On Even Fibonacci Numbers

Beginning with the integer s[0] = 1, construct a sequence of integers 
such that the nth term s[n] is the sum of  (k+1)*s[n-k]  for k=1 to 
n-1.  Thus we have

  n                                                   s[n]
 ---                                                  -----
  0      1                                              1
  1      1*s[0]                                         1
  2      1*s[1] + 2*s[0]                                3
  3      1*s[2] + 2*s[1] + 3*s[0]                       8
  4      1*s[3] + 2*s[2] + 3*s[1] + 4*s[0]             21
  5      1*s[4] + 2*s[3] + 3*s[2] + 4*s[1] + 5*s[0]    55

Interestingly, s[n] turns out to be just the Fibonacci number F[2n].  
(I've taken the liberty of setting F[0]=1 and F[1]=0, reversing the 
customary order of the initial values.)  So we have the identity

                           n
                F[2n] =  SUM  k F[2(n-k)]
                          k=1