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A122437 Allowable values of the "dropping time" of the Collatz (3x+1) iteration.

%I A122437 #53 Aug 17 2025 17:35:35
%S A122437 1,3,6,8,11,13,16,19,21,24,26,29,32,34,37,39,42,44,47,50,52,55,57,60,
%T A122437 63,65,68,70,73,75,78,81,83,86,88,91,94,96,99,101,104,106,109,112,114,
%U A122437 117,119,122,125,127,130,132,135,138,140,143,145,148,150,153,156,158,161
%N A122437 Allowable values of the "dropping time" of the Collatz (3x+1) iteration.
%C A122437 Only these numbers appear in A060445, which tabulates the "dropping time" of odd numbers. Note that all even numbers have a "dropping time" of 1.
%C A122437 a(n) is also the number of binary digits of 6^(n-1); for example, a(4)=8 since 6^(4-1)=216 in binary is 11011000, an 8-digit number. - _Julio Cesar de la Yncera_, Mar 28 2009
%C A122437 A positive integer (x) is an allowable value if and only if (x-1)/(1+log(2)/log(3)) - floor(x/(1+log(2)/log(3))) is not negative. - _K. Spage_, Oct 22 2009
%C A122437 Here the word "allowable" means that it is necessary for a sequence of iterates starting from odd value m to arrive at a value x = f^{floor(1+n+n*log(3)/log(2))}(m) < m, where n gives the number of odds in such a sequence including m, to have undergone precisely floor(1+n+n*log(3)/log(2)) iterations of f, where f(2*m)=m, f(2*m+1)=6*m+4. However, the formula for a(n+1) does not fully account for the order of odds and evens in such a sequence because it does not account for the effects of the "+1". Thus it is unknown whether it maximizes the value x for all values m. For example, fix m = 1 and the "+1" is enough to give the trivial cycle. So it is possible that for some m we have f^{floor(1+n+n*log(3)/log(2))}(m) >= m. - _Jeffrey R. Goodwin_, Aug 24 2011
%C A122437 The indices of the powers of 3 in A006899. - _Ruud H.G. van Tol_, Nov 02 2022
%H A122437 T. D. Noe, <a href="/A122437/b122437.txt">Table of n, a(n) for n = 1..1000</a>
%F A122437 a(1) = 1, a(n+1) = a(n) + A022921(n-1) + 1.
%F A122437 a(n+1) = floor(1 + n + n*log(3)/log(2)). - _T. D. Noe_, Sep 08 2006
%F A122437 a(n) = floor((1 + log(2)/log(3))*A020914(n-1)). - _K. Spage_, Oct 22 2009
%F A122437 a(n) = A020914(n-1) + n - 1. - _K. Spage_, Oct 23 2009 [corrected by _Ruud H.G. van Tol_, Nov 03 2022]
%F A122437 a(n) = a(n-1)+2 if 3^(n-1) < 2^(a(n-1)+2-(n-1)); a(n) = a(n-1)+3 otherwise. - _V. Barbera_, Aug 12 2025
%t A122437 Floor[1+Range[0,100]*(1+Log[2,3])] (* _T. D. Noe_, Sep 08 2006 *)
%t A122437 Map[Length[RealDigits[ #, 2][[1]]] &, Table[10^i, {i, 0, 50}]] (* _Julio Cesar de la Yncera_, Mar 28 2009 *)
%o A122437 (PARI) a(n)=logint(3^(n-1),2)+n \\ _Ruud H.G. van Tol_, Nov 04 2022
%Y A122437 Cf. A022921 (number of 2^m between 3^n and 3^(n+1)), A122442 (least k having dropping time a(n)).
%Y A122437 Cf. A006899.
%K A122437 nice,nonn
%O A122437 1,2
%A A122437 _T. D. Noe_, Sep 06 2006
%E A122437 Comment corrected and edited by _Jon E. Schoenfield_, Feb 27 2014