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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. 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, 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). - 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: A281941 A284278 A330080 * A137403 A082233 A330806 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 June 8 11:34 EDT 2023. Contains 363164 sequences. (Running on oeis4.)