2 puzzles and comments for A085688,A085689

[Dean Hickerson]

Farideh Firoozbakht wrote:

> A085688:  11,8,6,12,9,7,14,11,9,18,15,13,26,23,21,42,39,37,74,71,69,
...
> A085689:  4,2,4,12,6,12,36,18,36,108,54,108,324,162,324,972,486,972,
...             
> For A085688 :
>
> 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 
> 
> And for A085689 : 
>
>  a[1] = 4; a[n] := (2 + Mod[n, 3])*8^(-Floor[(1 + Mod[n, 3])/3])*a[n - 1]
>
> Can you find simpler formulas( without using if ) for A085688 and  A085689 ?

For A085688,

    a(n) = 2^floor((n-1)/3) + floor(21/(((n-1) mod 3)+2))

For A085689,

    a(n):=3^floor((n-1)/3) (4 - 2 floor((n mod 3)/2))


In Mathematica notation, these are

    a[n_]:=2^Floor[(n-1)/3]+Floor[21/(Mod[n-1,3]+2)]

and

    a[n_]:=3^Floor[(n-1)/3](4-2Floor[Mod[n,3]/2])

Dean Hickerson
dean at math.ucdavis.edu