[seqfan] New sequences from generalized Fibonacci sequence
Kerry Mitchell
lkmitch at gmail.com
Sun Nov 9 06:12:15 CET 2014
Hi all,
Here's something I've been playing with lately. Background: the Fibonacci
sequence is defined by the recurrence f(n+1) = f(n) + f(n-1). As n
increases, the ratio of consecutive terms approaches the golden ratio, Phi
~ 1.618. This is because, if we assume that f(n+1) = r x f(n), then the
recurrence becomes r^2 = r + 1, and Phi is the positive root of that
quadratic.
Generalizing, let f(n+1) = a f(n) + b f(n-1). Then, the recurrence
quadratic becomes r^2 = a r + b. The ratio of consecutive terms approaches
r = (a + sqrt(a^2 + 4b))/2. To create the sequences, let a and b be
positive integers. List a, b, and r and sort by increasing values of r.
In the case of a tie, sort by increasing values of sqrt(a^2 + b^2). The
first several sorted values of a, b, r, and sqrt(a^2+b^2) are:
1. 1, 1, 1.618, 1.414
2. 1, 2, 2, 2.236
3. 1, 3, 2.303, 3.162
4. 2, 1, 2.414, 2.236
5. 1, 4, 2.562, 4.123
6. 2, 2, 2.732, 2.828
7. 1, 5, 2.791, 5.099
8. 2, 3, 3, 3.606
9. 1, 6, 3, 6.083
10. 1, 7, 3.193, 7.071
The a sequence begins: 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 3, 1, 2, 1, 3, 2,
1, 3, 2, 1, 3, 2, 1, 1, 2, 3, 4, 1, 2, 3, 1, 4, 2, 1, 3, 2, 4, 1, 3, 2, 1,
4, 3, 2, 1, 4, 3, 2, 1, 1, 2.
The b sequence begins: 1, 2, 3, 1, 4, 2, 5, 3, 6, 7, 4, 1, 8, 5, 9, 2, 6,
10, 3, 7, 11, 4, 8, 12, 13, 9, 5, 1, 14, 10, 6, 15, 2, 11, 16, 7, 12, 3,
17, 8, 13, 18, 4, 9, 14, 19, 5, 10, 15, 20, 21, 16.
Neither sequence submits to upper trimming or lower trimming, as fractal
sequences do. Each sequence does count the other--the a term is how many
times the corresponding b term has occured in the b sequence, and vice
versa. The associative arrays of a and b are transposes of each other and
the first column of the b's associative array, 1, 4, 12, 28,55, 96, etc.,
seems to be A006000. The first column of a's associative array is not in
the OEIS.
Are these sequences interesting enough to submit to OEIS?
Thanks,
Kerry Mitchell
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