[seqfan] Re: Help with A002793 needed/Karol A. Penson

[Max Alekseyev]

The ogf also implies the explicit formula:

For n>1,
A002793(n) = \sum_{k=1}^n (k+1) * A058006(k-1) * binomial(n,k) * (n-1)! / (k-1)!

Max

On Mon, Jul 5, 2010 at 3:16 PM, Max Alekseyev <maxale at gmail.com> wrote:
> The exponential generating function for A002793 is
>
> ( E_1(1) - E_1(1/(1-x)) ) * exp(1/(1-x)) / (1-x)
> = ( Ei(-1/(1-x)) - Ei(-1) ) * exp(1/(1-x)) / (1-x)
>
> where E_1(x) and Ei(x) are the exponential integrals
> http://mathworld.wolfram.com/ExponentialIntegral.html
>
> Max
>
> On Mon, Jul 5, 2010 at 1:02 PM, Karol PENSON <penson at lptl.jussieu.fr> wrote:
>> Can anyone help with the following three questions concerning A002793:
>>
>>     1.  is the formula for a(n) known ?
>>     2.  is any generating function (ogf , egf ... etc. ) of a(n) known ?
>>     3.  in one of my calculations the following splitting of a(n)'s
>> appear :
>>
>>                   a(2)=4    =3+1
>>                   a(3)=20  =11+8+1
>>                   a(4)=124=50+58+15+1
>>                   a(5)=920=274+444+177+24+1,
>>                         etc.
>>         I would be happy to obtain the formula for the triangle.
>>
>>       Thanks in advance ,    Karol A. Penson
>>
>>
>>
>>
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