A281938 a(n) is the least k such that gcd(A006666(k), A006667(k)) = n.

2, 4, 8, 16, 32, 64, 128, 256, 512, 82, 129, 4096, 327, 16384, 32768, 1249, 35655, 159, 4926, 283, 377, 502, 603, 799, 1063, 1417, 1889, 2518, 3356, 4472, 5960, 7944, 10594, 14124, 18833, 25110, 33481, 44641, 59521, 79361, 105814, 141084, 188113, 250817, 334422

Offset: 1

Author

Michel Lagneau, Feb 02 2017

Keywords

• nonn

Comments

A006666: Number of halving steps to reach 1 in '3x+1' problem.

A006667: number of tripling steps to reach 1 in '3x+1' problem.

a(n) = 2^n for n = 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15.

The primes in the sequence are 2, 283, 1063, 1249, 1889, 44641, ...

Examples

			a(10) = 82 because gcd(A006666(82), A006667(82)) = gcd(70, 40) = 10, and there is no k < 82 such that gcd(A006666(k), A006667(k)) = 10.
		

Links

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

Crossrefs

Cf. A006577, A006666, A006667.

Programs

Maple

for n from 1 to 45 do:
ii:=0:
for k from 2 to 10^7 while(ii=0) do:
  m:=k:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
    if gcd(s1,s2)=n
     then
     ii:=1:printf(`%d %d \n`,n,k):
     else
    fi:
od:
od:

Mathematica

Function[w, First /@ Lookup[w, Function[k, If[k == {}, #, Take[#, First@ k]]]@ Complement[Range@ Max@ #, #]] &@ Keys@ w]@ KeySort@ PositionIndex@ Table[GCD[Count[NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &], ?(EvenQ[#] &)], Count[Differences[NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]], ?Positive]], {n, 2^16}] (* Michael De Vlieger, Feb 02 2017, Version 10, after Harvey P. Dale at A006666 and A006667 *)