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

12 December 2017 7,643 9 View

Interpretation for q * n * m^r?

I'm looking for examples of naturally-motivated interpretations of the expression q * n * m^r. Obviously, we can apply a word (string) interpretation here, or something similar. Looking for...

01 January 1970 5,237 1 View

Interpretation for q * n * m^(n-1)?

I'm looking for examples of naturally-motivated interpretations of the expression q * n * m^(n-1). Obviously, we can apply a word (string) interpretation here, or something similar. Looking for...

01 January 1970 9,899 0 View