# [seqfan] Re: Closed Form Solution to Special Diff Eq?

Paul D Hanna pauldhanna at juno.com
Mon Mar 2 07:33:19 CET 2009

```Max (and Seqfans),
You are correct - the differential equation is for A155200.
By expressing the g.f. as a curious diff. eq., I was hoping to provoke some nice insight.

I appreciated Andrew Hone's analysis in this regard.
It is clear that g.f. A(x) is divergent for all x, yet there may still be a nice formula, perhaps as a solution to some functional equation or something like a(n) = [x^n] F(x)^(2^n).

As an example, I point out a formula (cf. A014070) I found a year ago:
Sum_{n>=0} log(1+2^n*x)^n/n!  =  Sum_{n>=0} C(2^n,n)*x^n.
It was likewise surprising to me that this sum was a power series in x with integer coefficients.
This can be proved using Stirling numbers of the first kind (A008275), since:
"E.g.f. for m-th column (unsigned): ((-ln(1-x))^m)/m!"
leads to a nice formula C(2^n,n) for the coefficient of x^n.

The dual to the above series is the g.f. of A155200:
exp( Sum_{n>=1} 2^(n^2)*x^n/n ) = 1 + 2x + 10x^2 + 188x^3 +...
which I conjecture to be a power series with integer coefficients.
I had hoped for a nice term-wise formula for this series as well.

Max, to answer your question: no, I do not yet have a proof for the claim:
"for m integer, exp( Sum_{n>=1} m^(n^2)*x^n/n ) is a power series in x with integer coefficients."
Here is where I should have exercised more caution and stated this as a conjecture.
Though this series generates integer coefficients for hundreds of initial terms for various integer m, and is very likely true, it is not established by proof.
In the excitement of discovery, and with the desire to point out what is interesting about the series, I was not as rigorous in my wording as I should have been.
Unless I find a proof soon, I will need to edit the related sequences to indicate that the statements are conjectures.

So I am in pursuit of a proof.
If we look at the differential equation:
2*A(x)*A(4x) + 8*x*A(x)*A'(4x) - A'(x)*A(4x) = 0
one can derive the following statements.

2*A(x)*A(4x) + 8*x*A(x)*A'(4x) = 2*sum(k=0,n,4^k*(k+1)*a(k)*a(n-k))
and
A'(x)*A(4x) = sum(k=0,n,4^k*(n-k+1)*a(k)*a(n-k+1))

This leads to an alternate recurrence for A155200:

n*a(n) = 2*a(n-1) + Sum_{k=1..n-1} 4^k*a(k)*[2(k+1)*a(n-1-k) - (n-k)*a(n-k)]  for n>0, with a(0)=1.

However, this does not seem to simplify the problem.
a(n) = (1/n)*Sum_{k=1..n} 2^(k^2)*a(n-k) for n>0 with a(0)=1.

Should anyone see how to prove that the terms {a(n)} of A155200 are always integer, please let me know!

In the meantime, I will plan on making comments indicating that a proof is needed.
Thanks,
Paul

-- Max Alekseyev <maxale at gmail.com> wrote:
On Fri, Feb 27, 2009 at 5:40 AM, Paul D Hanna <pauldhanna at juno.com> wrote:
> Seqfans,
>      Is there a closed form for the solution A(x) to this differential equation?
>
> 2*A(x)*A(4x) + 8*x*A(x)*A'(4x) - A'(x)*A(4x) = 0
>
> A power series solution exists, but can it be expressed finitely in terms of known functions?

I guess you mean
http://www.research.att.com/~njas/sequences/A155200
that we already discussed in SeqFan a couple of weeks ago.

There is probably no closed-form expressing for the function A(x) but
its coefficients can be expressed in terms of complete Bell
polynomials:
a(n) = B_n( 0!*2^(1^2), 1!*2^(2^2), 2!*2^(2^3), ..., (n-1)!*2^(2^n) ) / n!

btw, could you please point out a proof for the following comment?
%C A155200 More generally, for m integer, exp( Sum_{n>=1} m^(n^2) *
x^n/n ) is a power series in x with integer coefficients.

Thanks,
Max

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