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My favorite integer sequence
(09/16/2014, 05:32 AM)jaydfox Wrote: I'm not sure if anyone else is still playing around with the binary partition function. I'm still learning lots of new things (about the function, or about complex analysis), so I keep poking at it.
Hi Jay, I'm in holiday mode and play around with less difficult things. It's really great what you are doing here, optimally it would now be compiled into one coherent treatize...
I hope I'll come back to this next week/in a couple of days.

Gottfried Helms, Kassel
(09/16/2014, 05:32 AM)jaydfox Wrote: I feel like I'm close to a pattern here that will really speed up calculations. I might be able to ditch the polynomials entirely. I've found some other strange recurrences, one of which uses that Thue-Morse sequence, which Gottfried previously described in post #10 on the first page of this thread. I'll try to detail those later this week.

Regarding the Thue-Morse connection, I first noticed it while trying to come up with a way to calculate the coefficients at powers of 2. I said previously:

(09/16/2014, 05:32 AM)jaydfox Wrote: However, what does seem to bear fruit is to look at the polynomials evaluated at powers of 2. The first few are:

1     1
1     2     1
1     4     4     1
1     8    16    10     1
1    16    64    84    36     1
1    32   256   680   656   202     1
1    64  1024  5456 10816  8148  1828     1

From left to right, these are the evaluations of the 0th, 1st, 2nd, 3rd, etc. order polynomials. From top to bottom, these are the polynomials evaluated at 2^k, where k is the row number, with the first row being k=-1 (to make the math simpler, so that the second row is evaluated at 2^0 = 1).

When you look at columns, you will note that the left-most column (the zeroeth column, as it corresponds to the 0-order polynomial) is always 1. The first column (2nd from the left) is a power of 2. The second column is a power of 4.

Well, if you look at the right side of the triangle, the diagonal is always 1. The first subdiagonal is A(2^k), i.e., it's a term in the sequence. Specifically, for the row which corresponds to A(2^n), the subdiagonal is A(2^(n-1)).

For example, in the 3rd row--the row corresponding to A(2^3), which is the row with 5 entries (1 8 16 10 1)--the subdiagonal entry is 10, which is A(2^2).

If you look at the second subdiagonal entry, this also follows a pattern. It helps to look a little further down the triangle. In the 5th row (1 32 256 680 656 202 1), the second subdiagonal is 656, which happens to be 692-36. Therefore, 656 = A(24)-A(8)

The third subdiagonal is 680. It took me a while to spot the pattern. I had to reverse engineer it, by comparing how the value was constructed from previous entries. But it turns out that it's equal to 1154-380+94. And 380 is 390 - 10. So it's 1154 - 390 - 94 + 10. This turns out to be A(28) - A(20) - A(12) + A(4).

P_{2^5},6 = 1 = A(0)
P_{2^5},5 = 202 = A(2^4)
P_{2^5},4 = 656 = A(3*2^3) - A(1*2^3)
P_{2^5},3 = 680 = A(7*2^2) - A(5*2^2) - A(3*2^2) + A(1*2^2)

(Here, P_{2^k}, 2^k means the row for A(2^k), and n means the term corresponding to the nth degree polynomial.)

When I noticed the last line there resembled the Thue-Morse sequence (+--+), which Gottfried had previously described, I decided to see if the pattern continues: (+--+-++-)

Predicted: P_5,2 = A(15*2) - A(13*2) - A(11*2) + A(9*2) - A(7*2) + A(5*2) + A(3*2) - A(1*2)

And sure enough, 1460 - 900 - 524 + 284 - 140 + 60 + 20 - 4 = 256, as predicted.

After some more investigation, I found that this pattern holds for any row in the table, not just at powers of 2.

For the row corresponding to A(37) = 3074:
1 37 342 1050 1180 450 14

P_37,6 = 14 = A(37-1*2^5)
P_37,5 = 450 = A(37-1*2^4) - A(37-3*2^4)
P_37,4 = 1180 = A(37-1*2^3) - A(37-3*2^3) - A(37-5*2^3) + A(37-7*2^3)

In general:

Here, t_j is the jth term in the Thue-Morse sequence, starting with t_0 = 1, t_1 = -1, and so on. We can take the infinite sums, because A(m) is 0 for any negative m. In practice, of course, we would stop the sum when m goes negative.

I suspect, but have not confirmed, that the reason for the Thue-Morse sequence in the first equation is due to the way that A is constructed in the second equation in terms of P, combined with the recurrence relation that defined A:
A(k) = A(k-1) + A([k/2]).
~ Jay Daniel Fox
Did you read post 55 ?

The polynomials have many structures.
Its very intresting.

Thanks for your time.


(10/02/2014, 11:24 PM)tommy1729 Wrote: Did you read post 55 ?

The polynomials have many structures.
Its very intresting.

Thanks for your time.



Hi Tommy,

Yes, I saw that post from my phone when you wrote it a few weeks back, didn't reply, and then got sidetracked.

There are so many patterns at work here. Some, upon reflection, are obvious (in hindsight, for me anyway). Some still seem mysterious, though of course this is only a lack of understanding on my part.

There are patterns in the coefficients of the polynomials. There are patterns in the evaluations of those polynomials. Some of the patterns are simple (powers of 2, small primorials, etc.), while other patterns are harder to see (like the connection with Thue-Morse). I'm trying to divide my time between studying these patterns, and studying the continuous function defined previously.
~ Jay Daniel Fox
I wanted to explain why the binary partition function f(x) and
Jay's J(x) satisfy f(x) ~ C J(x) for large x.

But mick beat me to it.

The proof can be constructed from these lemma's at MSE


The idea that difference and derivative start to approximate one another occurs often in complex dynamics and tetration.

For instance Levy used that idea and I used the idea when I wrote about parabolic fixpoints.

This is no surprise to me, but I wanted to state that clearly.


Did we already prove that all zero's of J(x) are real ?

Not sure.


Anyway, the binary partition function relates to fake function theory and tetration.

In math things are always related , it is just not clear from the beginning.

I Will partially explain in the fake function thread " searching for an asymptotic .. "


A funny and remarkable fact.

F(n) = F(n-1) + F(n/2).

G(n) = G(n-1) + G(n/2) + 1.

- n/2 is rounded -

How do you think F and G relate ?

Hint : they do !

Hint : if you take the difference of both sides you get a similar equation !

Hint : 2^(x+1) = 2 2^x.

Hint : Maybe change + 1 to + j to see it. ( use dummy variable ).




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