Conjecture of multiplicative persistence

Definitions and mathematical statement

Write a positive integer $n$ in decimal notation without leading zeros, $n=\sum_{j=0}^{r-1}d_j10^j$, with $d_j\in\{0,\ldots,9\}$ and $d_{r-1}\ne0$. Define $\Pi(n)=\prod_{j=0}^{r-1}d_j$ and put $\Pi(0)=0$. Every one-digit nonnegative integer is fixed by $\Pi$. Define its usual multiplicative persistence by \[ P(n)=\min\{k\ge0:\Pi^k(n)\in\{0,1,\ldots,9\}\}. \] The minimum exists because the digit product of a multi-digit positive integer is smaller than the integer. The conjecture is \[ P(n)\le11\qquad\text{for every positive integer }n. \] Thus no decimal integer should require twelve successive digit-product operations before reaching a one-digit fixed point.

The catalogue uses the least positive $k$ with $\Pi^k(n)=\Pi^{k+1}(n)$. This equals $\max\{1,P(n)\}$, so it gives exactly the same upper-bound question. This convention matters only for integers that already have one digit.

The base ten is fixed; changing the base produces a different persistence problem. Zeros in a multi-digit integer immediately give product zero, while the order of its digits does not affect its first product. These reductions help organize searches but do not bound all possible input lengths. A finite computation, however extensive, does not prove the universal bound without a theorem reducing arbitrary decimal lengths to the checked range.

Short English statement

Must repeated multiplication of decimal digits reach a one-digit number within eleven steps?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 4700031 (AMR-046-0031), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Gasull - Some open problems in low dimensional dynamical systems (2020), Problem environment 31: Conjecture of multiplicative persistence. Original source indexed by AMR-046-0031. https://arxiv.org/abs/2012.02524.