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 A085689 a(1) = 4; a(n) = if n == 2 mod 3 then a(n-1)/2, if n == 0 mod 3 then a(n-1)*2, if n == 1 mod 3 then a(n-1)*3. 0
 4, 2, 4, 12, 6, 12, 36, 18, 36, 108, 54, 108, 324, 162, 324, 972, 486, 972, 2916, 1458, 2916, 8748, 4374, 8748, 26244, 13122, 26244, 78732, 39366, 78732, 236196, 118098, 236196, 708588, 354294, 708588, 2125764, 1062882, 2125764, 6377292, 3188646, 6377292, 19131876 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Given as a puzzle: find the next term after 4, 12, 6, 12, 36, 18, 36! Thanks to Farideh Firoozbakht and Zak Seidov for the solution. LINKS Index entries for linear recurrences with constant coefficients, signature (0,0,3). FORMULA a[1] = 4; a[n] = (2 + Mod[n, 3])*8^(-Floor[(1 + Mod[n, 3])/3])*a[n - 1]. a(n) = 3^floor((n-1)/3) (4 - 2 floor((n mod 3)/2)). - Dean Hickerson, Jul 24, 2003 a(n) = 3*a(n-3). G.f.: -2*x*(2*x^2+x+2) / (3*x^3-1). - Colin Barker, Jul 31 2013 MAPLE a := proc(n) option remember; if n=1 then 4; elif n mod 3 = 2 then a(n-1)/2 elif n mod 3 = 0 then a(n-1)*2 else a(n-1)*3; fi; end; MATHEMATICA a[1] = 4; a[n_] := Switch[ Mod[n, 3], 0, 2a[n - 1], 1, 3a[n - 1], 2, a[n - 1]/2]; Table[ a[n], {n, 1, 43}] a[1] = 4; a[n_] := (2 + Mod[n, 3])*8^(-Floor[(1 + Mod[n, 3])/3])*a[n - 1] Do[Print[a[n], {n, 30}] CROSSREFS Sequence in context: A011382 A011302 A302603 * A134434 A261254 A168613 Adjacent sequences:  A085686 A085687 A085688 * A085690 A085691 A085692 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Jul 18 2003 STATUS approved

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Last modified October 23 19:37 EDT 2019. Contains 328373 sequences. (Running on oeis4.)