categorizing some sequences defined by third order recurrences and related to the fibonacci numbers etc.
Ross La Haye
rlahaye at new.rr.com
Sat Nov 6 18:30:13 CET 2004
Seqfans,
Looking for a broader understanding of the nature of A098703 recently, I
found that the sequence seemed to fit nicely into the following table --
a[i,j] 0 1 2 3 ...
0 A000045(n) A000045(n) A000045(n)
A000045(n)
1
A098703(n)
2 A001911(n)
3 A000285(n)
A094688(n)
4
5
6
...
Where a[i,j](n) denotes the the nth term of the a[i,j] sequence, the
structure of the sequences in the table can be defined using the following
third order linear recurrence --
a[i,j](0) = 0, a[i,j](1) = 1, a[i,j](2) = i + 1; for n > 2, a[i,j](n) = (j +
1)a[i,j](n - 1) - (j - 1)a[i,j](n - 2) - ja[i,j](n - 3)
The charateristic polynomial, then, is x^3 - (j + 1)x^2 - (j - 1)x - j.
The table above is sparsely populated; the following table, which includes
"fitting" *and* "almost fitting" sequences, fills it in quite a bit more --
a[i,j] 0 1 2 3 ...
0 A000045(n) A000045(n) A000045(n)
A000045(n)
1 A000045(n + 1)* A000071(n + 2) A027934(n - 1) A098703(n)
2 A000032(n)* A001911(n) A008466(n + 1)
3 A000285(n) A027961(n)
A094688(n)
4 A022095(n - 1) A053311(n - 1) A027974(n - 1)
5 A022096(n - 1)
6 A022097(n - 1)
...
* for n > 0
For all the sequences I've examined, the following relations hold --
a[i,j](n) = Convolution of a[0,i] with i^n, for i = j and i > 0, i.e. the
entries of the main diagonal after a[0,0](n) consist of the convolution of
the Fibonacci numbers with 1^n, 2^n, 3^n etc...
a[i,j](n) = Sum [a[i - 1,j](k), {k=0...n}], for i = j and i > 0
a[i,j](n) = a[i - 1,j](n) + a[j,j](n - 1), for i > 0
a[i,1](n) = Sum[a[i - 1,0](k) {k=0...n}], for i > 0
Additionally,
a[0,3](n) = ((Phi^(n-1) + (-Phi)^(1-n) + Phi^(n+1) + (-Phi)^(-1-n)) / 5
a[1,3](n) = (3^n + Phi^(n-1) + (-Phi)^(1-n)) / 5
a[2,3](n) = (2(3^n) + Phi^(n-1) + (-Phi)^(1-n) - Phi^(n+1) - (-Phi)^(-1-n))
/ 5
a[3,3](n) = (3^(n+1) - Phi^(n+2) - (-Phi)^(-2-n)) / 5
(where Phi denotes the golden mean)
...and...
a[1,3](n) = (3^1)(a[1,3](n-1) - a[0,0](n-3)
a[1,3](n) = (3^2)(a[1,3](n-2) - a[3,0](n-4)
a[2,0](n) = a[2,1](n) - a[2,1](n-3)
a[2,3](n) = a[1,3](n) + a[3,3](n-1)
a[2,3](n) = a[0,3](n) + 2a[3,3](n-1)
I imagine many other relations can be derived. At any rate, I just thought
I'd note the foregoing in case it turned out to be of interest. Please let
me know if there are any errors that I've missed etc.
Ross
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