Let sum(k>=0, k^n/(2*k+1)!) = (x(n)*e + y(n)/e)/z(n), where x(n) and z(n) are >0, then a(n)=y(n).

A080094

Let sum(k>=0, k^n/(2*k+1)!) = (x(n)*e + y(n)/e)/z(n), where x(n) and z(n) are >0, then a(n)=y(n).