/5)=2 elevens. - Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 08 2006

Complement of A010874, since A010874(n)+5*a(n)=n. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 01 2007

Index entries for sequences related to linear recurrences with constant coefficients

Floor(n/5), n>=0.

G.f.: x^5/((1-x)(1-x^5)).

a(n)= -1 + Sum_{k=0..n} {[8*(sin(2*Pi*k/5))^2-5]^2-5}/20, with n>=0. a(n)= -1 + Sum_{k=0..n} 1/50*{-9*[k mod 5]+[(n+1) mod 5]+[(n+2) mod 5]+[(n+3) mod 5]+11*[(n+4) mod 5]}, with n>=0. - Paolo P. Lava (ppl(AT)spl.at), May 15 2007

a(n)=(n-A010874(n))/5. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), May 29 2007

Also, floor((n^5-1)/5*n^4) will produce this sequence. Moreover, floor((n^5-n^4)/(5*n^4-4*n^3)) (n>=1) will produce this sequence as well. - Mohammad K. Azarian (azarian(AT)evansville.edu), Nov 08 2007

This sequence is also the sequence Floor(n*e^(-(1+sqrt(5))/2))(n>=1). - Mohammad K. Azarian (azarian(AT)evansville.edu), May 13 2008

(Other) sage: [floor(n/5) - 1 for n in xrange(5, 88)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Dec 01 2009]

Cf. A008648.

a(n)=A010766(n, 5).

Cf. A004526, A002264, A002265, A010761, A010762, A110532, A110533.

Partial sums: A130520. Other related sequences: A004526, A010872, A010873, A010874.

Sequence in context: A104355 A092278 A105512 this_sequence A075249 A008648 A154099

Adjacent sequences: A002263 A002264 A002265 this_sequence A002267 A002268 A002269

nonn,easy,new

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