Alright, an upper bound on $m(6)$ is $$ B = 2^{11} 3^{11} 5^7 7^7 11^7 13^5 17^5 19^5 23^5 29^3 31^3 37^3 41^3 43^3 47^3 53^2 59^2 61^2 67^2 71^2 73^2 79^2 83^2 89^2 $$ which is sort of large, granted.

Next, find the first primorial (product of the consecutive primes beginning with 2) that exceeds $\sqrt B.$ Call the largest prime factor of that primorial $Q = p_r,$ where $p_1=2, p_2=3,$ and so on. I imagine $r < 100$ and maybe $r < 50.$

Finally, run the $r$-tuple loop with nonincreasing exponents on the primes $2,3,\ldots,p_r$ such that each resulting number $N$ given by that prime factorization is less than $B \cdot e^{10}$ by using logarithms, that is $\log N < \log B + 10.0.$

For each such $N$ that satisfies $p(p(p(p(p(p(N)))))) = 1$ but $p(p(p(p(p(N))))) \neq 1,$ print out a line beginning with $\log N$ followed by the $r$-tuple of exponents. Sort. Alternatively, print out nothing, but save $\log N$ and its $r$-tuple in a datatype of some kind, and keep replacing every time a smaller $\log N$ appears. In the end, print out that information. Or, as a sort of hybrid that I like, print out every time $\log N$ decreases. In the beginning, improvements come thick and fast, then slow down as you near the winner. Nice to see some progress reports, you see.

I claim this can actually be done, successfully. Good exercise for certain types of programming class, although the math part may need explaining. Getting a 50 variable multiple loop with certain bounds built in is likely a bit of work...

FRIDAY: At least I was able to find the bound on primes. The big number $B \approx 1.2 \cdot 10^{113},$ and $\sqrt B \approx 3.4 \cdot 10^{56}.$ It suffices to use the first 35 primes in the multiple loop, as the "primorial" $$ P_{35} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots 137 \cdot 139 \cdot 149 \approx 1.5 \cdot 10^{57} $$ is larger than $\sqrt B.$ Oh, I always use logs base $e \approx 2.718287828,$ and $$ \log B \approx 260.37 $$

Alright, an upper bound on $m(6)$ is $$ B = 2^{11} 3^{11} 5^7 7^7 11^7 13^5 17^5 19^5 23^5 29^3 31^3 37^3 41^3 43^3 47^3 53^2 59^2 61^2 67^2 71^2 73^2 79^2 83^2 89^2 $$ which is sort of large, granted.

Next, find the first primorial (product of the consecutive primes beginning with 2) that exceeds $\sqrt B.$ Call the largest prime factor of that primorial $Q = p_r,$ where $p_1=2, p_2=3,$ and so on. I imagine $r < 100$ and maybe $r < 50.$

Finally, run the $r$-tuple loop with nonincreasing exponents on the primes $2,3,\ldots,p_r$ such that each resulting number $N$ given by that prime factorization is less than $B \cdot e^{10}$ by using logarithms, that is $\log N < \log B + 10.0.$

For each such $N$ that satisfies $p(p(p(p(p(p(N)))))) = 1$ but $p(p(p(p(p(N))))) \neq 1,$ print out a line beginning with $\log N$ followed by the $r$-tuple of exponents. Sort. Alternatively, print out nothing, but save $\log N$ and its $r$-tuple in a datatype of some kind, and keep replacing every time a smaller $\log N$ appears. In the end, print out that information. Or, as a sort of hybrid that I like, print out every time $\log N$ decreases. In the beginning, improvements come thick and fast, then slow down as you near the winner. Nice to see some progress reports, you see.

I claim this can actually be done, successfully. Good exercise for certain types of programming class, although the math part may need explaining. Getting a 50 variable multiple loop with certain bounds built in is likely a bit of work...

