A389305 Numbers that can terminate in 3 by successively dividing a contiguous substring of digits by 2 but cannot terminate in 1.

3, 6, 12, 14, 18, 22, 24, 28, 36, 42, 44, 48, 56, 66, 72, 74, 78, 82, 84, 88, 96, 106, 206, 406, 806

Offset

1,1

Comments

From Michael S. Branicky, Sep 29 2025: (Start)

Also, numbers k such that A389140(k) = 3.

There are no further terms than the 25 numbers listed. Proof. First, all odd numbers k > 3 cannot have A389140(k) = 3. In one step, x --> y with x >= y >= x/2 and A389140(x) = 3 iff A389140(y) = 3. Explicit calculation shows there are no terms A389140(k) = 3 for k in 1024..2047, so there can be no higher ones. (End)

Links

Table of n, a(n) for n=1..25.

Example

806 is a term because 806-->406-->206-->106-->56-->28-->14-->12-->6-->3 and 806 cannot terminate in 1.

Crossrefs

Cf. A389140, A386551.

Sequence in context: A062053 A274652 A339552 * A102040 A232485 A232486

Adjacent sequences: A389302 A389303 A389304 * A389306 A389307 A389308

Keyword

nonn,base,fini,full

Author

Ali Sada, Sep 29 2025

Status

approved