# [seqfan] Re: Integer Sequence Analysis in Mathematica 7

Mitch Harris maharri at gmail.com
Thu Nov 20 19:43:45 CET 2008

```First, thanks, Eric, for responding so quickly and informatively.

On Thu, Nov 20, 2008 at 12:01 PM, Eric W. Weisstein <eww at wolfram.com> wrote:
> On Thu, 20 Nov 2008, Mitch Harris wrote:
>
>>> make it possible to take lists of sequence elements and
>>> systematically find large classes of closed-form Mathematica formulas
>>
>> systematically -guess-, right? (for a finite list)
>
> Technically yes, it is a guess.  But:
>
> 1) Note that FindSequenceFunction begins with Find*, and all Find*
> functions in Mathematica are intended to be used on inputs that do not
> necessarily have a unique (or any) solution
>
> 2) A result is only returned if it can be validated on some number terms
> beyond those that were used to infer the functional form:

Oh. Probably mentioned in the docs, and I never caught the common
factor in practice.

>> Sorry to turn back to the negative comments but (maybe Eric can pass
>> this along to the mma
>
> (Just a brief note for anyone interested that for certain historical
> reasons, people in my neck of the woods prefer writing out Mathematica or
> using M- (or M--, I actually can't remember which ;) instead of the
> abbreviations mma or Mma.)

OK. (Of course I just looked for this and unfortunately google -can't-
find "M--" or "M-" because you can't search for non-alphabetic

>> designers), look at that first example for the determinant of a Hilbert
>> matrix. How would one normally write that? With products of factorials.
>> But the mma solution uses these special functions, Glaisher, BarnesG.
>
> I assume you are referring to this example?
>
> In[2]:= l = Table[1/Det[HilbertMatrix[n]], {n, 10}]
> Out[2]= {1, 12, 2160, 6048000, 266716800000, 186313420339200000, \
> 2067909047925770649600000, 365356847125734485878112256000000, \
> 1028781784378569697887052962909388800000000, \
> 46206893947914691316295628839036278726983680000000000}
>
> In[3]:= FindSequenceFunction[l][n]
> Out[3]= (2^(-(1/12) - n + 2*n^2)*Glaisher^3*BarnesG[1/2 + n]*
>    BarnesG[3/2 + n])/(Pi^n*(E^(1/4)*BarnesG[1 + n]^2))

exactly.

>> First time I've -ever- seen these.
>
> They're actually quite handy; see e.g.,

OK. I should have googled before complaining!

>> And lots of other manipulations that by hand would result in factorials
>> and binomials, through mma result in opaque Gammas with sqrt(pi)-like
>> coefficients. I understand that mma might be designed in the direction
>> of people having more facility with analysis than me, but, hey, if a
>> summation/product is not closed-form, then replacing it with a special
>> function name is not really either,
>
> Actually, it's quite beautiful that Mathematica now has Hyperfactorial and
...
> concise/shorter Mathematica representations that could be added).

OK. OK. I'm convinced!

> A symbolic math program can't (or at least doesn't want to) return
> unevaluated products because you can't *do* anything with them; they're
> almost dead to further symbolic analysis.

I sort of agree sometimes, and then again sometimes not. A discussion
for another place...

>> and the summation/product will say more about the combinatorics anyway.
>
> That I'm not so sure about, though I have no trouble believing there could
> be situations where it was the case.  Can you by any chance give such an
> example?

Knowing that something is a summation or factorial will tell you more
than the opaque function name, unless you already know the meaning of
that function name.

The classic example is the Fibonacci function. Consider its definition
as a recurrence, which can be expanded into a simple (recurrent)
summation. That recurrence (or summation) explains the combinatorics
right away, you can read it off almost directly: binary strings with
no adjacent 1's - if the last char is 0, then F_{n-1}, if the last is
1, the second to last must be 0, so F_{n-2}.

But if you're given the Binet solution, defining F_n = (phi^n -
1/phi^n)/sqrt(5), then you don't know what it -means- (but, yes, you
can get asymptotics and other properties easier).

(if it helps I'm really coming from a combinatorics perspective -
Graham-Knuth-Patashnik, Concrete Mathematics,  or even better,
Benjamin-Quinn, Proofs that Count)

> I do understand your points.  But if I haven't been able to shed some
> light onto why symbolic "super"-functions are/could be worth the cost of
> having to learn about them, I'm certainly happy to discuss further.

I have no problem in principle with replacing a convoluted form with a
name and short list of params. Maybe it was just the shock of an
unrecognized notation, that I was embarrassed to not know.

--
Mitch Harris

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