A179118 Number of Collatz steps to reach 1 starting with 2^n + 1.

1, 7, 5, 19, 12, 26, 27, 121, 122, 35, 36, 156, 113, 52, 53, 98, 99, 100, 101, 102, 72, 166, 167, 168, 169, 170, 171, 247, 173, 187, 188, 251, 252, 178, 179, 317, 243, 195, 196, 153, 154, 155, 156, 400, 326, 495, 496, 161, 162, 331, 332, 408, 471, 410, 411, 337, 338, 339, 340, 553

Offset: 0

Author

Mitch Harris, Jan 04 2011

Keywords

• nonn
• easy

Comments

There are many long runs of consecutive terms that increase by 1 (see second conjecture in A277109). For n < 40000, the longest run has 1030 terms starting from a(33237) = 244868 and ending with a(34266) = 245897. - Dmitry Kamenetsky, Sep 30 2016

Examples

			a(1)=7 because the trajectory of 2^1+1=3 is (3,10,5,16,8,4,2,1).
		

Links

• Dmitry Kamenetsky, Table of n, a(n) for n = 0..9999 (terms n = 1...1000 from Mitch Harris)
• MathOverflow, Introduced on Mathoverflow by Henrik Rüping

Crossrefs

Cf. A000051, A006577, A070976, A074472, A075486, A193688 (starting with 2^n-1), , A179118, A277109.

Programs

Mathematica

CollatzNext[n_] := If[Mod[n, 2] == 0, n/2, 3 n + 1]; CollatzPath[n_] := CollatzPath[n] = Module[{k = n, l = {}}, While[k != 1, k = CollatzNext[k]; l = Append[l, k]]; l]; Collatz[n_] := Length[CollatzPath[n]]; Table[Collatz[2^n+1],{n,1,50}]
f[n_] := Length@ NestWhileList[If[OddQ@ #, 3 # + 1, #/2] &, 2^n + 1, # > 1 &] - 1; Array[f, 60] (* Robert G. Wilson v, Jan 05 2011 *)
Array[-1 + Length@ NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, 2^# + 1, # > 1 &] &, 60, 0] (* Michael De Vlieger, Nov 25 2018 *)

PARI

nbsteps(n)= s=n; c=0; while(s>1, s=if(s%2, 3*s+1, s/2); c++); c;
a(n) = nbsteps(2^n+1); \\ Michel Marcus, Oct 28 2018

Python

def steps(a):
  if a==1:     return 0
  elif a%2==0: return 1+steps(a//2)
  else:        return 1+steps(a*3+1)
for n in range(60):
  print(n, steps((1<
				

Formula

a(n) = A006577(2^n+1) = A006577(A000051(n)).

a(n) = A075486(n) - 1. - T. D. Noe, Jan 17 2013

Extensions

a(0)=1 prepended by Alois P. Heinz, Dec 12 2018