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 2 of 7: Close the recurrence for x=1x=1
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f(1,y)=y+2f(1,y) = y+2
Detailed analysis

The relation f(1,y+1)=f(1,y)+1f(1,y+1)=f(1,y)+1 with initial value f(1,0)=2f(1,0)=2 is solved by induction on yy: f(1,y)=y+2f(1,y)=y+2 for all y≥0y\ge0.