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A068508 a(n)=round[(a(n-1)+a(n-2))/a(n-3)] starting with a(1)=a(2)=a(3)=1. 2
1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1, 1, 1, 2, 3, 5, 4, 3, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

COMMENTS

While this sequence has period 8, the unrounded version b(n)=(b(n-1)+b(n-2))/b(n-3) seems to have a quasi-period of about 8.7 for this particular starting point.

Terms of the simple continued fraction of 1198/[sqrt(5368485)-1563]. [From Paolo P. Lava, Aug 06 2009]

The unrounded version b(n) = A185332(n) / A185341(n) as given in A205303 has 8.694171... quasi-period. - Michael Somos, Oct 22 2018

LINKS

Table of n, a(n) for n=1..105.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,1).

FORMULA

a(n)=a(n-8)

a(n)=1/56*{19*(n mod 8)+12*[(n+1) mod 8]+12*[(n+2) mod 8]-9*[(n+3) mod 8]-2*[(n+4) mod 8]-2*[(n+5) mod 8]+5*[(n+6) mod 8]+5*[(n+7) mod 8]} with n>=0 - Paolo P. Lava, Nov 27 2006

EXAMPLE

a(7)=round[(a(6)+a(5))/a(4)]=round[(5+3)/2]=4

CROSSREFS

Cf. A048112, A185332, A185341, A205303.

Sequence in context: A021981 A281941 A284278 * A137403 A082233 A058981

Adjacent sequences:  A068505 A068506 A068507 * A068509 A068510 A068511

KEYWORD

nonn,easy

AUTHOR

Henry Bottomley, Mar 25 2002

STATUS

approved

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Last modified July 17 18:47 EDT 2019. Contains 325109 sequences. (Running on oeis4.)