# [seqfan] Re: A Remarkable Eigenfunction - A179497

Paul D Hanna pauldhanna at juno.com
Mon Aug 2 08:16:48 CEST 2010

```SeqFans,
Here is a related series expression for A(x) using nested derivatives.
Using abbreviations A=A(x), A_n = n-th iteration of A(x), and operator D(F)=d/dx F,
then A satisfies the series:

A = x + A^2 + A^2*D(A^2)/2! + A^2*D(A^2*D(A^2))/3! + A^2*D(A^2*D(A^2*D(A^2)))/4! +...

Likewise, the n-th iteration of A may be expressed in a similar fashion using n as an argument:

(**)   A_n = x + n*A^2 + n^2*A^2*D(A^2)/2! + n^3*A^2*D(A^2*D(A^2))/3! + n^4*A^2*D(A^2*D(A^2*D(A^2)))/4! +...

which holds for real n as well as intergers.  I find this appealing.

Here is some rudimentary PARI code for (**):
A_(k,A)=local(D=x);x+sum(n=1,#A,k^n*(D=A^2*deriv(D))/n!)
where A_(k,A) returns the k-th iteration of A.

>From the above series (**), is it any easier now to see why the relation
[A_{n+1}(x)]^2 = A(x)^2 * A_n'(x)
holds?

We may write:
A^2*A_n' = A^2 + n*A^2*D(A^2) + n^2*A^2*D(A^2*D(A^2))/2! + n^3*A^2*D(A^2*D(A^2*D(A^2)))/3! + n^4*A^2*D(A^2*D(A^2*D(A^2*D(A^2))))/4! +...

but it is still not readily apparent to me that this sum should equal [A_{n+1}(x)]^2.

At any rate, does the series (**) lead to any nice formulas?

Thanks,
-- Paul

---------- Original Message ----------
From: "Paul D Hanna" <pauldhanna at juno.com>
To: seqfan at list.seqfan.eu
Subject: [seqfan]  A Remarkable Eigenfunction - A179497
Date: Sun, 1 Aug 2010 13:58:48 GMT

SeqFans,

Consider the eigenfunction described by:

E.g.f. satisfies: A(A(x))^2 = A(x)^2 * A'(x)

[http://www2.research.att.com/~njas/sequences/A179497].

Explicitly, the series begins:

A(x) = x + 2*x^2/2! + 18*x^3/3! + 312*x^4/4! + 8240*x^5/5! + 297000*x^6/6! + 13705776*x^7/7! + 776778688*x^8/8! + 52511234688*x^9/9! + 4143702216960*x^10/10! +..

Let A_n(x) denote the n-th iteration of A(x) where A_{n+1}(x) = A_n(A(x)) with A_0(x)=x;

then the pattern continues:

[A_3(x)]^2 = A(x)^2 * A_2'(x),

[A_4(x)]^2 = A(x)^2 * A_3'(x),

[A_5(x)]^2 = A(x)^2 * A_4'(x), ...

so that the iterations of A(x) satisfy:

(*) [A_{n+1}(x)]^2 = A(x)^2 * A_n'(x)   for all n.

I find formula (*) to be remarkable, even though the proof for n>0
is an easy exercise in induction and the chain rule.

Surely A(x) must have some other nice properties.

What else can be said about A(x)?

Paul

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