A277367 a(n) = gcd(A006666(n), A006667(n)) where A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

0, 1, 1, 2, 1, 2, 1, 3, 1, 1, 2, 1, 1, 1, 1, 4, 3, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 5, 2, 1, 1, 3, 3, 3, 1, 1, 1, 1, 1, 4, 4, 4, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 10, 10, 1, 1, 3, 3

Offset: 1

Author

Michel Lagneau, Oct 11 2016

Keywords

• nonn

Examples

			a(17) = 3 because gcd(A006666(17), A006667(17)) = gcd(9, 3) = 3.
		

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006666, A006667, A277068.

Programs

Maple

nn:=100:
for n from 1 to nn do:
m:=n:it0:=0:it1:=0:
   for j from 1 to 1000 while(m<>1) do:
    if irem(m,2)=0
     then
     m:=m/2:it0:=it0+1:
     else
     m:=3*m+1:it1:=it1+1:
    fi:
   od:
   q:=gcd(it0,it1):printf(`%d, `,q):
  od:

Mathematica

Table[GCD[Count[NestWhileList[If[OddQ@ #, 3 # + 1, #/2] &, n, # > 1 &], ?EvenQ], Count[Differences[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # > 1 &]], ?Positive]], {n, 87}] (* Michael De Vlieger, Oct 13 2016, after Harvey P. Dale at A006666 and A006667 *)

PARI

a(n) = {my(se = 0); my(so = 0); while (n!=1, if (n % 2, so++; n = 3*n+1, se++; n = n/2);); gcd(se, so);} \\ Michel Marcus, Oct 13 2016

Formula

a(2^m) = m.