Timeline for A lower bound for the largest prime divisor of an integer

12 events

when toggle format	what		by	license	comment
Nov 28 at 4:57	vote	accept	MAY
Nov 28 at 4:55	comment	added	GH from MO		@MAY Square-freeness is used in the equation $\log n=\sum_{p\mid n}\log p$, which is equivalent to $n=\prod_{p\mid n}p$. This equation holds if and only if $n$ is square-free.
Nov 28 at 4:52	comment	added	MAY		One more question, where is the square-freeness used in your estimation?
S Nov 28 at 4:40	history	suggested	mathworker21	CC BY-SA 4.0	I added a comma that makes a big difference.
Nov 28 at 4:33	review	Suggested edits
S Nov 28 at 4:40
Nov 28 at 4:32	vote	accept	MAY
Nov 28 at 4:53
Nov 28 at 4:22	history	edited	GH from MO	CC BY-SA 4.0	deleted 2 characters in body
Nov 28 at 4:16	comment	added	GH from MO		@MAY I don't understand what you mean. I just gave you a very precise lower bound for the largest prime divisor $P$ of $n$, assuming $n$ is square-free: $P>0.99999998\log n$. If $n$ is not square-free, then there is no lower bound apart from the trivial one: $P\geq 2$.
Nov 28 at 4:15	history	edited	GH from MO	CC BY-SA 4.0	added 67 characters in body
Nov 28 at 4:04	comment	added	MAY		Thank you! Do you know of any refinements or corrections to improve the lower bound for the largest prime divisor of $n $?
Nov 28 at 4:02	history	edited	GH from MO	CC BY-SA 4.0	edited body
Nov 28 at 3:55	history	answered	GH from MO	CC BY-SA 4.0