FRIDAY: At least I was able to find the bound on primes. The big number $B \approx 1.2 \cdot 10^{113},$ and $\sqrt B \approx 3.4 \cdot 10^{56}.$ It suffices to use the first 35 primes in the multiple loop, as the "primorial" $$ P_{35} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots 137 \cdot 139 \cdot 149 \approx 1.5 \cdot 10^{57} $$ is larger than $\sqrt B.$ Oh, I always use logs base $e \approx 2.718287828,$ and $$ \log B \approx 260.37 $$ Given that the loop has 35 variables rather than 100, I can probably do this myself, but certain parts do need to be rewritten in GMP. I also suspect that running time need not be huge, although such predictions sometimes disappear in the face of reality.

Alright, an upper bound on $m(6)$ is $$ B = 2^{11} 3^{11} 5^7 7^7 11^7 13^5 17^5 19^5 23^5 29^3 31^3 37^3 41^3 43^3 47^3 53^2 59^2 61^2 67^2 71^2 73^2 79^2 83^2 89^2 $$ which is sort of large, granted.

Next, find the first primorial (product of the consecutive primes beginning with 2) that exceeds $\sqrt B.$ Call the largest prime factor of that primorial $Q = p_r,$ where $p_1=2, p_2=3,$ and so on. I imagine $r < 100$ and maybe $r < 50.$

Finally, run the $r$-tuple loop with nonincreasing exponents on the primes $2,3,\ldots,p_r$ such that each resulting number $N$ given by that prime factorization is less than $B \cdot e^{10}$ by using logarithms, that is $\log N < \log B + 10.0.$

For each such $N$ that satisfies $p(p(p(p(p(p(N)))))) = 1$ but $p(p(p(p(p(N))))) \neq 1,$ print out a line beginning with $\log N$ followed by the $r$-tuple of exponents. Sort. Alternatively, print out nothing, but save $\log N$ and its $r$-tuple in a datatype of some kind, and keep replacing every time a smaller $\log N$ appears. In the end, print out that information. Or, as a sort of hybrid that I like, print out every time $\log N$ decreases. In the beginning, improvements come thick and fast, then slow down as you near the winner. Nice to see some progress reports, you see.

I claim this can actually be done, successfully. Good exercise for certain types of programming class, although the math part may need explaining. Getting a 50 variable multiple loop with certain bounds built in is likely a bit of work...

Alright, an upper bound on $m(6)$ is $$ B = 2^{11} 3^{11} 5^7 7^7 11^7 13^5 17^5 19^5 23^5 29^3 31^3 37^3 41^3 43^3 47^3 53^2 59^2 61^2 67^2 71^2 73^2 79^2 83^2 89^2 $$ which is sort of large, granted.

Next, find the first primorial (product of the consecutive primes beginning with 2) that exceeds $\sqrt B.$ Call the largest prime factor of that primorial $Q = p_r,$ where $p_1=2, p_2=3,$ and so on. I imagine $r < 100$ and maybe $r < 50.$

Finally, run the $r$-tuple loop with nonincreasing exponents on the primes $2,3,\ldots,p_r$ such that each resulting number $N$ given by that prime factorization is less than $B \cdot e^{10}$ by using logarithms, that is $\log N < \log B + 10.0.$

For each such $N$ that satisfies $p(p(p(p(p(p(N)))))) = 1$ but $p(p(p(p(p(N))))) \neq 1,$ print out a line beginning with $\log N$ followed by the $r$-tuple of exponents. Sort. Alternatively, print out nothing, but save $\log N$ and its $r$-tuple in a datatype of some kind, and keep replacing every time a smaller $\log N$ appears. In the end, print out that information. Or, as a sort of hybrid that I like, print out every time $\log N$ decreases. In the beginning, improvements come thick and fast, then slow down as you near the winner. Nice to see some progress reports, you see.

I claim this can actually be done, successfully. Good exercise for certain types of programming class, although the math part may need explaining. Getting a 50 variable multiple loop with certain bounds built in is likely a bit of work...

FRIDAY: At least I was able to find the bound on primes. The big number $B \approx 1.2 \cdot 10^{113},$ and $\sqrt B \approx 3.4 \cdot 10^{56}.$ It suffices to use the first 35 primes in the multiple loop, as the "primorial" $$ P_{35} = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots 137 \cdot 139 \cdot 149 \approx 1.5 \cdot 10^{57} $$ is larger than $\sqrt B.$ Oh, I always use logs base $e \approx 2.718287828,$ and $$ \log B \approx 260.37 $$