[seqfan] Re: I need help with defining these 3 sequences
M. F. Hasler
seqfan at hasler.fr
Tue Mar 3 23:55:37 CET 2020
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 be
a(6) = a(5) + a(13) = ***8*** + 13 = F(6) + F(7) = F(8) = 21
(is this correct?)
and then
a(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 later
a(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
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