@MartinSleziak I waited a bit to see whether somebody will respond in this room. Now I voted to close as a duplicate. (I will see from the results of the review whether other users agree with my assessment from the result of close vote review. I will then take a closer look at the posts once again and decide whether to flag for merging or not.)

@MartinSleziak The questions were closed as duplicates. I have flagged for merging - we will see whether the moderators will consider merging suitable or not.

Do you suggest reopening for some reason?

I think this should be reopened because gamma function is way to different

@Ramanujan And what about the other duplicate which was suggested there: Is there a way to solve for an unknown in a factorial?

Gamma function is for complex ( negative or fractional) and factorial is just for whole number

So you suggest reopening that question?

Yes

Or do you suggest just changing the duplicate target to this one: Is there a way to solve for an unknown in a factorial?

Which is basically the same as reopening it and closing it as a duplicate, but of the other question (not the one about inverse of $\Gamma(z)$.)

@DanielFischer Sorry for bothering you again - you have the bad luck that you are the only moderator pingable in this room.

What do you think about Ramanujan's suggestion - which if I understand correctly is reopening Is there a way to reverse factorials? and then closing is as a duplicate of Is there a way to solve for an unknown in a factorial?.

Let me take a look.

He has a point that it might be more suitable duplicate target than Is there an Inverse Gamma $\Gamma^{-1} (z) $ function ?. You may notice that there are several other posts which are closed as duplicate of the one linked by @Ramanujan: math.stackexchange.com/questions/linked/171882

@DanielFischer Of course, if you prefer that this should be better decided by the community rather than by a moderator's action, Ramanujan might make a post about this in the reopen request thread.

@MartinSleziak No, I'll change the dupe-target, this one is indeed much better.

@DanielFischer Thanks for that! There are many posts linked there. But personally I do not have time or energy at the moment to check whether some of them seem as a suitable candidates for merging.

I'll take a look at at least some of them.

@Ramanujan As you can see, the post you linked is now closed as a duplicate of a different question, not the one about Gamma function.

@MartinSleziak OK

When was this sort introduced? :P

@Ramanujan Linked questions were introduced in 2010: stackoverflow.blog/2010/04/new-linked-posts

Although probably not exactly in this format.

Judging by this post, this originally worked using something like math.stackexchange.com/search?q=url:%22http://… and it was only later changed to the current format math.stackexchange.com/questions/linked/171882

I did not find exactly when. (And it is probably not that important when it was chenged.)

@Ramanujan If you go to any question which has many linked question and you click on "see more linked questions", you can see that you get url with this format.

The link is shown in the sidebar only if there are many linked questions, but the same format works for other questions, too.

I see

So is it shown only to me?

And different questions for others?

@Ramanujan No, if anybody goes to the url math.stackexchange.com/questions/linked/30732 they see the same questions. The same is true for math.stackexchange.com/questions/linked/171882 which I linked above.

The only difference is that the first link is shown in the sidebar (as the "see more linked questions" link). To get the second one, I had to know that format is http://math.stackexchange.com/questions/linked/id and modify the link manually by putting there the question id.

OK,thanks for sharing information

@MartinSleziak The questions are merged now. Thanks to Daniel Fischer.

duplicates I voted to close Sum of binomial coefficients when lower suffices is same in the series: ${m \choose m}+{m+1 \choose m}+{m+2 \choose m}+...+{n \choose m}$ as a duplicate of Proof of the Hockey-Stick Identity: $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$.

It seems that the users in the review quote disagree with me. Is there some substantial difference between the two post that I am missing?