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A085688 a(1) = 11; a(n) = if n == 2 mod 3 then a(n-1)-3, if n == 0 mod 3 then a(n-1)-2, if n == 1 mod 3 then a(n-1)*2. 3
11, 8, 6, 12, 9, 7, 14, 11, 9, 18, 15, 13, 26, 23, 21, 42, 39, 37, 74, 71, 69, 138, 135, 133, 266, 263, 261, 522, 519, 517, 1034, 1031, 1029, 2058, 2055, 2053, 4106, 4103, 4101, 8202, 8199, 8197, 16394, 16391, 16389, 32778, 32775, 32773, 65546, 65543, 65541, 131082 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Sequence (in reversed order) was given as a puzzle: find the next term after 18, 9, 11, 14, 7, 9, 12! Thanks to Farideh Firoozbakht and Zak Seidov for the solution.

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-2).

FORMULA

a[1]=11; for k=>1, a[3k-1]=7+2^(3k-1), a[3k]=5+2^(3k-1), a[3k+1]=10+2^(3k). - Zak Seidov, Jul 24 2003

a(n) = 2^floor((n-1)/3) + floor(21/(((n-1) mod 3)+2)). - Dean Hickerson, Jul 24 2003

G.f. -x*(-11-8*x-6*x^2+21*x^3+15*x^4+11*x^5) / ( (x-1)*(2*x^3-1)*(1+x+x^2) ). - R. J. Mathar, Oct 20 2013

MAPLE

a := proc(n) option remember; if n=1 then 11 elif n mod 3 = 2 then a(n-1)-3 elif n mod 3 = 0 then a(n-1)-2 else a(n-1)*2; fi; end;

MATHEMATICA

a[1] = 11; a[n_] := (3 - (-1)^mod[n, 3])/2*a[n - 1] - (1 + (-1)^mod[n, 3])/2* Floor[mod[n, 3]/2] - (-1)^mod[n, 3] - 1 (from Farideh Firoozbakht, Jul 23 2003)

nxt[{n_, a_}]:=Module[{c=Mod[n+1, 3]}, {n+1, Which[c==2, a-3, c==0, a-2, c==1, 2a]}]; NestList[nxt, {1, 11}, 60][[All, 2]] (* or *) LinearRecurrence[ {0, 0, 3, 0, 0, -2}, {11, 8, 6, 12, 9, 7}, 70] (* Harvey P. Dale, Mar 14 2020 *)

CROSSREFS

Sequence in context: A090841 A085757 A003567 * A164059 A306494 A068974

Adjacent sequences:  A085685 A085686 A085687 * A085689 A085690 A085691

KEYWORD

nonn

AUTHOR

N. J. A. Sloane, Jul 18 2003

STATUS

approved

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Last modified July 14 23:27 EDT 2020. Contains 335752 sequences. (Running on oeis4.)