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While I was working on some graph theory problem I encounter the following recurrence relation $$a_{n+1}=a_{n-1}+6$$ where $a_0=3.$

Note: I have rewritten the recurrence relation as recommended.

  • It always helps to compute the first few numbers: 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 51, 55, 57, 61, 63, 67, 69, 73, 75, 79, 81, 85, 87, 91, 93, 97, 99, ... – Yuval Filmus Mar 24 '15 at 13:12

2 Answers2

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You're alternately adding $2$ and $4$ to the last term of your sequence. Thus every two terms you add $6$, which allows you to easily work out a formula for $a_{2n}$ and from that, deduce a formula for $a_{2n+1}$.

Jack M
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$a_1=a_0+4$;

$a_2=a_1+2=a_0+4+2;$

.

$a_4=a_0+4+2+4+2$

$a_5=a_0+4+2+4+2+4$

thus

if $n$ is odd $a_n=a_0+\frac{n+1}{2}4+\frac{n-1}{2}2$

if $n$ is even $a_n=a_0+\frac{n}{2}(2+4)$

math
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