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I have this equation $a_{n+2}=a_{n+1}+6a_n$ for $a_0=0,a_1=1$.

I should $a_{0+2}=a_{0+1}+6a_0$ (1) relation.The case $a_1=1$ it should be $a_{1+2}=a_{1+1}+6a_1$.

I just replace it. What should I have done?

md2perpe
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ek.Sek
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  • Welcome to Math.SE. I've formatted the first line of your post to show how math notation can be use (as introduced in this note). Please check that I did not unintentionally change your meaning, and edit the rest of your post using similar syntax. I don't follow your meaning. – hardmath Jun 21 '18 at 14:13
  • same as above for my editing, which superimposed in time with the above (sorry) – G Cab Jun 21 '18 at 14:15
  • Welcome to Math.SE! Could you clarify what your question is? As it stands, I'm not sure I understand what you are asking. –  Jun 21 '18 at 14:17
  • hi, it is retroactive – ek.Sek Jun 21 '18 at 14:18
  • This is not clear. Taking $n=0$ your recursion gives $a_2=a_1+6a_0$. From the initial conditions, that would tell us that $a_2=1+6\times 0=1$. Is that what you are asking? – lulu Jun 21 '18 at 14:25
  • Can you clarify your question? As you can see from the comments, nobody is sure what you are asking. – lulu Jun 21 '18 at 14:32
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    What do you want to do? If the task is to calculate for example $a_2$, $a_3$ and $a_4$ then you are on a correct path:

    $$a_2 = { \text{ take $n=0$ } } = a_{0+2} = a_{0+1} + 6 a_0 = a_1 + 6 a_0 = 1 + 6 \times 0 = 1.$$

    $$a_3 = { \text{ take $n=1$ } } = a_{1+2} = a_{1+1} + 6 a_1 = a_2 + 6 a_1 = 1 + 6 \times 1 = 7.$$

    $$a_4 = { \text{ take $n=2$ } } = a_{2+2} = a_{2+1} + 6 a_2 = a_3 + 6 a_2 = 7 + 6 \times 1 = 13.$$

     – md2perpe Jun 21 '18 at 15:06
  • @md2perpe thank you a lot :) – ek.Sek Jun 21 '18 at 16:13

1 Answers1

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A closed-form of the n-th term of this sequences is in http://oeis.org/A015441 .

R. J. Mathar
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