One way to approach this is to transform your difference equation in the single variable y to a system of two difference equations, which you can simply march forward in time.
So neglecting, for the moment the special condition when n = mk, you could define y1(n) = y(n) and y2(n) = y1(n-1), noting also that a(n), b(n) and c(n) are in fact not functions of n, but simple constants, and substituting for h(u) would give the system of equations
y1(n+1) = a*y1(n) - b*y2(n) - c*y2(n)^3
y2(n+1) = y1(n)
Assuming you have initial conditions for y you could then just march forward in a loop
a = -3/2
b = 1
c = 1/4
y1(1)= y1ic
y2(1)= y2ic
nFinal = 11
preallocate
y1 = zeros(nFinal+1,1)
y2 = zeros(nFinal+1,1)
for n = 1:nFinal
y1(n+1) = a*y1(n) - b*y2(n) - c*y2(n)^3
y2(n+1) = y1(n)
end
plot(y1)
You will have to further modify your code to handle the special condition where n = m*k. I actually can't understand the details of what your are doing there from the description, for example I don't see where I_k is defined.
Hopefully this gives you a start on how to approach this.
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