cp's OEIS Frontend

A265099 Least k such that floor(2^A006666(k)/3^A006667(k)) - k = n.

%I A265099 #12 Oct 21 2019 02:19:31
%S A265099 1,6,9,19,18,27,33,37,36,50,43,56,59,66,57,74,78,72,97,87,86,98,112,
%T A265099 119,118,134,123,115,114,130,149,148,157,135,179,144,153,187,220,174,
%U A265099 173,172,197,196,255,224,238,219,236,203,249,268,247,246,230,229,228
%N A265099 Least k such that floor(2^A006666(k)/3^A006667(k)) - k = n.
%C A265099 A006666 and A006667 are the number of halving and tripling steps to reach 1 in 3x+1 problem.
%C A265099 Conjecture: k exists for all n.
%C A265099 In other words, given an integer n, there always exists at least an integer k and a pair of integers (a, b) such that n + k = 2^a/3^b where a is the number of halving steps to reach 1, and b is the number of tripling steps to reach 1, in the 3x+1 problem.
%e A265099 a(0) = 1 because A006666(1) = 0 and A006667(1) = 0 => floor(2^0/3^0) - 1 = 1 - 1 = 0;
%e A265099 a(1) = 6 because A006666(6) = 6 and A006667(6) = 2 => floor(2^6/3^2) - 6 = floor(64/9) - 6 = 7 - 6 = 1.
%t A265099 lst={};Do[Collatz[k_]:=NestWhileList[If[EvenQ[#],#/2,3 #+1]&,k,#>1&];nn=500;t={};k=0;While[Length[t]<nn,k++;c=Collatz[k];ev=Length[Select[c,EvenQ]];od=Length[c]-ev-1;If[Floor[2^ev/3^od]-k==n,AppendTo[lst,k];Break[]]],{n,0,60}];lst
%Y A265099 Cf. A006666, A006667, A075680, A211981, A225089.
%K A265099 nonn
%O A265099 0,2
%A A265099 _Michel Lagneau_, Dec 01 2015