Iron-Fisted Recurrence-Summation

\[\begin{eqnarray} f(0,k) &=& \begin{cases} 1 , k = 0 , 1 \\ 0 , \text{otherwise} \end{cases} \\ f(n,k) &=& f(n-1,k) + f(n-1,k-2n), \qquad n = 1,2,3,\ldots \end{eqnarray}\]

A function \(f:\mathbb{N}_0\times \mathbb{Z}\mapsto\mathbb{Z}\) satisfies the conditions above.

If \( \large\displaystyle \sum_{k=0}^{\binom{2009}2} f(2008,k) \) can be expressed as \(a^b\), where \(a\) and \(b\) are positive integers with \(a\) prime, find \(b-a\).

\[\]Notation: \( \dbinom MN \) denotes the binomial coefficient, \( \dbinom MN = \dfrac{M!}{N!(M-N)!} \).

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