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%I #26 Sep 14 2017 03:53:51
%S 3,3,423,3,81,423,75,3,0,81,11003,423,155,75,35,3,239,0,151,81,23,
%T 11003,21,423,21,155,341,75,201,35,75,3,21,239,15,0,113,151,21,81,635,
%U 23,1131,11003,2017,21,75,423,1267,21,75,155,253,341,151,75,7931,201,75,35,69,75,213,3,1073,21,423,239,61,15
%N Smallest number k (different from a power of 2) such that A006577(n*k) = A006577(n) + A006577(k), or 0 if no such number exists.
%C A006577 is the number of halving and tripling steps to reach 1 in the '3x+1' problem. If n is a power of 2, a(n) = 3.
%C If k is a power of 2, we obtain trivial results, for example A006577(n*2^m) = A006577(2^m) + A006577(n) = m + A006577(n) => the smallest k is 1.
%C It appears that a(n) = 0 for n of the form 9*2^a = 9, 18, 36, 72, ...
%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%e a(3) = 423 because A006577(3*423) = A006577(1269) = 39, and A006577(3) + A006577(423) = 7 + 32 = 39.
%p lst:={ }:C:= proc(n) a := 0 ; x := n ; while x > 1 do if type(x, 'even') then x := x/2:a:=a+1: else x := 3*x+1 ; a := a+1 ; end if; end do; a ; end proc:
%p for m from 0 to 40 do:lst:=lst union {2^m}:od:for n from 1 to 73 do: ii:=0:for k from 2 to 50000 while(ii=0) do:z:=n*k : if {k} intersect lst = {} and C(z)=C(n)+C(k) then ii:=1: printf ( "%d %d \n",n,k):else fi:od: if ii=0 and {n} intersect lst = {} and {k} intersect lst = {} then printf ( "%d %d \n",n,0):else fi:od:
%Y Cf. A006577, A057716.
%K nonn
%O 1,1
%A _Michel Lagneau_, Feb 25 2013
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