A348006 Largest increment in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.

0, 0, 11, 0, 11, 11, 35, 0, 35, 11, 35, 11, 27, 35, 107, 0, 35, 35, 59, 11, 43, 35, 107, 11, 59, 27, 6155, 35, 59, 107, 6155, 0, 67, 35, 107, 35, 75, 59, 203, 11, 6155, 43, 131, 35, 91, 107, 6155, 11, 99, 59, 155, 27, 107, 6155, 6155, 35, 131, 59, 203, 107

Offset: 1

Author

Paolo Xausa, Oct 02 2021

Keywords

• nonn
• easy

Comments

The largest increment occurs when the trajectory reaches its largest value via a 3x+1 step.

All nonzero terms are odd, since they are of the form 2k+1, for some k >= 5.

Examples

			a(3) = 11 because the trajectory starting at 3 is 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, and the largest increment (from 5 to 16) is 11.
a(4) = 0 because there are only halving steps in the Collatz trajectory starting at 4.
		

Links

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

Crossrefs

Cf. A006370, A025586, A070165, A087232.

Programs

Mathematica

nterms=100;Table[c=n;mr=0;While[c>1,If[OddQ[c],mr=Max[mr,2c+1];c=3c+1,c/=2^IntegerExponent[c,2]]];mr,{n,nterms}]

PARI

a(n)=n>>=valuation(n,2); my(r); while(n>1, my(t=2*n+1); n+=t; n>>=valuation(n,2); if(t>r, r=t)); r \\ Charles R Greathouse IV, Oct 25 2022

Python

def A348006(n):
    c, mr = n, 0
    while c > 1:
        if c % 2:
            mr = max(mr, 2*c+1)
            c = 3*c+1
        else:
            c //= 2
    return mr
print([A348006(n) for n in range(1, 100)])

Formula

If n = 2^k (for k >= 0), a(n) = 0; otherwise a(n) = 2*A087232(n)+1 = (2*A025586(n)+1)/3 = A025586(n)-A087232(n).