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A337349 To get a(n), take 3*n+1 and divide out any power of 2; then multiply by 3, subtract 1 and divide out any power of 2. 4

%I #19 Aug 27 2020 21:26:13

%S 1,1,5,7,19,1,7,1,37,5,23,25,55,7,1,17,73,19,41,43,91,1,25,13,109,7,

%T 59,61,127,1,17,35,145,37,77,79,163,5,43,11,181,23,95,97,199,25,13,53,

%U 217,55,113,115,235,7,61,31,253,1,131,133,271,17,35,71,289,73,149

%N To get a(n), take 3*n+1 and divide out any power of 2; then multiply by 3, subtract 1 and divide out any power of 2.

%C When a(x) is iterated, what are the limit cycles? Are there any besides (1) and (17 -> 19 -> 43 -> 97 -> 109 -> 61)?

%C Up to 1000000000 every number eventually reaches one of those two cycles. In this range, the longest trajectory starts with n=458788881 and takes 193 steps to reach 1. - Christian Boyer (cboyer(AT)club-internet.fr), Sep 16 2006

%H Hugo Pfoertner, <a href="/A337349/b337349.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n) = A075677(A067745(n+1)).

%F a(2*n+1) = A102421(n).

%p A337349 := proc(n)

%p local a;

%p a := 3*n+1;

%p while modp(a,2) = 0 do

%p a := a/2 ;

%p end do:

%p a := 3*a-1 ;

%p while modp(a,2) = 0 do

%p a := a/2 ;

%p end do:

%p a ;

%p end proc: # _R. J. Mathar_, Aug 24 2020

%t a[n_] := Module[{k = 3n+1}, k = k/2^IntegerExponent[k, 2]; k = 3k-1; k = k/2^IntegerExponent[k, 2]; k];

%t a /@ Range[0, 100] (* _Jean-François Alcover_, Aug 27 2020 *)

%Y Cf. A102421 (bisection), A102423.

%K nonn,easy

%O 0,3

%A _N. J. A. Sloane_, based on email from Dan Asimov (dasimov(AT)earthlink.net), Sep 15 2006

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Last modified April 26 16:30 EDT 2024. Contains 372003 sequences. (Running on oeis4.)