Timeline for Product of exponents of prime factorization

8 events

when toggle format	what		by	license	comment
Jul 30, 2016 at 22:19	comment	added	Will Jagy		@PeterKošinár thank you. I sort of remember how this went; after Felipe's complaint, i do not believe i will do anything to bump this to the front page. It seems to me that we never found $m(6),$ but did find $m(5).$ If you find $m(6)$ and can prove it, that would be worth a new answer.
Jul 30, 2016 at 22:07	comment	added	Peter Košinár		Quick computer search for values $n$ (using the observation that exponents in prime factorization are non-increasing) for which $p(n)=m(6)$ yielded $m(6)\leq 2^{22}3^{11}(5\cdot7\cdot11)^7(13\cdot17\cdot19\cdot23)^5(29\cdot31\cdot37\cdot41)^4\cdot(43\cdot47\cdot53\cdot59\cdot61\cdot67)^3$.
Oct 4, 2013 at 19:21	history	edited	Will Jagy	CC BY-SA 3.0	added 451 characters in body
Oct 4, 2013 at 19:15	history	edited	Will Jagy	CC BY-SA 3.0	added 451 characters in body
Oct 4, 2013 at 4:58	comment	added	Will Jagy		@TheMaskedAvenger, I think I understand; reduce the bound $B$ and you get faster run time. Indeed, after dong that, there is nothing to prevent replacing $B$ by $N$ every time we get some $N < B,$ and perhaps drop the $10.0$
Oct 4, 2013 at 4:34	comment	added	The Masked Avenger		Note that p^4 can sometimes replace p^2 q^2 and for p(6) 89^2 is larger than 2^11, so I think only primes less than 70 are needed.
Oct 3, 2013 at 23:22	comment	added	Joseph O'Rourke		"which is sort of large, granted." :-)
Oct 3, 2013 at 22:15	history	answered	Will Jagy	CC BY-SA 3.0