MathLabs

Problem 6

The function f(x,y)f(x,y) satisfies f(0,y)=y+1,f(0,y) = y+1, f(x+1,0)=f(x,1),f(x+1,0) = f(x,1), f(x+1,y+1)=f(x,f(x+1,y))f(x+1,y+1) = f\big(x, f(x+1,y)\big) for all non-negative integers x,yx,y. Determine f(4,1981)f(4,1981).
Step 3 of 7: Solve the x=2x=2 level
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f(2,0)=f(1,1)=3,f(2,y+1)=f(1,f(2,y))=f(2,y)+2  ⟹  f(2,y)=2y+3f(2,0)=f(1,1)=3,\quad f(2,y+1)=f(1,f(2,y))=f(2,y)+2 \implies f(2,y)=2y+3
Detailed analysis

By rule (2), f(2,0)=f(1,1)=1+2=3f(2,0)=f(1,1)=1+2=3 using Step 2. By rule (3), f(2,y+1)=f(1,f(2,y))=f(2,y)+2f(2,y+1)=f(1,f(2,y))=f(2,y)+2 using Step 2 again. Solving this recurrence with initial value 33 gives f(2,y)=2y+3f(2,y)=2y+3.