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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. 2
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

Table of n, a(n) for n=1..52.

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)

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 March 25 16:26 EDT 2019. Contains 321470 sequences. (Running on oeis4.)