Seq Transformation Based On Cont Fractions

Leroy Quet qq-quet at mindspring.com
Fri Mar 9 18:24:22 CET 2007 

sum{k=1 to n} 1/k, the nth harmonic number.
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Subject: Re: Seq Transformation Based On Cont Fractions
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Date: Fri, 9 Mar 2007 12:59:38 -0500
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Seqfans, 
     After rewriting my program, I confirm that my original observation
was correct after all.
Please reconsider this nice problem.  
 
It is quite surprising that A030266: 
"Shifts left under COMPOSE transform with itself."
[0, 1, 1, 2, 6, 23, 104, 531, 2982, 18109, 117545, ...]
should have the generating function A(x) defined by (1) below! 
 
I have confirmed that the g.f. of A030266 agrees with x*A(x)  
up to 150 initial terms. 
It would be nice to have a proof! 
 
Problem:
-------------------------------------------------- 
(1) 
Find the unique function A(x) that satisfies: 
A = 1 + xAB 
B = 1 + xABC 
C = 1 + xABCD 
D = 1 + xABCDE 
E = 1 + xABCDEF 
...
where A=A(x), B=B(x), C=C(x), ... are an infinite set 
of functions with integer coefficients. 
-------------------------------------------------- 
 
It can be easily shown that the following expressions 
are equivalent to the original problem (1).  
(2)
A = 1/(1 - xB) 
AB = 1/(1 - xB - xC) 
ABC = 1/(1 - xB - xC - xD) 
ABCD = 1/(1 - xB - xC - xD - xE) 
ABCDE = 1/(1 - xB - xC - xD - xE - xF) 
...
(3)
A = 1 + xB/(1 - xB) 
B = 1 + xC/(1 - xB - xC) 
C = 1 + xD/(1 - xB - xC - xD) 
D = 1 + xE/(1 - xB - xC - xD - xE) 
E = 1 + xF/(1 - xB - xC - xD - xE - xF) 
...
(4)
B = (1 - 1/A)/x*(1) 
C = (1 - 1/B)/x*(1 - xB) 
D = (1 - 1/C)/x*(1 - xB - xC) 
E = (1 - 1/D)/x*(1 - xB - xC - xD) 
F = (1 - 1/E)/x*(1 - xB - xC - xD - xE) 
...
(5)
A = 1 + x/(1 - xB - xC) 
B = 1 + x/(1 - xB - xC - xD) 
C = 1 + x/(1 - xB - xC - xD - xE) 
D = 1 + x/(1 - xB - xC - xD - xE - xF) 
...
This last set of expressions (5) forms a glorified continued fraction,
and admits a unique solution to A(x). 
 
Perhaps some variation of these expressions would produce 
other significant sequences... 
  
