[seqfan] Re: I need help with defining these 3 sequences

[Ali Sada]

 Thank you very much Dr. Hasler. I really appreciate your response.
Again, my language didn't help me convey what I wanted to say.
After we fill the first five terms, we add a(4) + a(5) = 13. So, a(13) =13.Now, we have 7 gaps between a(5) and a(13).Since a(13) is on the right, we fill the first gap on the left: a(6) = a(13) + a(5) = 18.Then, we fill the first gap on the right: a(12)= a(6) + a(13) = 18+13 = 31.We fill the first gap on the left: a(7) = a(6) + a(12) = 18+31 = 49.We fill the first gap on the right: a(11) = a(7) + a(12) = 49+ 31 = 80.We fill the first gap on the left: a(8) = a(7) + a(11) = 49+80 =129.We fill the first gap on the right: a(10)= a(8) + a(11) = 129+80 = 209.We fill the first gap on the  left: a(9) = a(8) + a(10) = 129+209 = 338. 
Now we don't have any more gaps, so we continue. We add a(12) + a(13) = 13+31= 44. So, a(44) = 44, and we repeat the same procedure above.
It is much easier to show it on a blackboard! Please see this image. https://justpaste.it/3pflgIn each row, we add the two green numbers to get the yellow number.
Thank you again for your help.
Best,
Ali








    On Tuesday, March 3, 2020, 5:56:08 PM EST, M. F. Hasler <seqfan at hasler.fr> wrote:  
 
 On Sun, Mar 1, 2020 at 2:30 AM Ali Sada wrote:

Sequence 2: 
We start with a(1) = 1 and a(2) = 2. 
a(3) = a(1) + a(2) = 3. We don’t have any gap here, so we continue. 
We add a(2) and a(3) and we get 5. So, a(5) = 5. To find a(4), we add a(5) + a(3) = 8.
We filled the gap between 3 and 5, so we continue.
We add a(4) + a(5) = 5+8= 13, so a(13) =1 3. Now, we need to fill the gaps between 6 and 12.
a(6) = a(5) + a(13) = 5+13 = 18
a(12) = a(6) + a(13) = (...)


This is a variant of the Fibonacci numbers F(n) = F(n-1)+F(n-2).While there is a "gap" to fill [say, between a(n) and a(n+m)], you place alternatively such sums to the left and to the right of the gap.then you continue with a(X) = X = a(n+m) + a(n+m-1) which is changed w.r.t. Fibonacci formula because here a(n+m-1) = a(n+m) + a(n+1) = 2 a(n+m) + a(n). 
Maybe I did n't understand correctly, but I think you made an error in the 2nd-to-last line above (so all subsequent terms are wrong):I think it should bea(6) = a(5) + a(13) = ***8*** + 13 = F(6)+ F(7) = F(8) = 21 (is this correct?)and thena(12) = a(6) + a(13) = 21 + 13 = F(8) + F(7) = F(9) = 34(followed by a(7) = a(6)+a(12) = F(8)+F(9) = F(10), a(11) = a(7)+a(12) = F(10)+F(9) = F(11), etc)and latera(13) + a(12) = F(7) + F(9) = 13 + 34 = 47 =: a(47)
I think this is the first term which is not a Fibonacci number(because of the gap being larger than 1 as for (a(3), a(5)), and therefore it has F(k+2) as last term and not just F(k+1)).
If we used, to continue once a gap is filled, a(n+m) and a(n+1) instead of a(n+m-1), then the sequence would consist only of Fibonacci numbers:a(13)+a(6) = F(7)+F(8) = F(9) = 34 =: a(34)a(34)+a(14) = F(9)+F(10) = F(11) = 89 =: a(89)a(89)+a(35) = F(11)+F(12) = F(13) = 233 =: a(233) etc.
- Maximilian