From Osher Doctorow
Let's consider the equation:
1) an - an-1 = (1/2)(an-1 - an-2)
where n, n-1, n-2 are subscripts in sequence (an) or (a_n), n = 1, 2,
3, ....
Although I obtained (1) by looking at 2, 4, 5, there are some rather
curious features of (1). Let's solve (1) for an in terms of an-1 and
an-2):
2) an = (1/2)(3an-1 - an-2)
Let's consider the integers 1, 2, 3, ..., 10, although any sequence of
increasing positive integers (by one each time) would do. Trying
various initial values counting downward from 10 for an, which is
presumably larger than an or an-1, brings us finally to:
3) an = 5
Now count downward again for an-1, not repeating 5. The first number
that we get to is:
4) an-1 = 4
Inserting into (2) yields:
5) 5 = (1/2)(12 - an-2) = 6 - (1/2)an-2
or in other words:
6) (1/2)an-2 = 1
which has solution:
7) an-2 = 2
which is correct for the sequence 5, 4, 2 corresponding to (1 +
sqrt(5))/2, 1 + sqrt(4), 1 + sqrt(2) which we've been studying for
several recent postings.
Equations (1) and (2) are equivalent to:
8) 2an - 3an-1 + an-2 = 0
This is an alternating (finite) series, quite important in Probable
Influence/Causation (PI), with respective coefficients of terms -3, 2,
1, or their magnitudes 3, 2, 1. If we write it as:
9) 2a3 - 3a2 + a1 = 0
then a1 is almost an "analog" of a generalized Lie bracket 3a^2 -
2a^3.
I'll try to continue this later.
Osher Doctorow
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