Thanks,
     Paul
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2600.0" name=3DGENERATOR></HEAD>
<BODY bottomMargin=3D0 leftMargin=3D3 topMargin=3D0 rightMargin=3D3>
<DIV></DIV>
<DIV>Seqfans, <BR>     After rewriting my program, I co=
nfirm=20
that my original observation was correct after all.</DIV>
<DIV>Please reconsider this nice problem.  </DIV>
<DIV> </DIV>
<DIV>It is quite surprising that A030266: </DIV>
<DIV>"Shifts left under COMPOSE transform with itself."</DIV>
<DIV>
<DIV>[0, 1, 1, 2, 6, 23, 104, 531, 2982, 18109, 117545, ...]</DIV>
<DIV>should have the generating function A(x) defined by (1) belo=
w!=20
</DIV>
<DIV> </DIV></DIV>
<DIV>I have confirmed that the g.f. of A030266 agrees with=
=20
x*A(x)  </DIV>
<DIV>up to 150 initial terms. </DIV>
<DIV>It would be nice to have a proof! </DIV>
<DIV> </DIV>
<DIV>Problem:</DIV>
<DIV>-------------------------------------------------- </DIV>
<DIV>(1) </DIV>
<DIV>
<DIV>Find the unique function A(x) that satisfies: </DIV>
<DIV>A =3D 1 + xAB <BR>B =3D 1 + xABC <BR>C =3D 1 + xABCD <BR>D =3D 1 + xAB=
CDE <BR>E =3D 1=20
+ xABCDEF </DIV>
<DIV>...</DIV>
<DIV>where A=3DA(x), B=3DB(x), C=3DC(x), ... are an inf=
inite set=20
</DIV>
<DIV>of functions with integer coefficients. </DIV>
<DIV>-------------------------------------------------- </DIV>
<DIV> </DIV>
<DIV>It can be easily shown that the following expressions </DIV>
<DIV>are equivalent to the original problem (1).  </DIV>
<DIV>(2)<BR>A =3D 1/(1 - xB) <BR>AB =3D 1/(1 - xB - xC) <BR>ABC =3D 1/(1 - =
xB - xC -=20
xD) <BR>ABCD =3D 1/(1 - xB - xC - xD - xE) <BR>ABCDE =3D 1/(1 - xB - xC - x=
D - xE -=20
xF) <BR>...<BR>(3)<BR>A =3D 1 + xB/(1 - xB) <BR>B =3D 1 + xC/(1 - xB - xC) =
<BR>C =3D 1=20
+ xD/(1 - xB - xC - xD) <BR>D =3D 1 + xE/(1 - xB - xC - xD - xE) <BR>E =3D =
1 + xF/(1=20
- xB - xC - xD - xE - xF) <BR>...<BR>(4)<BR>B =3D (1 - 1/A)/x*(1) <BR>C =3D=
 (1 -=20
1/B)/x*(1 - xB) <BR>D =3D (1 - 1/C)/x*(1 - xB - xC) <BR>E =3D (1 - 1/D)/x*(=
1 - xB -=20
xC - xD) <BR>F =3D (1 - 1/E)/x*(1 - xB - xC - xD - xE) <BR>...<BR>(5)<BR>A =
=3D 1 +=20
x/(1 - xB - xC) <BR>B =3D 1 + x/(1 - xB - xC - xD) <BR>C =3D 1 + x/(1 - xB =
- xC - xD=20
- xE) <BR>D =3D 1 + x/(1 - xB - xC - xD - xE - xF) <BR>...</DIV></DIV>
<DIV>This last set of expressions (5) forms a glorified continued=20
fraction,</DIV>
<DIV>and admits a unique solution to A(x). </DIV>
<DIV> </DIV>
<DIV>Perhaps some variation of these expressions would produce </DIV>
<DIV>other significant sequences... </DIV>
<DIV>  </DIV>
<DIV>Thanks,</DIV>
<DIV>     Paul</DIV></BODY></HTML>

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----__JWM__J41c6.167eS.2781M

Seqfans,
Also, empirically verified, I found the formula:
B(x) =3D A(x*A(x))=20
C(x) =3D B(x*A(x))
D(x) =3D C(x*A(x))
E(x) =3D D(x*A(x))
...
Can the above be easily proven from the property:
where x*A(x) is the g.f. of A030266?
=20
Paul

-- "Paul D. Hanna" <pauldhanna at juno.com> wrote:
Seqfans,=20
Please reconsider this nice problem.   It is quite surprising that A030266:=
ined by (1) below!  I have confirmed that the g.f. of A030266 agrees with x=
*A(x)  up to 150 initial terms. It would be nice to have a proof!  Problem:=
-------------------------------------------------- (1) Find the unique func=
tion A(x) that satisfies: A =3D 1 + xAB=20
B =3D 1 + xABC=20
C =3D 1 + xABCD=20
D =3D 1 + xABCDE=20
E =3D 1 + xABCDEF ...where A=3DA(x), B=3DB(x), C=3DC(x), ... are an infinit=
e set of functions with integer coefficients. -----------------------------=
---------------------  It can be easily shown that the following expression=
A =3D 1/(1 - xB)=20
AB =3D 1/(1 - xB - xC)=20
ABC =3D 1/(1 - xB - xC - xD)=20
ABCD =3D 1/(1 - xB - xC - xD - xE)=20
ABCDE =3D 1/(1 - xB - xC - xD - xE - xF)=20
...
(3)
A =3D 1 + xB/(1 - xB)=20
B =3D 1 + xC/(1 - xB - xC)=20
C =3D 1 + xD/(1 - xB - xC - xD)=20
D =3D 1 + xE/(1 - xB - xC - xD - xE)=20
E =3D 1 + xF/(1 - xB - xC - xD - xE - xF)=20
...
(4)
B =3D (1 - 1/A)/x*(1)=20
C =3D (1 - 1/B)/x*(1 - xB)=20
D =3D (1 - 1/C)/x*(1 - xB - xC)=20
E =3D (1 - 1/D)/x*(1 - xB - xC - xD)=20
F =3D (1 - 1/E)/x*(1 - xB - xC - xD - xE)=20
...
(5)
A =3D 1 + x/(1 - xB - xC)=20
B =3D 1 + x/(1 - xB - xC - xD)=20
C =3D 1 + x/(1 - xB - xC - xD - xE)=20
D =3D 1 + x/(1 - xB - xC - xD - xE - xF)=20
...This last set of expressions (5) forms a glorified continued fraction,an=
d admits a unique solution to A(x).  Perhaps some variation of these expres=
----__JWM__J41c6.167eS.2781M

