# [seqfan] Re: need algebra proof

quantum pkt qntmpkt at yahoo.com
Thu Mar 18 03:20:17 CET 2010

```thx for your help!
To summarize, in the context of partitions p(x)  and ruler sequences, we have
p(x) = A(x)/ A(x^2), when A(x) = Euler transform of the ruler sequence for k=2 (A001511) : (1, 2, 1, 3, 1, 2, 1, 4,...);
p(x) = B(x) / B(x^3) when B(x) = Euler transform of the ruler sequence for k = 3...; then generally,
p(x) = C(x)/C(x^k) , when C(x) = Euler transform of ruler sequence for k = 2, 3, 4,...and so on.; where also we have the corollary
A(x) = p(x) * p(x^2) * p(x^4) * p(x^8) * p(x^16) * ..; etc: analogous power series. for any k.
...
The companion operation p(x) = A(x) * A(x^2) relating to the ruler sequence for k=2 seems to be (conjecture - I have difficulty calculating enough terms with a pocket calculator - maybe somebody's program would help):....that
p(x) = A(x) * A(x^2) when A(x) = Euler transform of A035263: (1, 0, 1, 1, 1, 0, 1, 0, 1,...) i.e. "First Feigenbaum Symbolic or period doubling sequence", where A035263 = parity of the ruler sequence (1, 2, 1, 3, 1, 2, 1, 4,...), where 1's in A035263 = Tower of Hanoi moves CW, otherwise CCW.  The basic setup for k=2 Tower of Hanoi operations would be  a multiplication table with the odd integer heading and powers of 2 as first column, getting:
1,...3,...5,...7,...9,....11,...13,...
2,...6,..10,.14,.18,...22,...26,...
4,.12,..20,.28, 36,...44..  36..
8,.24,..40,.56,.72,...88 104,....
16,...
...
Such that given rows are labeled (1,2,3,...) = # of TOH disc moved in n-th move.  We bisect the array taking odd numbered rows only, = indices in A035263 = 1's = CW moves where even rows = 0 = CCW moves.,
Check: Indices of (1, 0, 1, 1, 1, 0, 1, 0, 1,...): 1's = 1, 3, 4, 5, 7, 9,...where these terms are in odd numbered rows.
(matching the power series by rows: 1/2 + 1/4 + 1/8  + 1/16 + ....= 1, with the series pointing to a parsed (1, 2, 3,...) such that (1, 3, 5, 7,...) = 1/2 of the terms, (2, 6, 10, 14,...) = 1/4 of the terms, etc and the odds corresponding to (1/2 + 1/8 + 1/32 + ...) = 2/3 corresponding to the fact that 2/3 of the TOH moves are CW and 1/3 are CCW.
If p(x) = A(x) * A(x^2) when A(x) = Euler transform of (1, 0, 1, 1, 1, 0, 1,...), an analogous operation for
p(x) = B(x) * B(x^2) and any k may exist in the generalized k TOH  setups with heading ("not multiples of k") and left column powers of k.
But I haven't found that generalization yet.
Gary

On http://list.seqfan.eu/pipermail/seqfan/2010-March/004072.html

we had earlier correspondences, which -- it's 11:30 pm and I'm getting lazy --

rjm> From: Richard Mathar <mathar at strw.leidenuniv.nl>
rjm> Date: Tue, 09 Feb 2010 23:29:30 +0100
rjm> Organization: Leiden Observatory, The Netherlands
rjm> To: qntmpkt at hotmail.com
rjm> Subject: Re: new transform - Gary
rjm> Cc: maximilian.hasler at gmail.com, mathar at strw.leidenuniv.nl
rjm> ..
rjm>
rjm> IN generatign function terms, you are searching the solution
rjm> to S(x)=A(x)/A(x^2), where A(x) is the unknown. The example
rjm> of starting with S(x) = sum_i i*x^(i-1), a shifted version of A000027,
rjm> gives A(x) in A171238. The standard physicist's way of finding A(x)
rjm> is to start with A(x) = S(x) as an ansatz, and then
rjm> to iterate  A(X)  := S(x)*A(x^2) repeatedly. I think this
rjm> is called a B"urmann-Lagrange series...what Kepler did as you know,
rjm> but no correct mathematician would ever accept. Paul Barry seems to
rjm> provide a lot of sequences defined through this type of explicit
rjm> requirement on A(x^2) as a function of A(x). You may ask him what
rjm> the history of this is..

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From: Richard Mathar <mathar at strw.leidenuniv.nl>
To: seqfan at seqfan.eu
Sent: Wed, March 17, 2010 3:32:01 PM
Subject: [seqfan] Re: need algebra proof

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