A365260 Number of steps for n to stop, according to the "multiply with zero" rules explained in A365994.

2, 6, 2, 8, 8, 3, 2, 4, 2, 4, 4, 2, 2, 3, 5, 8, 7, 5, 2, 6, 2, 9, 3, 5, 2, 3, 2, 11, 4, 7, 2, 6, 2, 11, 4, 2, 4, 5, 2, 10, 3, 2, 12, 5, 2, 5, 4, 6, 2, 5, 2, 4, 3, 9, 6, 3, 3, 5, 5, 16, 2, 13, 2, 2, 10, 7, 5, 7, 2, 4, 3, 2, 5, 5, 11, 16, 7, 10, 4, 8, 2, 4, 7, 3, 10, 4, 2, 11, 3, 3, 2, 3, 2, 4, 3, 5, 2, 5, 5, 7

Offset

1,1

Comments

a(n) = 60 has the longest trajectory (16 steps) for n < 300.

Links

Table of n, a(n) for n=1..100.

Eric Angelini, Multiply with zero, personal blog "Cinquante signes" Sept. 2023.

Example

The trajectory of 1 is "1, 101, stop", 2 steps;
The trajectory of 2 is "2, 102, 1706, 20853, 210993, 3981053, stop", 6 steps;
The trajectory of 3 is "3, 103, stop", 2 steps;
The trajectory of 4 is "4, 104, 1308, 20654, 230898, 2654087, 35907393, 1196913103, stop", 8 steps; etc.

Mathematica

Table[Length@Most@NestWhileList[(d=Divisors@#;
Min[Select[FromDigits@Flatten[IntegerDigits/@Riffle[#, 0]]&/@Table[{d[[k]], d[[Length@d-k+1]]}, {k, Length@d}], Count[IntegerDigits@#, 0]<2&]])&, k, IntegerQ], {k, 100}]

Prog

(Python)
from sympy import divisors
def a(n):
    k, steps = n, 1
    while True:
        s = set()
        for d in divisors(k, generator=True):
            v, w = str(d), str(k//d)
            if "0" not in v and "0" not in w:
                s.add(int(v + "0" + w))
        if len(s) == 0: return steps
        k, steps = min(s), steps + 1
print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Sep 26 2023

Crossrefs

Cf. A365993, A365994, A366059, A366060.

Sequence in context: A091944 A069936 A043294 * A368522 A033641 A153190

Adjacent sequences: A365257 A365258 A365259 * A365261 A365262 A365263

Keyword

base,nonn

Author

Giorgos Kalogeropoulos and Eric Angelini, Sep 25 2023

Status

approved