2 puzzles and comments for A085688,A085689

[f.firoozbakht at sci.ui.ac.ir]

Dear Neil and dear seqfans ,

Hello ,

See A085688 and A085689 ,

 A085688:  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
Name:      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.


 A085689:  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

Name:      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.

             
I found the following formulas for A085688 and A085689 ( without using If ) :


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 ?



Regards,

Farideh



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