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I have come across a sequence and am wondering if anyone knows of a closed form expression for it. I am a bit too lazy to figure it out on my own seeing as it is such a minor part of what I am doing. Any help would be greatly appreciated.

The sequence is $c_{jk}$ such that $c_{(j+1)(k+1)} = c_{(j+1)k} - c_{jk}$ and $c_{j0} = 0$ for $j \neq 0$ and $c_{0k} = 1$ for $k\ge 0$

JmsNxn
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    -1: That's motivating, you are too lazy to do it so you want someone here to do it. – copper.hat Nov 25 '13 at 20:20
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    I'm too lazy to post an answer seeing as its such a minor part of this site. – Fly by Night Nov 25 '13 at 20:20
  • Yeah @JmsNxn : that lazy bit in your post isn't the most intelligent thing to say to someone you're asking for help. In fact, I can't think of many dumber things than that one to say when asking for help....hmmm. – DonAntonio Nov 25 '13 at 21:20
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    The initial conditions that you’ve given are insufficient to determine the array. You need to specify $c_{j0}$ for $j>1$, and if you want values of $c_{jk}$ for $j<k$ you also need to specify $c_{0k}$ for $k>0$. – Brian M. Scott Nov 25 '13 at 21:33
  • Sorry, sorry. That makes a lot of sense, I wrote it in a rush. I made the necessary adjustments. – JmsNxn Nov 25 '13 at 21:51

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Making a small table of values gives the game away:

$$\begin{array}{c|rr} j\backslash k&0&1&2&3&4&5&6&7\\ \hline 0&1&1&1&1&1&1&1&1\\ 1&0&-1&-2&-3&-4&-5&-6&-7\\ 2&0&0&1&3&6&10&15&21\\ 3&0&0&0&-1&-4&-10&-20&-35\\ 4&0&0&0&0&1&5&15&35\\ 5&0&0&0&0&0&-1&-6&-21\\ 6&0&0&0&0&0&0&1&7\\ 7&0&0&0&0&0&0&0&-1 \end{array}$$

Clearly $$c_{jk}=(-1)^j\binom{k}j\;.$$

And indeed

$$(-1)^{j+1}\binom{k}{j+1}-(-1)^j\binom{k}j=(-1)^{j+1}\left(\binom{k}{j+1}+\binom{k}j\right)=(-1)^{j+1}\binom{j+1}{k+1}\;,$$

verifying the conjecture.

Brian M. Scott
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