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$$\begin{align} a_0 &= 1 \\ a_1 &= 1 \\ a_k &= 2~a_{k-1} + 3~a_{k-2} \quad \text{ for } k \ge 2 \end{align}$$

Proof by Strong Induction: For all non-negative integers $n$, $a_n$ is an odd integer.

Proof by Strong Induction:

$$\begin{align} P(n) &= a_n \text{ is an odd integer } \\ P(0) &= a_0 \text{ is an odd integer } \\ \end{align}$$

Assume $P(n)$, show $P(n+1)$

Basis Step: $P(0)$, $P(1)$, $P(2)$, $P(3)$, $P(4)$, $P(5)$

Inductive Step: Want to show for all integers $n \ge 5$, if $P(i)$ is true for all $0 < i < n$, then $P(n)$ is true

Let $c > 5$, be arbitrary and fixed

IH: Assume $P(i)$ for any $0<i<c$

I don't know where to go in the proof from here. Can anybody help me show that $P(i)$ is always odd, I am confused with strong induction, too many variables are introduced.

Charles
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Michael
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    The idea is that since $a_{k-2}$ is odd, $2a_{k-1}+3a_{k-2}$ is an even number plus $3$ times an odd number, so it is odd. – André Nicolas Apr 04 '15 at 15:34
  • Strong induction does not have a base case. It is simply the $n=0$ case of the inductive step. The $c$ in the inductive hypothesis should be an $n$, and stating the inductive hypothesis redundant, you already stated it when you wrote the "inductive step". – DanielV Apr 04 '15 at 16:06

3 Answers3

2
  • IH the induction hypothesis is $p(i)$ is true for any :$0< i< c$
  • Goal we want to prove that $P(c)$ is true

We can use : $$a_c=\underbrace{2a_{c-1}}_{\text{even}}+\underbrace{3a_{c-2}}_{\text{odd}} $$

and IH implies that both $a_{c-2}$ and $a_{c-1}$ are odd so $a_{c}$ is odd hence $P(c)$ is true.

Elaqqad
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For example, $\,a_0 = 1 = a_1\,$ are both odd so $\,a_2 = 2a_1 + 3a_0\,$ = even + odd = odd. That is precisely how the general induction step is proved, i.e. for $\,k\ge 2\,$ prove in the same way that $\,a_k\,$ is odd, given that, by (strong) induction hypothesis, both $\,a_{k-1}\,$ and $\,a_{k-2}\,$ are odd.

Remark $\ $ Insight is gained by viewing things mod $2.\,$ Then the recurrence reduces to simply $\, a_k \equiv a_{k-2},\,$ so by induction $\,a_{2n} \equiv \color{#c00}{a_0}\,$ and $\,a_{2n+1}\equiv \color{#c00}{a_1}.\,$ But both base cases $\,\color{#c00}{a_0,a_1}$ are $\equiv 1.$

If instead $\,a_0\equiv 0,\ a_1\equiv 1\,$ then, since $\,a_{n}\equiv a_{n\ {\rm mod}\ 2},\,$ we get $\,a_n\,$ is even $\iff n\,$ is even

Bill Dubuque
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You verified the base step and you have the induction assumption. Here's how you can use it: $$ a_{k+1}=\underbrace{2a_{k}}_{\text{even times odd is even}}+\underbrace{3a_{k-1}}_{\text{odd times odd is odd}} $$ Since an even number plus an odd number is an odd number, this proves your result. Note that strong induction was necessary to conclude that $a_{k-1}$ was odd. That's really all you need.

Daniel W. Farlow
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