MathLabs

Problem 3

Let 0<f(1)<f(2)<f(3)<…0<f(1)<f(2)<f(3)<\ldots be a sequence with all its terms positive integers. The nn-th positive integer which doesn't belong to the sequence is f(f(n))+1f(f(n))+1. Find f(240)f(240).
Step 4 of 5: Use the sibling branch to bridge the gap
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In plain words

The second branch shifts the chain just enough to reach an index immediately below 240.

f(56)=90⇒f(91)=147⇒f(148)=239f(56)=90 \Rightarrow f(91)=147 \Rightarrow f(148)=239
Detailed analysis

From f(56)=90f(56)=90, the second relation gives f(91)=56+90+1=147f(91)=56+90+1=147. Applying the second relation again to f(91)=147f(91)=147 gives f(148)=91+147+1=239f(148)=91+147+1=239.