generalized n-term sequences

I wrote,

<< What's missing from the three-term series is the built-in ordering
rule supplied by the single generator of the two-term series, any
ideas? >>

Paul Erlich wrote,

<< Perhaps the fact that the interval of duplication is different at
the different stages of construction is what's making this problem
rather difficult? >>

OK, even though I haven't worked out the kinks in what I came up with
at work last week, I'm all but sure it's the right idea...

If you look at a two-term series as a two-stepsize index, [a,b] where
the alphabetized variables are small to large stepsizes, then a+b is
the subset or scale of b+(a+b) where the size -- not to be confused
with the amount or index -- of a is 1, the size of b is 2, and b+(a+b)
is an equal division of the octave.

This of course would be true at any point in the sequence... witness
Yasser's famous example.

So a three-term Tribonacci series is analogous to the Fibonacci series
in this sense where a+b+c is the corresponding subset of the sum of
the 3rd, 4th, and 5th terms and the size -- again, not to be confused
with the amount or "index" -- of a=2, b=3, c=4, and the sum of the
3rd, 4th, and 5th terms is an equal division of the octave.

So if the "3, 7, 12, 22, 41" example that Paul gave was backed up (so
to speak) so that the three-stepsize index were [2,2,3], you'd then
have 2, 2, 3, 7, 12, 22, ...

This is the generalized fib to tri (and higher term) parallel in the
Yasser sense.

So a two-stepsize [2,5] two-term series example is:

2, 5, 7, 12, 19, ...

And a+b is the subset of the sum of the 2nd and 3rd terms where a=1
and b=2. This is the familiar 7-out-of-12.

A three-stepsize [2,2,3] three-term series example would be:

2, 2, 3, 7, 12, 22, 41, ...

And a+b+c is the subset of the sum of the 3rd, 4th, and 5th terms
where a=2, b=3, and c=4. This would be the familiar syntonic diatonic
as a 7-out-of-22.

A four-stepsize [2,2,1,2] four-term series example would be:

2, 2, 1, 2, 7, 12, 22, 43, 84, ...

And a+b+c+d is the subset of the sum of terms 4, 5, 6, and 7, and a=4,
b=6, c=7, and d=8. This would be a 7-out-of-43 scale.

I can't see much reason to extend this idea much beyond this four-term
point, though one of course could by using the same methods.

Generalizing n-term series so that they have a built-in ordering rule
analogous to the single generator of the two-term series seems to me
to hinge on generalizing the M-out-of-N idea. In other words, as a
two-term series results in an M-out-of-N, a three-term series should
result in an L-out-of-M-out-of-N, and so forth and so on up the n-term
series chain...

But if each new series is viewed as a parallel to the two-term, then
they also carry a stepsize condition that does not always agree with
the generalized the M-out-of-N idea.

For example if you take a [1,1,3] index, you'd have a 5-tone subset of
17 equal where a=2, b=3, and c=4. But a[1,1,3] L-out-of-M-out-of-N
would give a 0 4 8 9 13 17 scale of 4+4+1+4+4. No good.

However, taking L-out-of-M-out-of-N back to the two-generator idea
would give an 0 1 5 9 14 17 scale of 1+4+4+5+3:

5----14
/ \ / \
/ \ / \
0-----9-----1

This in turn could then be readdressed with the three-term series rule
of a=2, b=3, and c=4 as a 2+4+4+4+3.

This, while not nearly as "elegant" as I had hoped, seems to work.

If so, then this should allow for a generalized "scaled by P rule"
which is what I was after, but had given up on (and now I remember
why!), even before Paul brought all this up.

Here's the basic idea again...

Let stepsizes be small to large alphabetized variables so that
two-stepsize = [a,b], three-stepsize = [a,b,c] and so forth and so on
where the alphabetized variables are any whole numbers.

Then assign each variable an uppercase fixed size:

A = (LOG(2)-LOG(1))*(1200/LOG(2))
B = (LOG(3)-LOG(1))*(1200/LOG(2))
C = (LOG(4)-LOG(1))*(1200/LOG(2))

etc.

This will be our fixed interval template, and will allow for a
convenient (and aurally agreeable) distribution of stepsizes amongst
any [a,b,...] index:

a/b = log(3)/log(2)
c/b = log (4)/log(3)
d/c = log(5)/log(4)

etc.

Next you scale a given [a,b,...] index by a given periodicity (P).
This will give us a percentage (X) to scale the fixed A,B,... interval
template with.

So for a two-stepsize index you'd have:

(a*A)+(b*B)/P = X

For a three-stepsize index:

(a*A)+(b*B)+(c*C)/P = X

And so forth and so on...

Now A/X, B/X,... will give a given [a,b,...] index corresponding
stepsizes scaled to P.

Here's the previous [1,1,3] "Tribonacci" example where a/b =
log(3)/log(2), c/b = log (4)/log(3), and d/c = log(5)/log(4):

0 140 419 699 978 1200
0 280 559 839 1060 1200
0 280 559 781 920 1200
0 280 501 641 920 1200
0 222 361 641 920 1200

Anyway, this is the basic idea... but what I'm still looking for is a
less wobbly generalization of the M-out-of-N idea; too much voodoo is
required as it stands now.

Any thing come to mind?

--Dan Stearns