# Proving $\forall d\in\mathbb{Z_+},\ \min\{n:\lambda^n(10^d)=1\}=d+2,$ where $\lambda()$ is the Carmichael function.

(I'm posting this question in the spirit of this advice and will post an answer if no one else does so.)

[This paper] proves the following:

If $$k, x$$ and $$a_1,a_2,a_3\dots$$ are positive integers, then $$a_1\uparrow a_2\uparrow\dots a_{s}\uparrow x\pmod k$$ is independent of the value of $$x$$, provided $$s\gt h(k).$$ Here $$h(k):=\min\{n:\lambda^n(k)=1\}$$, where $$\lambda()$$ is the Carmichael function.

Thus, if we're considering the last $$d$$ decimal digits of such a tower, the proviso involves $$k=10^d.$$ I've verified by computer that $$h(10^d)=d+2$$ for $$1\le d\le 10^3$$.

Q: Is it correct that $$\ h(10^d)=d+2\$$ for every positive integer $$d$$? How to prove it? For other bases $$b$$ as well, are there correspondingly simple formulas for $$\ h(b^d)?$$

EDIT: Since my answer would essentially coincide with that of @Onir, let me instead just mention here that the same method gives the following explicit formula for $$\lambda^n(10^d), \ \ n,d\in\mathbb{Z_+}:$$

$$\lambda^n(10^d)=\begin{cases} 1&\text{if 1\le d\le n-2}\\ 2&\text{if d=n-1}\\ 2^2\,5^{d-n}&\text{if n\le d\le 2n-1}\\ 2^{d-2n}\,5^{d-n}&\text{if d\ge 2n+2} \end{cases}$$

• for general $b$ it seems a bit hard, for example, if $b=p$ it would require being able to factor $p-1$, and then for each prime $q$ dividing $p-1$ we would also have to factor $q-1$ and so on Commented May 13, 2021 at 0:56
• @Onir Yes, I see. What I had in mind was that there might be a few special bases of interest that might be worth the trouble. Commented May 13, 2021 at 0:58
• I think for bases in which the prime divisors of $b$ are all $2$ or fermat primes we can do something similar Commented May 13, 2021 at 1:00

for $$a\geq 4$$ and $$b\geq 1$$ we have

$$\lambda(2^a 5^b) = \operatorname{lcm}(2^{a-2},4\cdot 5^{b-1}) = 2^{a-2}5^{b-1}$$.

Let $$k = \lceil d/2 \rceil$$.

Then $$\lambda^k(10^d) = 2^25^{d-k}$$

From here each application of $$\lambda$$ just reduces the exponent of $$5$$ by one, so we have:

$$\lambda^d(10^d) = 2^2$$

$$\lambda^{d+1}(10^d) = 2$$

$$\lambda^{d+2}(10^d) = 1$$