[seqfan] Re: I need help with defining these 3 sequences
M. F. Hasler
seqfan at hasler.fr
Wed Mar 4 15:07:16 CET 2020
On Tue, Mar 3, 2020 at 8:40 PM Ali Sada wrote:
> 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.
>
Oh yes, I was misled by:
" We add a(4) + a(5) = 5+8 = ... " in the earlier mail.
This made me think that a(5) = 8 instead of 5. Sorry for this mistake!
I should have realized it since the borders of the intervals are always
a(m) = m...
So yes, indeed, the deviation from Fibonacci's occurs already with a(6) =
18.
In (PARI):
a=Vec(m=1,44); while(#a >= m = a[n=m]+a[n-(n>1)], a[m]=m; for(k=1,m-n-1, a[
if( k%2, n+k\/2, m-k\2 )] = a[n+k\2]+a[m-k\/2+1] ));a
= [1, 2, 3, 8, 5, 18, 49, 129, 338, 209, 80, 31, 13, 57, 158, 417, 1093,
2862, 7493, 19617, 51358, 134457, 352013, 921582, 2412733, 6316617,
16537118, 43294737, 70052356, 26757619, 10220501, 3903884, 1491151, 569569,
217556, 83099, 31741, 12124, 4631, 1769, 676, 259, 101, 44, ...]
The sequence of "new upper ends" m' = a(m-1) + a(m) is
[1, 2,] 3, 5, 13, 44, 145, 479, 1582, 5225, 17257, 56996, ...:
As written earlier, after the gap (3,5) with only one "hole", we have
a(m-1) = a(n+1) + a(m) = a(n) + a(m)*2
(where n is the left end of the gap = predecessor of m), and thus m' =
a(m-1)+a(m) = a(n) + a(m)*3,
i.e., m(i) = 3 m(i-1) + m(i-2) after m(1..5) = (1, 2, 3, 5, 13).
(PARI) for(n=1+#a=[1,2,3,5,13],#a=Vec(a,100), a[n]=a[n-1]*3+a[n-2]);a
- Maximilian
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