B(n+2) - 2*B(n+1) = Sum[k*S(n+1,k+1),{k,0,n)]; B(n) --> nth Bell number, S(n,k) --> Stirling number of the second kind?This appears to be true by computation, but I am not aware of either an algebraic or combinatorial proof. It appears possible to derive a similar identity -- B(n+1) - B(n) = Sum[(k+1)*S(n,k+1),{k,0,n}] -- from S(n,k) = k*S(n-1,k) + S(n-1,k-1), but it's unclear to me how to get to B(n+2) - 2*B(n+1) = Sum[k*S(n+1,k+1),{k,0,n)]. Moreover, is there a pattern here that is known, or, if unknown, would be worth investigating further?StirlingAlgebraComputationShare Facebook Twitter LinkedIn Reddit

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