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What you have done is not induction, but rather a proof by contradiction (which does the job nicely by the way). However, there is still the issue of your formula's deduction which is what I'm interested in.

After reading the psi20's formula, I was reminded of something. It is a well known fact that gcd(m,n) = am + bn for some integers a and b (where m and n are both non-negative integers). Given a number of the form pm + qn, p and q being both non-negative integers, the "next" number can be constructed by adding gcd(m,n), i.e. pm + qn + gcd(m,n) = pm + qn + am + bn = (p + a)m + (q + b)n. As long as p + a and q + b are non-negative integers, then all is well. The first "trick" is in finding integers a and b closest to 0 satisfying gcd(m,n) = am + bn. Let m = 8 and n = 3. gcd(8,3) = 1 = (-1)8 + (3)3, so a = -1 and b = 3. The second "trick" is to use the following formula: (|b| - 1)n + |a|m = (2)3 + (1)8 = 14, to obtain the lowest value (or price) sought. I check this for a couple of cases and it seems to work (maybe somebody can prove me wrong). I don't why/how it works. Maybe someone can shed a light on this.

Demosaicing (as I understand it) extrapolates information from an image to contruct a new image with more information (particularly the RGB info. of each pixel). Hence, the process is not destructive, so you should (in theory) be able to contruct the original picture. Look at the demosaicing algorithms and determine whether they are reversible.

I'm not sure what you are asking exactly. Do you want to obtain the original non-demosaiced image? If demosaicing is a reversable process, then I guess it can be done. And if it can be done, I'm pretty sure someone has already made software to do so. Google around. By the way, why is this in the applied maths. forum?

Very good counter-example. I retract what I wrote earlier. Here is another counter-example: z + 1 = 5(x + 10) + 7(y - 7). I don't see any method of solving your problem other than by brute force (i.e. plugging, chugging, and looking).

Aeschylus' method is all you need. z + 1 = n(x + p) + m(y - q) implies that 1 = mp - nq. This means either mp is odd and np is even, or mp is even and np is odd, which further implies that either m is odd, or n is odd. The example with m = 4 and n = 10 won't work because they are both even.

Judging by your response, it seems you are not familiar with induction and inductive proofs. I made an error in the argument of the inductive step, which I rectified. Completing the inductive step completes the proof. I now need someone to proofread my argument in the inductive step so I can have some closure on the solution to the problem.

I guess you are referring to the last two summations. Note that [math]\sum_{k=0}^{n+1} ns_{n,k} = \sum_{k=0}^{n} ns_{n,k} + ns_{n,n+1}[/math] and since s[n,n + 1] = 0 (Every table must have at least one person, but with n + 1 tables, this is impossible so the answer is zero. In general s[n,k] = 0 if k > n) the summation reduces to n!n. The summation [math]\sum_{k=0}^{n+1} s_{n,k-1}[/math] is (without using sigma notation) s[n,-1] + s[n,0] + s[n,1] + ... + s[n,n]. The expression s[n,-1] is pure nonsense (i.e. -1 tables), so I effectively made it zero. Hence the resulting summation. I must admit the proof isn't rigorous. I should prove the recurrence relation first before proving the problem in question, but for my purposes this is enough.

I figured out the problem. Back to this question: If a new person with shirt number x = n + 1 comes along, in how many ways can I sit him given a particular seating arrangment? Given a particular arrangement a[1], a[2], a[3], ..., a[n] in s[n,k], I can add x behind a[1] to get a new arrangement, or behind a[2], or ..., or behind a[n] (and this is the same as adding x in front of a[1] since the tables are circular). This gives ns[n,k] combinations. However, I can also sit x on his/her own table (which is not covered in the previous combinations) and have the rest sit in the k - 1 remaining tables, which can be done in s[n,k - 1] ways. So, the answer to the question is ns[n,k] + s[n,k - 1]. In other words, s[n + 1,k] = ns[n,k] + s[n,k - 1]. This means [math]\sum_{k=0}^{n+1} s_{n+1,k} = \sum_{k=0}^{n+1} (ns_{n,k} + s_{n,k - 1})[/math] Noting that [math]\sum_{k=0}^{n+1} ns_{n,k} = \sum_{k=0}^{n} ns_{n,k} = n!n[/math] since s[n,n + 1] = 0 and [math]\sum_{k=0}^{n+1} s_{n,k - 1} = \sum_{k=0}^{n} s_{n,k} = n![/math] since s[n,-1] makes no sense (so it is effectively 0). Hence the expression simplifies to n!n + n! = (n + 1)n! = (n + 1)!. I think is correct. What do you think? [edit]I made an error with the last summation (I had k - 1 instead of k), but now it's correct.[/edit]

I found an error. The equation s[n + 1,k] = (n + 1)s[n,k] fails for n = 1. Actually, it fails for all n > 0. I knew it could be this easy. Back to the drawing board.

