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Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example, $$p(5184) = p(2^6 3^4) = 24 \;,$$ $$p(65536) = p(2^{16}) = 16 \;.$$ Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example, $P(5184) = 3$ because $$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$ and $P(65536)=4$ because $$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$ Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program. It was previously posed on MSE.)

It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.

---

Update. Will Jagy showed that almost certainly $m(5) = 2^9 3^6 5^4 7^3 11^2 =9681819840000 \approx 10^{13}$. As it seems that an explicit expression for $m(k)$ is not in the offing, I will accept his resolution of Q2 and leave Q1 open.

Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example, $$p(5184) = p(2^6 3^4) = 24 \;,$$ $$p(65536) = p(2^{16}) = 16 \;.$$ Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example, $P(5184) = 3$ because $$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$ and $P(65536)=4$ because $$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$ Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program. It was previously posed on MSE.)

It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.

---

Update. Will Jagy showed that almost certainly $m(5) = 2^9 3^6 5^4 7^3 11^2 =9681819840000 \approx 10^{13}$. As it seems that an explicit expression for $m(k)$ is not in the offing, I will accept his resolution of Q2 and leave Q1 open.

Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example, $$p(5184) = p(2^6 3^4) = 24 \;,$$ $$p(65536) = p(2^{16}) = 16 \;.$$ Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example, $P(5184) = 3$ because $$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$ and $P(65536)=4$ because $$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$ Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program. It was previously posed on MSE.)

It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.

---

Update. Will Jagy showed that almost certainly $m(5) = 2^9 3^6 5^4 7^3 11^2 \approx 10^{13}$. As it seems that an explicit expression for $m(k)$ is not in the offing, I will accept his resolution of Q2 and leave Q1 open.

Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example, $$p(5184) = p(2^6 3^4) = 24 \;,$$ $$p(65536) = p(2^{16}) = 16 \;.$$ Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example, $P(5184) = 3$ because $$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$ and $P(65536)=4$ because $$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$ Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program. It was previously posed on MSE.)

It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.

---

Update. Will Jagy showed that almost certainly $m(5) = 2^9 3^6 5^4 7^3 11^2 =9681819840000 \approx 10^{13}$. As it seems that an explicit expression for $m(k)$ is not in the offing, I will accept his resolution of Q2 and leave Q1 open.

Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example, $$p(5184) = p(2^6 3^4) = 24 \;,$$ $$p(65536) = p(2^{16}) = 16 \;.$$ Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example, $P(5184) = 3$ because $$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$ and $P(65536)=4$ because $$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$ Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program. It was previously posed on MSE.)

It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.

Let $p(n)$ be the product of the exponents of the prime factorization of $n$. For example, $$p(5184) = p(2^6 3^4) = 24 \;,$$ $$p(65536) = p(2^{16}) = 16 \;.$$ Define $P(n)$ as the number of iterations of $p(\;)$ to reduce $n$ to $1$. For example, $P(5184) = 3$ because $$p(5184)=24, \;p(24) = p(2^3 3^1) = 3, \;p(3)=1 \;;$$ and $P(65536)=4$ because $$p(65536) = 16, \;p(16)=p(2^4)=4, \;p(4)=p(2^2)=2, \; p(2)=1 \;.$$ Finally, define $m(k)$ to be the minimum value of $n$ such that $P(n) = k$.

Q1. What is $m(k)$?

(I ask this question out of curiosity, not because it is part of a research program. It was previously posed on MSE.)

It is easy to see that $m(1)=2$, $m(2)=4$, and $m(3)=16$, the latter because $16=2^{2^2}$. But, thanks to Calvin Lin's insight, $m(4)$ is not a power of $2$, but instead is $m(4)=1296= 2^4 3^4$. I do not know the value of $m(5)$.

Q2. More specifically: What is $m(5)$?

I do know that $m(5) > 2 \times 10^8$.

---

Update. Will Jagy showed that almost certainly $m(5) = 2^9 3^6 5^4 7^3 11^2 \approx 10^{13}$. As it seems that an explicit expression for $m(k)$ is not in the offing, I will accept his resolution of Q2 and leave Q1 open.