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From: TCL on 17 Jun 2010 15:34
Let n be an odd integer. Let m be the largest integer such
that 2^m divides 3n+1. Define

T(n) = (3n+1)/(2^m).

One has T(6465)=4849, T(4849)=3637.

Kind of interesting I would say.
From: hagman on 17 Jun 2010 16:36
On 17 Jun., 21:34, TCL <tl...(a)cox.net> wrote:
> Let n be an odd integer. Let m be the largest integer such
> that 2^m divides 3n+1. Define
>
> T(n) = (3n+1)/(2^m).
>
> One has  T(6465)=4849, T(4849)=3637.
>
> Kind of interesting I would say.

Gees, it took me a while why you would find this interesting
I guess you're aiming at these numbers being of the form
n = 101*k+1

Cf. also 1001*l+1:
512513 -> 384385 -> 288289 -> 216217 -> 162163 -> 243245 (almost!)
or more generally sequences starting with 2^a*(10^b+1)+1 with 2^a<10^b

hagman
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