Dobinski Formula

Author

Eric W. Weisstein
January 17, 2009

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Combinatorics/DobinskisFormula.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/DobinskisFormula.html.

©2009 Wolfram Research, Inc. except for portions noted otherwise

Sums

exp(-x) (∑_(k=0)^∞)(((k^n) (x^k))/(k!))==(BellB2[n, x])

Exp[-x]Sum[((k^n)/(k!))(x^k),{k,0,∞}]

(1/e)(∑_(k=0)^∞)((k^n)/(k!))==(BellB[n])

(1/E)Sum[((k^n)/(k!)),{k,0,∞}]

(∑_k^n)((StirlingS2[n, k])/((m-k)!))==((m^n)/(m!))

Sum[StirlingS2[n,k](1/((m-k)!)),{k,n}]

FullSimplify[Table[Sum[StirlingS2[n,k](1/((m-k)!)),{k,n}],{n,5}],m∈Integers&&m>1]

Table[((m^n)/(m!)),{n,5}]//FullSimplify

(∑_k^n)(λ^k) (StirlingS1[n, k])==(BellB2[n, λ])

Sum[StirlingS1[n,k](λ^k),{k,n}]Sum[((λ^j)/(j!)),{j,0,∞}]

Sum[StirlingS1[n,k]x^k,{k,n}]

Simplify[Table[Sum[x^k*StirlingS2[n, k], {k, n}] - BellB[n, x], 
   {n, 5}]]

exp(-λ) (∑_m^∞)(((m^n) (λ^m))/(m!))==(BellB2[n, λ])

Exp[-λ]Sum[((m^n)/(m!))(λ^m),{m,∞}]