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I had this problem in my discrete math/modular arithmatic course where I had to find the first 10 terms of a series F(r), starting from F(3).

The given information is:

F(3)=1
F(4)=13
F(10) % (10^9+7) is 719666144
F(r) is defined and exists for all values of r>=3

Is it possible to solve such problems? How do we approach these? Is there anyway we can actually find the general term?

EDIT: One of my friends claimed he solved the complete problem fron this much data. I just wanted to check If there is someone that bright actually present or that he was just bragging.

The additional information is:

We define an onto function from $[n] \times [n]$ to $[n-2] \cup \{0\}$ as follows, where $[n] = \{1,2,3,\ldots ,n\}$,

$$f : [n] \times [n] \rightarrow [n-2] \cup \{0\}.$$

1) $f(x,x) = 0$.

2) $f(x,y) = f(y,x) > 0$, for $y ≠ x$.

3) $f(x,y) \leq \max\{f(x,z),f(z,y)\}$ for all $x,y,z$ belonging to $[n]$.

F(r) is the number of ways in which f(x,y) can be defined for n=r.

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2  
What's stopping you from setting $F(r) = 0$, $r \ne 3,4,10$? There have to be more conditions? –  Cocopuffs Jun 12 '13 at 19:09
    
We just saw this one here where it was closed as not a real question. –  Ross Millikan Jun 12 '13 at 19:12
5  
Based on my mind-reading skills, I believe this might be the actual problem. –  Peter Košinár Jun 12 '13 at 19:14
    
@PeterKošinár Great find! –  Cocopuffs Jun 12 '13 at 19:16
1  
I request the administrator to kindly block/delete this question as this is from a contest on a programming site which is still going on.codechef.com/JUNE13/problems/SPMATRIX here is the link.Thankyou. –  swapedoc Jun 12 '13 at 21:07
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1 Answer

Many here will tell you (correctly) that there are infinitely many choices. Usually, if you are given finitely many terms, you are expected to find the "simplest" rule that results in those terms. Nothing such stands out for these, which is why the other one was closed.

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