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Let $g(n)$ be defined by $g(1) = 0,g(2) = 1, g(n+2) = g(n)+g(n+1)+1$. Prove that if the Fibonacci sequence is defined by $F_n = F_{n-1}+F_{n-2}$ where $F_1 = 1, F_2 = 1$, then $F(n+1) = g(n)+1$ for $n \geq 1$.

I was thinking about proving this using induction, but I was wondering if there was an easier way to make sense of it.

Puzzled417
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  • Induction in this case should be mechanical. – André Nicolas Jul 30 '16 at 21:56
  • @AndréNicolas Assuming it is true for some $n$, then $F(n+2) = F(n+1)+F(n) = g(n)+1+F(n)$. What do we do from here? – Puzzled417 Jul 30 '16 at 22:06
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    $=g(n)+1+g(n-1)+1=(g(n)+g(n-1)+1)+1=g(n+1)+1$. – André Nicolas Jul 30 '16 at 22:23
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    Note that we are using strong induction. From the induction assumption that the result holds everywhere at or below $n$, we show it holds at $n+1$. – André Nicolas Jul 30 '16 at 22:24
  • I would prefer slightly to write it as $F(n+1)+F(n)=g(n)+1+g(n-1)+1=(g(n)+g(n-1)+1)+1=g(n+1)+1$. We need to check also the two base cases, $n=1$ and $n=2$. – André Nicolas Jul 30 '16 at 22:27
  • Another way to think of this is that if $h(n)=g(n-1)+1$ for $n\ge2$, then $F_n$ and $H(n)$ are sequences that have the same values for $n=2$ and $n=3$ and satisfy the same recurrence relation, so they must be equal. (This follows easily, using strong induction.) – user84413 Jul 30 '16 at 22:46

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Hint $\ f(n) := g(n\!-\!1)+ 1\,$ satisfies the same recurrence as $\,F_n\,$ and has the same initial conditions $\,f(2) = 1,\, f(3) = 2\,$ so $\,f(n) = F_n\,$ for all $\,n>1\,$ by the uniqueness theorem for recurrences (which has an extremely short and easy proof by induction)

Remark $ $ You could, of course, do the inductive proof directly, skipping the language of the uniqueness theorem, but the slight amount of extra effort to understand it from the perspective of uniqueness will prove very useful for solving similar problems.

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Bill Dubuque
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