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

1,3

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.

Links

Table of n, a(n) for n=1..60.

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

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).

Example

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.

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}]

Prog

(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)])
(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

Crossrefs

Cf. A006370, A025586, A070165, A087232.

Sequence in context: A342123 A187553 A003621 * A338827 A055963 A127805

Adjacent sequences: A348003 A348004 A348005 * A348007 A348008 A348009

Keyword

nonn,easy

Author

Paolo Xausa, Oct 02 2021

Status

approved