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Stirling number
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==As inverse matrices== The Stirling numbers of the first and second kinds can be considered inverses of one another: :<math>\sum_{j=k}^n s(n,j) S(j,k) = \sum_{j=k}^n (-1)^{n-j} \biggl[{n \atop j}\biggr] \biggl\{{\!j\! \atop \!k\!}\biggr\} = \delta_{n,k}</math> and :<math>\sum_{j=k}^n S(n,j) s(j,k) = \sum_{j=k}^n (-1)^{j-k} \biggl\{{\!n\! \atop \!j\!}\biggr\} \biggl[{j \atop k}\biggr]= \delta_{n,k},</math> where <math>\delta_{nk}</math> is the [[Kronecker delta]]. These two relationships may be understood to be matrix inverse relationships. That is, let ''s'' be the [[lower triangular matrix]] of Stirling numbers of the first kind, whose matrix elements <math>s_{nk}=s(n,k).\,</math> The [[matrix inverse|inverse]] of this matrix is ''S'', the [[lower triangular matrix]] of Stirling numbers of the second kind, whose entries are <math>S_{nk}=S(n,k).</math> Symbolically, this is written :<math>s^{-1} = S\,</math> Although ''s'' and ''S'' are infinite, so calculating a product entry involves an infinite sum, the matrix multiplications work because these matrices are lower triangular, so only a finite number of terms in the sum are nonzero.
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