# n-th self discrete convolution

Lets define discrete $f_N(i) = 1,\space i = 1...N$

I need to find $G_N^m = \underbrace {f_N * f_N * ... * f_N}_{m}$

For example $G_6^3$ have value (1,3,6,10,15,21,25,27,27,25,21,15,10,6,3,1) , which is palyndromic and have $\binom{i}{2}$ as first 5 values.

So it seems to be common rule, every $G_N^m$ is palyndromic, have $mN - m + 1$ values total and have $\binom {i} {m-1}$ as first $N-1$ values.

But i wonder if there general formula for $G_N^m(i)$ ?

Additional information about discrete convolution: values of $G_N^m$ are coefficients of $(\frac{x^N - 1}{x -1})^m$ polynomial

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Privet! What $*$ operations means? I guess it regular multiplication, \cdot - $\cdot$? –  Salech Alhasov Mar 5 '12 at 1:02
@SalechAlhasov No it's discrete convolution goo.gl/I8bJ2 –  Odomontois Mar 5 '12 at 1:17
Interesting! Thanks! –  Salech Alhasov Mar 5 '12 at 1:24
These numbers are OEIS A109439; the entry gives no formula for them other than the generating function, which you already know, so I suspect that no nicer formula is known. –  Brian M. Scott Mar 5 '12 at 9:57