I think I've figured it out. Here is the outline Basis Step (n = 1). Trivial. Inductive Step. Assume [math]\sum_{k=0}^n s_{n,k} = n![/math] is true. Again suppose each of the n persons is wearing a t-shirt with a number on it between 1 and n where no two persons have the same number. Every particular seating arrangement in s[n,k] can be written as some number string a[1], a[2], a[3], ..., a[n] where a is some number in {1, ..., n} and a = a[j] only if i = j. If a new person with shirt number x = n + 1 comes along, in how many ways can I sit him given a particular seating arrangment? The answer is n + 1 (I can put x in front of a[1] to get a new seating arrangement, or in front of a[2], or ..., or in front of a[n], or behind a[n]). Therefore s[n + 1,k] = (n + 1)s[n,k]. Hence [math]\sum_{k=0}^{n+1} s_{n+1,k} = (n+1)\sum_{k=0}^n s_{n,k} = (n+1)![/math] Hence the desired result. I don't see any flaw in this proof, but who knows. I will be glad when someone gives me a "thumbs up" on this.

Check out this link: http://mathworld.wolfram.com/StirlingNumberoftheFirstKind.html I'm glad to see someone else interested in this. There is another problem I'm working on right now with no progress: Prove that [math]\sum_{k=0}^n s_{n,k} = n![/math] I'm trying to use induction with this one, but I can't complete the proof.

It has been a while, but now I've been given enough hints to solve this problem. Suppose each of the n persons is wearing a t-shirt with a number on it between 1 and n where no two persons have the same number. If I ask the person with t-shirt number n (hereby referenced as dave) to sit in one table and the rest to sit at the other table, how many seating arrangements are there? Simple: s[n - 1,1] = (n - 2)! = (n - 1)!/(n - 1). Say I want another person to sit with dave. I can pick from n - 1 persons to sit with dave. The rest of the n - 2 persons will sit at the other table and can do so in s[n - 2,1] = (n - 3)! ways. The total combinations is therefore (n - 1)(n - 3)! = (n - 1)!/(n - 2). Say I want two persons sitting with dave. I can choose two persons in (n - 1)(n - 2) ways. The rest of the n - 3 people can sit at the other table in s[n - 3,1] = (n - 4)! ways. The total combinations is therefore (n - 1)(n - 2)(n - 4)! = (n - 1)!/(n - 3). . . . Say I want n - 2 persons to sit with dave. Choosing n - 2 person can be done in (n - 1)! ways. The other person can sit at the other table in only one way. The total combinations is therefore (n - 1)!. Adding all the above combinations gives the desired equation. So deviously simple...

Cryptology is just applied number theory. You'll also need a lot of discrete maths. I suppose.

Two things: 1. Make sure port forwarding is turned on in your router for the port on the computer running the ftp server. 2. Make sure the firewall is not blocking incomming ftp requests from outside your network.

Interesting question. I've never seen a physical theory that does not make use of time so...

Anybody out there running a cluster in their home?

Not to be petulant or anything, but I don't think (judging from the responses) anybody will satisfy your requirements Doron Shadmi. You say "No theory exist without us so if we can understand them we have to do the best we can in order to use their insights to help us to survive...". How does the Pythagorean Theorem, say, help me find food to eat? It doesn't. How does morality help me find food to eat? It doesn't either. I don't see where this is going.

Time is a concept, an idea, a metaphysical thing, like 5164 said. It only exists in your mind. Might as well ask if space exists.

Figure what out?

What about ZF? There are other set theories which have included-middle axioms, particularly those based on intuitionistic set theory. I know of a set theory called IKP which maybe just what you're looking for. The journal of symoblic logic is a good source for these sort of things. Have you ever tried publishing your ideas formally?

Emptiness and fullness are intricately tied to one another, i.e. you can't define one without knowledge of the other and vice versa. It's like trying to define "true" without "false" and vice versa.

I've been told that the most lucrative field in mathematics is statistics. If your concentration is in statistics, then you're golden (if you don't believe me, ask your local statistician). Another less lucrative field (but still good) is operations research.