EDITED on A096773.

[Gottfried Helms]

Am 21.08.04 00:33 schrieb Robert G. Wilson v:
> Neil,
> 
> 	The title was "a(2m-1) = (2^(2m) -1)/3, a(2m) = (5*2^(2m-1)-1)/3." but that does 
> not work so I used the recursive formula for the title. I also put in the 
> corrected direct formula in the %Formula line. Also I added the Mathematica
> coding, extended the sequence and added the cross ref line.
> 
> Thanx, Bob.
> 
> 
> %I A096773
> %S A096773 0,3,1,13,5,53,21,213,85,853,341,3413,1365,13653,5461,54613,21845,
> %T A096773 218453,87381,873813,349525,3495253,1398101,13981013,5592405,55924053,
> %U A096773 22369621,223696213,89478485,894784853,357913941,3579139413,1431655765
> %N A096773 a(n+2) = a(n)*4 + 1; a(1)=0, a(2) = 3.
> %C A096773 Remainders for classes m of integers n (mod 2^(m+1)), which segment the 
> integers in 2 groups after one Collatz (3x+1)-transformation.
> %F A096773 a(2n) = (5*2^(n-1) - 1)/3 & a(2n-1) = (2^(n - 1) - 1)/3 with a(1)=0 & 
> a(2) = 3.
> %t A096773 a[1] = 0; a[2] = 3; a[n_] := a[n] = 4a[n - 2] + 1; Table[ a[n], {n, 
> 35}] (from RGWv Aug 20 2004)
> %Y A096773 The bisections are A002450 & A072197.
> %K A096773 easy,nonn,new
> %O A096773 1,2
> %A A096773 Gottfried Helms (helms(AT)uni-kassel.de), Aug 15 2004
> %E A096773 Edited and extended by RGWv (rgwv at rgwv.com), Aug 20 2004
> 
> 
> 
Hi Robert,

 unfortunately I made a typo. Clearly the recursion formula can be expressed
 as a direct one. Seems I have missed the "2" with the n in the exponent:

 a(2n) = (5*2^(2n-1) - 1)/3       a(2n-1) = (2^(2n-2)-1)/3
 ----------------------------------------------------------
                                     a(1)  =  (2^0-1)/3 = 0
  a(2) = (5*2^1 - 1) / 3 = 3         a(3)  =  (2^2-1)/3 = 1
  a(4) = (5*2^3 - 1) / 3 =13         a(5)  =  (2^4-1)/3 = 5
  a(6) = (5*2^5 - 1) / 3 =53         a(7)  =  (2^6-1)/3 =21
 ...
Ah, so it works.

Sorry for that inconvenience!

I think the direct formula is useful. Should it be in the titel again?

Regards -

Gottfried Helms