STATUS
reviewed
approved
reviewed
approved
proposed
reviewed
editing
proposed
a(n) = Sum_{i=0..floor((n-2)/4)} binomial(n,4i+2)*(4i+2)!/(2^(2i+1)*(2i+1)!). - Luis Manuel Rivera Martínez, May 22 2018 [The same as Ralf Stephan's formula above.]
proposed
editing
editing
proposed
a(n) = Sum_{i=0..floor((n-2)/4)} binomial(n,4i+2)*(4i+2)!/(2^(2i+1)*(2i+1)!). - Luis Manuel Rivera Martínez, May 22 2018 [The same as _Ralf Stephan_'s formula above.]
reviewed
editing
proposed
reviewed
editing
proposed
a(n) = Sum_{i=0..floor((n-2)/4)} binomial(n,4i+2)*(4i+2)!/(2^(2i+1)*(2i+1)!). - Luis Manuel Rivera Martínez, May 22 2018
reviewed
editing
proposed
reviewed