# Mystery sequence generator - who can explain ?

ranjan sinha
Posers and Puzzles 25 Jan '05 18:29
1. 25 Jan '05 18:291 edit
Consider the sequence of integers s(1), s(2), s(3), .... s(n)....... defined by s(1)= 1; s(n+1) = 3*s(n) +1:etc..
Now it so turns out that 2*s(n) +1 = 3^n.
I have veriffied this to be true for n= 1,2,3,4, etc.

Can someone prove as to why is this so? And is this true for all n?
2. DoctorScribbles
BWA Soldier
25 Jan '05 21:29
Originally posted by ranjan sinha
Consider the sequence of integers s(1), s(2), s(3), .... s(n)....... defined by s(1)= 1; s(n+1) = 3*s(n) +1:etc..
Now it so turns out that 2*s(n) +1 = 3^n.
I have veriffied this to be true for n= 1,2,3,4, etc.

Can someone prove as to why is this so? And is this true for all n?
I can prove that it is true for all n.
3. 26 Jan '05 05:52
Originally posted by DoctorScribbles
I can prove that it is true for all n.
I think I too have the proof. Can you give detailed steps of YOUR proof?
4. DoctorScribbles
BWA Soldier
26 Jan '05 07:04
Originally posted by sarathian
I think I too have the proof. Can you give detailed steps of YOUR proof?
Induction does the trick.
5. 26 Jan '05 08:56
Yep, Induction is the easiest way.
6. TheMaster37
Kupikupopo!
26 Jan '05 10:24
Originally posted by ranjan sinha
Consider the sequence of integers s(1), s(2), s(3), .... s(n)....... defined by s(1)= 1; s(n+1) = 3*s(n) +1:etc..
Now it so turns out that 2*s(n) +1 = 3^n.
I have veriffied this to be true for n= 1,2,3,4, etc.

Can someone prove as to why is this so? And is this true for all n?
Say above holds for a number N, define M = N+1

Then
2*s(M) + 1 =
2*(3*s(N) + 1) + 1 =
2*(s(N) + 2*s(N) + 1) + 1 =
2*(s(N) + 3^N) + 1 =
2*s(N) + 2*3^N + 1 =
3^N + 2*3^N =
3*3^N =
3^M

So if above holds for N, it also holds for M. Above holds for 1.
7. 26 Jan '05 12:481 edit
Originally posted by TheMaster37
Say above holds for a number N, define M = N+1

Then
2*s(M) + 1 =
2*(3*s(N) + 1) + 1 =
2*(s(N) + 2*s(N) + 1) + 1 =
2*(s(N) + 3^N) + 1 =
2*s(N) + 2*3^N + 1 =
3^N + 2*3^N =
3*3^N =
3^M

So if above holds for N, it also holds for M. Above holds for 1.
Yep...That's it.
ðŸ™„ðŸ™„ðŸ˜²ðŸ˜²
8. TheMaster37
Kupikupopo!
26 Jan '05 13:53
I was jsut feeling like writing it out to stay in practise. I generally find it very lazy to simply say &quot;I have the proof&quot;.
9. 26 Jan '05 19:58
Originally posted by TheMaster37
I was jsut feeling like writing it out to stay in practise. I generally find it very lazy to simply say "I have the proof".
You're definitely out of practise ðŸ˜›

Much shorter (only one substitution) is:

1 + 2*s(M) =
1 + (6*s(N) + 2) =
3 * (2*s(N) + 1) =
3 * 3^N =
3^M