A033491 a(n) is the smallest integer that takes n halving and tripling steps to reach 1 in the 3x+1 problem.

1, 2, 4, 8, 16, 5, 10, 3, 6, 12, 24, 48, 17, 34, 11, 22, 7, 14, 28, 9, 18, 36, 72, 25, 49, 98, 33, 65, 130, 43, 86, 172, 57, 114, 39, 78, 153, 305, 105, 203, 406, 135, 270, 540, 185, 361, 123, 246, 481, 169, 329, 641, 219, 427, 159, 295, 569, 1138, 379, 758, 283, 505

Offset: 0

Author

Jeff Burch

Keywords

• nonn
• nice
• look

Comments

a(n) is the smallest term in n-th row of A127824. - Reinhard Zumkeller, Nov 29 2012

Interestingly, there are many n such that a(n) = 2*a(n-1). - Dmitry Kamenetsky, Feb 11 2017

a(n) is the position of the first occurrence of n in A006577. - Sean A. Irvine, Jul 07 2020

Links

• T. D. Noe, Table of n, a(n) for n = 0..1924 (from Eric Roosendaal's data) [Roosendaal's table is now complete through 2007 - _N. J. A. Sloane_, Oct 21 2012]
• Eric Roosendaal, 3x+1 Class Records
• Eric Weisstein's World of Mathematics, Collatz Problem
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A126727 (missing numbers).

Cf. A006577, A127824.

Programs

Haskell

a033491 = head . a127824_row  -- Reinhard Zumkeller, Nov 29 2012

Mathematica

f[ n_ ] := Module[ {i = 0, m = n}, While[ m != 1, m = If[ OddQ[ m ], 3m + 1, m/2 ]; i++ ]; i ]; a = Table[ 0, {75} ]; Do[ m = f[ n ]; If[ a[[ m + 1 ]] == 0, a[[ m + 1 ]] = n ], {n, 1, 1250} ]; a
With[{c=Table[Length[NestWhileList[If[OddQ[#],3#+1,#/2]&,n,#!=1&]],{n,2000}]}, Flatten[Table[Position[c,i,1,1],{i,70}]]] (* Harvey P. Dale, Jan 06 2013 *)

PARI

a(n)=if(n<0,0,k=1; while(abs(if(k<0,0,s=k; c=1; while((1-(s%2))*s/2+(s%2)*(3*s+1)>1,s=(1-(s%2))*s/2+(s%2)*(3*s+1); c++); c)-n-1)>0,k++); k)

Python

import numpy
nupto = 62
A033491 = numpy.zeros(nupto, dtype=object)
k, counter = 1, 0
while counter < nupto:
    kk, n = k, 0
    while n <= nupto and kk != 1:
        if kk % 2 == 0:
            kk //= 2
        else:
            kk = (kk*3+1)//2
            n += 1
        n += 1
    if n < nupto and not A033491[n]:
        A033491[n] = k
        counter += 1
    k += 1
print(list(A033491)) # Karl-Heinz Hofmann, Feb 11 2023

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001