The mysterious result while searching a formula for A165343

[Thomas Scheuerle]

Dear SeqFan list,

The sequence A165343  is the Hankel transform of  (n!)^2.

We se the comment: 

"It would be highly desirable to obtain a closed form"

This motivated me to look into this.

I have a method which was successful in many cases when I search for an underlying formula in a Hankel transform based sequence.

The first step is to find the Stieltjes continued fraction expansion of (n!)^2.

With help of PARI I see here:

1/(1-x/(1-3*x/(1-(20/3)*x/(..)))

1, 3, 20/3, 164/15, 3537/205, 127845/5371, ...

The next step is to multiply these numbers in pairs and take the denominators of the resulting fractions. If these are all square numbers it means you are with 

this method on the right track and should look on the square roots of these.

Indeed we see square numbers (be careful not far checked yet!). We take the roots.

This gives us the mystery sequence:

1, 1, 3, 5, 41, 131, 947, 6197, 116499, 1622749, 78523849, ...

The next step is finding a recursion that defines this sequence.

This can be a surprising and nonlinear thing like for example Somos sequences

but also something completely different ...

Maybe you have ideas how to find such recurrences here ?  

Here at this point I decided to ask this list as I can imagine to learn new methods this way, may search for some recurrence was not yet successful with some limited effort yet.

Finally the expectation would be that the numbers in A165343 can be factored

with help of the terms from this sequence.

This seems to be indeed the case:

1, 3, 656, 58910976, 

we see the factors

   3;  2, 41; 2,3,947;  ....    

but there are also sometime others, this is unfortunately a red flag and not a good sign for this method of formula guessing ...

I wish all here a happy and successful 2026

All the best

Thomas

  

[N A]

I went for a completely different route in trying to find a related sequence for n!.

I thought of that as a generating function, and I'll show it here so hopefully it's useful to you:

Here I'm calling the derivative "D" and working with a formal power series in R[[x]].

Start by thinking of each term as n!x^n.

The goal is to find a closed form for f(x) = Sum{n>=0}n!x^n

Take a_n = n!, then a_(n+1) = (n+1)! = (n+1)n!.

So a_(n+1) = a_n(n+1)

Notice you can multiply by x^n on both sides to get

a_(n+1)x^n = a_n(n+1)x^n.

Sum from n = 0 to infinity to get:

Sum{n>=0}a_(n+1)x^n = Sum{n>=0}a_n(n+1)x^n

On the left, this is a_1 + a_2*x + a_3*x^2 + ...

If you add in (a_0 - a_0), you'd get f(x)/x. So LHS = (f(x) - 1)/x.

On the right, this looks like the derivative of (f(x) * x), so RHS = D(x(f(x)) = xf'(x) + f(x).

You now have (f(x)-1)/x = xf'(x) + f(x), a differential equation! From here on it's standard ODE solving.

I imagine for you, it's going to involve the extra step of taking the Hankel transform of whatever function you found.

[Neil Sloane]

When you post a message to the list, if your name isn't obvious from your email address, would you please sign your name, just as a matter of common courtesy?

Best regards

Neil 

Neil J. A. Sloane, Chairman, OEIS Foundation.

Also Visiting Scientist, Math. Dept., Rutgers University, 

Email: njas...@gmail.com

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[N A]

Ohh sorry about that, its just my initials that made a joke of being "Not Availlable".

Signed,
Nour Abouyoussef

[Thomas Scheuerle]

Thank you Nour,

I agree with you.

My main problem here is that I do not know a simple rule how to apply the Hankel transform onto a generating function when it is in standard form.

But if the generating function describes the function by continuous fraction expansion, then indeed I know how to transform it.

Now I must admit inside A001044 is a form of generating function provided by Sergei N. Gladkovskii which would almost do the trick here.

But frankly I simply failed to transform it into such a form that I could easily work further by spending small effort here.

So maybe I simply was looking for a shortcut here and should work on this provided generating function instead.

All the best

Thomas

[Thomas Scheuerle]

I also should mention here the relation.

If the ordinary generating function is the continued fraction expansion 1/(1-b*x/(1-c*x/(1-d*x/(...))))

then its Hankeltransform is:

1,  c*b,  e*d*c^2*b^2,  g*f*e^2*d^2*c^3*b^3, ... 

This means if you can give a recurrence relation for this continued fraction expansion  than 

you can give also a recurrence for its Hankeltransform.

[N A]

Fascinating... I expect that since the structure is rather nice that there will be a continued fraction just like for n!. But if there wasn't, might I suggest this:

Call f(x) = whatever the generating function for Sum{n>=0} n!^2*x^n is.

J_1(kx) = Bessel Function of the First kind of order zero evaluated at kx.

j(n) = nth term of the formal power series of J_1(kx).

With the goal of F(k) = Integral{0, Inf} (f(x)J_1(kx)xdx),

Maybe you could try expanding J_1(kx)*f(x) as another formal power series:
Integral{0, Inf} (Sum{n>=0} j(n)f(x)xdx), and work from there to find a closed form.

What might trip us up here is how we handle the kx evaluation in trying to get the power series of J_1(kx).

If that doesn't work, message me and we could grind through the math together?

Please let me know if this is correct.

Nour Abouyoussef

[N A]

Also, I noticed that while trying to generalize for n!^z for natural z,
There's a rather elegant differential equation:
f_z(x) = whatever the generating function for Sum{n>=0} n!^z*x^n is

f_z(x) - 1 = (xD)^z(xf_z(x))
By way of noticing that the recurrence relation:

a_(n+1) = (n+1)^z*a_n,

which with the same sort of building the differential equation as before yields an iterated Euler operation on f_z(x)*x on the RHS and the same sort of (f_z(x)-1)/x on the LHS.

So I wouldn't expect the solution for z to be too bad to find. Transforming them though, that is another matter.

[N A]

f_z(x) - 1 = x*(xD)^z(xf_z(x))

Missed an x* here.

[Thomas Scheuerle]

It is difficult to say anything about patterns in the continued fraction expansion of a transcendental function.

While for algebraic functions there is always hope to describe it,

for transcendental function it could be anything between very easy and beyond human understanding.

As a rule of thumb the difficulty of describing the continued fraction expansion is on the same level

than describing rational points of the underlying function or curve. This shows that slight changes in the function

can highly impact the difficulty of describing it.

    

This motivates me now to give up and go for other problems first.

Thank you very much for your help here.

All the best

Thomas

[jp allouche math]

Dear all

The OEIS is referring (e.g., for sequence A302117) to a preprint of Travis Sherman
entitled "Summation of Glaisher- and Apery-like Series".

Though the initial webpage where it was accessible does not work any longer,
it is still possible to obtain the preprint at, e.g,
https://web.archive.org/web/20201001000000*/https://math.arizona.edu/~rta/001/sherman.travis/series.pdf

But Page 12 of that Preprint is incomplete (misprinted). Does anyone have a full version?
or does anyone know how to contact the author?

Many thanks in advance
best 
jean-paul

[Thomas Scheuerle]

It looks like all PDF versions out there have the same problem.

I searched for tex or postscript but did not find any such file.

Maybe the e-mail address in last page of https://arxiv.org/pdf/1606.07178 is valid as this is more recent.

[jp allouche math]

Thank you very much Maximilian and Thomas!
I will try this (somehow strange?) email address.

best wishes
jean-paul

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