<html><P>Seqfans,<BR>    I have verified that the first 200 =
terms (at least) agree.<BR>Also, empirically verified, I found the formula:=
</P>
<P>B(x) =3D A(x*A(x)) <BR>C(x) =3D B(x*A(x))<BR>D(x) =3D C(x*A(x))<BR>E(x) =
=3D D(x*A(x))<BR>...<BR>Can the above be easily proven from the property:<B=
R>  A(x) =3D 1 + x*A(x)*A(x*A(x))<BR>where x*A(x) is the g.f.&nbs=
p;of A030266?<BR> <BR>Paul<BR><BR>-- "Paul D. Hanna"&nb=
p;  After rewriting my program, I confirm that my original observation=
<DIV>Please reconsider this nice problem.  </DIV>
<DIV> </DIV>
<DIV>It is quite surprising that A030266: </DIV>
<DIV>"Shifts left under COMPOSE transform with itself."</DIV>
<DIV>
<DIV>[0, 1, 1, 2, 6, 23, 104, 531, 2982, 18109, 117545, ...]</DIV>
<DIV>should have the generating function A(x) defined by (1) belo=
w! </DIV>
<DIV> </DIV></DIV>
<DIV>I have confirmed that the g.f. of A030266 agrees with x=
*A(x)  </DIV>
<DIV>up to 150 initial terms. </DIV>
<DIV>It would be nice to have a proof! </DIV>
<DIV> </DIV>
<DIV>Problem:</DIV>
<DIV>-------------------------------------------------- </DIV>
<DIV>(1) </DIV>
<DIV>
<DIV>Find the unique function A(x) that satisfies: </DIV>
<DIV>A =3D 1 + xAB <BR>B =3D 1 + xABC <BR>C =3D 1 + xABCD <BR>D =3D 1 + xAB=
CDE <BR>E =3D 1 + xABCDEF </DIV>
<DIV>...</DIV>
<DIV>where A=3DA(x), B=3DB(x), C=3DC(x), ... are an inf=
inite set </DIV>
<DIV>of functions with integer coefficients. </DIV>
<DIV>-------------------------------------------------- </DIV>
<DIV> </DIV>
<DIV>It can be easily shown that the following expressions </DIV>
<DIV>are equivalent to the original problem (1).  </DIV>
<DIV>(2)<BR>A =3D 1/(1 - xB) <BR>AB =3D 1/(1 - xB - xC) <BR>ABC =3D 1/(1 - =
xB - xC - xD) <BR>ABCD =3D 1/(1 - xB - xC - xD - xE) <BR>ABCDE =3D 1/(1 - x=
B - xC - xD - xE - xF) <BR>...<BR>(3)<BR>A =3D 1 + xB/(1 - xB) <BR>B =3D 1 =
+ xC/(1 - xB - xC) <BR>C =3D 1 + xD/(1 - xB - xC - xD) <BR>D =3D 1 + xE/(1 =
- xB - xC - xD - xE) <BR>E =3D 1 + xF/(1 - xB - xC - xD - xE - xF) <BR>...<=
BR>(4)<BR>B =3D (1 - 1/A)/x*(1) <BR>C =3D (1 - 1/B)/x*(1 - xB) <BR>D =3D (1=
(1 - 1/E)/x*(1 - xB - xC - xD - xE) <BR>...<BR>(5)<BR>A =3D 1 + x/(1 - xB -=
<DIV>This last set of expressions (5) forms a glorified continued fraction,=
</DIV>
<DIV>and admits a unique solution to A(x). </DIV>
<DIV> </DIV>
<DIV>Perhaps some variation of these expressions would produce </DIV>
<DIV>other significant sequences... </DIV>
<DIV>  </DIV>
<DIV>Thanks,</DIV>
<DIV>     Paul</DIV></html>

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