
sigma reccurence
f:N>N
f(0) = 1
$\displaystyle 1+\sum_{j=0}^{n1} jf(j)$, n>=1
i.e.
1 + ((0*f*0) + (1*f*1) + (2 * f *2) ....(n1)*f*(n1)) < I'm sure this isn't right
then f(4) =
1+ ((0*4*0) + (1*4*1) + (2 * 4 *2) + (3 * 4 *3))
= 1 + 0 + 4 + 16 + 36
= 55
Have I worked this correctly?
Thanks for your time.

No doesn't appear correct
How about this recurrence based on your defintion
f(n) = n.f(n1)

Am I correct in assuming f stands for function, f is not a numeric variable. only the n inside the function(n1) is substituted for a number.
f(n) = n.f(n1) <  ok, I see, that makes sense. The the final value goes up to n(n1)
so, how about
1 + ( 1(f(n1)) + (2(f(n1)) ....(n(f(n1)))
then f(4) =
1+ 1(11) + 2(21) + 3(31) + 4(41)
= 1+ 0 + 2 + 6 + 12
= 21