$f(1)=1$ Let $f(2)=a,f(4)=b$ $m=k=2: f(3)=2a-1$ $m=5,k=2: f(5)=f(4)+f(3)-f(2)=a+b-1$ $m=k=5: f(7)=2f(5)-f(3)=2b-1$ $m=k=7: f(11)=2f(7)-f(3)=4b-2a-1$ $m=11,k=2: f(10)=f(11)+f(2)-f(3)=4b-3a=f(2)f(5)=a(a+b-1)$ $m=11,k=5: f(13)=f(11)+f(5)-f(3)=5b-3a-1$ $m=13,k=2: f(12)=f(13)+f(2)-f(3)=5b-4a=f(3)f(4)=b(2a-1)$ So we have $4b-3a=a(a+b-1)$ and $b(2a-1)=5b-4a$ $b=\frac{2a}{3-a}$ $a^2-3a+2=0$ so $(a,b)=(1,1),(2,4)$ and $f(11)=1$ or $11$, both values are possible for functions $f(n)=1$ or $f(n)=n$