A Function With Integer Derivatives

[Joshua Zucker]

Mathematica gives the following array, for m and n ranging from 0 to 7:
{0, -4, 24, -156, -1280, 149310, -7843752, 353544128}, 
{-4, 0, -12, -1120, 75630, -3463488, 144427136, -4623524064}, 
{-24, -12, 0, 35190, -1189944, 50257088, -1355463936, -49084034760}, 
{-156, 1120, 35190, 0, 16857344, -299496096, -28958514120, 6219833558400}, 
{1280, 75630, 1189944, 16857344, 0, -14751996840, 2034251572320,
-287451033341208},
{149310, 3463488, 50257088, 299496096, -14751996840, 0,
-102849760407768, 8785417569185088},
{7843752, 144427136, 1355463936, -28958514120, -2034251572320,
-102849760407768, 0, -244837625707566528},
{353544128, 4623524064, -49084034760, -6219833558400,
-287451033341208, -8785417569185088, -244837625707566528, 0}

maybe that will give some useful raw material to someone looking for a
shortcut formula.  Mathematica seems to be chugging along recursively:
it takes it a LOT longer to get 7 terms than 5.

This is a(m,n) before doing the division that Leroy Quet suggests.

Here's a bit more: rows m 0 up to 10 and n 0 up to m-1 (I'm pretty
sure that if n=m we'll get 0, from the experience above.  Hm, I think
a little symmetry, x <-> -x, proves it, too).

{{}, 
 {-4}, 
 {-24, -12}, 
 {-156, 1120, 35190}, 
 {1280, 75630, 1189944, 16857344}, 
 {149310,  3463488, 50257088, 299496096, -14751996840},
 {7843752, 144427136, 1355463936, -28958514120, -2034251572320,
-102849760407768},
 {353544128, 4623524064, -49084034760, -6219833558400,
-287451033341208, -8785417569185088, -244837625707566528},
 {12951788544, -39979518120, -15260471661600, -834919888996248,
-31428255639545856, -825255100620000960, -2613396170121287040,
1960943389880952625920},
 {97448310360, -32445271457280, -2163949216098648, -94194093379299264,
-2872745555987396544, -25147264734463576320, 5015138004975012837120,
489113374200154302213120, 43366631911519919988898368},
 {-60643679896800, -5042617359465048, -252713330556291072,
-8947611784014671040, -131236732533674033280, 13382644936551140908800,
1719656925830167623106560, 132992759957958273958026048,
7204339562836877669403666432, 363560895137179119151555401600}}

Feel free to submit any parts of this that are useful to OEIS, Leroy,
and if you want more terms from Mathematica, let me know.  It's
getting pretty slow to calculate this far, though!

--Joshua Zucker


On Apr 4, 0005 11:05 AM, Leroy Quet <qq-quet at mindspring.com> wrote:
> Consider the function, defined for -1 < x < 1,
> 
> f_m,n(x) =
> 
> (1-x)^((1-x)^(-m)) (1+x)^((1+x)^(-n)),
> 
> where m and n are positive integers.
> 
> Now, the derivatives of f at x = 0 are all integers.
> 
> What I am interested in is the array, {a(m,n)}, where
> a(m,n) = the (m+n+1)th order derivative of f_m,n(x) at x = 0.
> 
> The reason I am interested in this derivative is that
> a(m,n) is always divisible by (m+n).
> 
> (Actually, more generally, the (j+1)th order derivative of f_m,n(x) is
> always divisible by
> GCD(m+n,j).)
> 
> So, could someone please calculate/submit the sequence of {a(m,n)} (read
> off by diagonals) and also submit {b(m,n)}, where b(m,n) = a(m,n)/(m+n)?
> 
> Is there a direct non-recursive way of calculating {a(m,n)} and {b(m,n)}
> without using calculus?
> (I have not actually calculated the terms myself, so this may be trivial
> to impossible.)
> 
> thanks much,
> Leroy Quet
> 
>