Sum of reciprocals of double factorials

[Jonathan Post]

I am a big fan of my Caltech co-alumnus Eric W. Weisstein, have told
him so face-to-face, and of his employer Steve Wolfram.  I have made
some contribution to 19 pages of MathWorld.

So I am delighted that my 4 Aug 2008 email to seqfans on sums of
reciprocals of double factorials duplicated what Mathematica
developers were independently doing in gerater depth.  I suppose that
Max Alekseyev independently arrived at the lovely formula.

Is it now useful for me to submit the two sequences of numerators and
denominators of convergents, hotlinking to the new MathWorld page?

On Wed, Aug 6, 2008 at 4:26 PM, Max Alekseyev <maxale at gmail.com> wrote:
> Eric,
>
> I think, at MathWorld it's better to give series identities rather
> than just evaluation at x=1.
>
> That is:
> sum_(n=0)^infty x^n/n!! = exp(x^2/2) [1 + sqrt(pi/2) erf(x/2sqrt(2))]
> etc.
>
> Maybe, it also makes sense to split it into subcases:
>
> sum_(n=0)^infty x^(2n)/(2n)!! = exp(x^2/2)
> sum_(n=0)^infty x^(2n+1)/(2n+1)!! = exp(x^2/2) sqrt(pi/2) erf(x/2sqrt(2))
> etc.
>
> Regards,
> Max
>
> On Wed, Aug 6, 2008 at 4:02 PM, Eric W. Weisstein <eric at weisstein.com> wrote:
>> On Wed, 6 Aug 2008, Jonathan Post wrote:
>>
>>> Perhaps it is now worth submitting the two sequences of numerators and
>>> denominators of the convergents to this real number, given that Eric
>>> Wesstien has added to MathWorld and OEIS with Max Alekseyev's lovely
>>> formula:
>>>
>>> Weisstein, Eric W. "Double Factorial." From MathWorld--A Wolfram Web
>>> Resource. http://mathworld.wolfram.com/DoubleFactorial.html
>>>
>>> "The sum of reciprocal double factorials can be given in closed form as
>>> sum_(n=0)^infty1/(n!!)=sqrt(e)[1+sqrt(pi/2)erf(1/2sqrt(2))]"
>>> and
>>>
>>> (Sloane's A143280).
>>
>> Turns out the development version of Mathematica gets that one out of the
>> box.
>>
>>> This leaves open the analogous:
>>>
>>> sum_(n=0)^infty1/(n!!!)
>>
>> That's a bit trickier.  Mathematica doesn't get it quite out of the box, but
>> with some human help, the result is
>>
>> 1/3 E^(1/3) (3 + 3^(2/3) Gamma[2/3] + 3 3^(1/3) Gamma[4/3] -
>>   3^(1/3) Gamma[1/3, 1/3] - 3^(2/3) Gamma[2/3, 1/3])
>>
>> (probably can be simplified somewhat)
>>
>> == 3.2989135380884190034012178082614697690778036956832...
>>
>>> sum_(n=0)^infty1/(n!!!!)
>>>
>>> sum_(n=0)^infty1/(n!!!!!)
>>>
>>> etcetera, and their convergents.
>>
>> Ah; what's very nice about the above formula for the n == 3 case is that you
>> can immediately see the pattern for the general case:
>>
>> ReciprocalFactorialSumConstant[n_] :=
>>  1/n Exp[1/n] (n + Sum[n^(k/n) Gamma[k/n, 0, 1/n], {k, n - 1}])
>>
>> In[2]:= ReciprocalFactorialSumConstant[2] // FullSimplify
>> Out[2]= Sqrt[E] + Sqrt[(E \[Pi])/2] Erf[1/Sqrt[2]]
>>
>> In[3]:= ReciprocalFactorialSumConstant[3]
>> Out[3]= 1/3 E^(1/3) (3 + 3^(1/3) Gamma[1/3, 0, 1/3] +
>>   3^(2/3) Gamma[2/3, 0, 1/3])
>>
>> In[4]:= N[%, 20]
>> Out[4]= 3.2989135380884190034
>>
>> In[5]:= ReciprocalFactorialSumConstant[4]
>> 1/4 E^(1/4) (4 + Sqrt[2] Gamma[1/4, 0, 1/4] + 2 Gamma[1/2, 0, 1/4] +
>>   2 Sqrt[2] Gamma[3/4, 0, 1/4])
>>
>> In[6]:= N[%, 20]
>> Out[6]= 3.4859449774535577452
>>
>> etc.
>>
>> Cheers,
>> -Eric
>>
>>> On Mon, Aug 4, 2008 at 12:41 PM, Jonathan Post <jvospost3 at gmail.com>
>>> wrote:
>>>>
>>>> Very elegant, Max Alekseyev!
>>>>
>>>> The generalization to k-tuple factorials is similar, and the
>>>> convergents require a modulo k adjustment?
>>>>
>>>> In my earlier email, of course I meant "What are the real numbers to
>>>> which the sum of reciprocals of k-tuple
>>>> factorials converges?"
>>>>
>>>> With your formulae, is this now worth submitting, as definitive rather
>>>> than merely an analogue of existing seqs?
>>
>




Although marked as "easy", I am failing to understand A118960,
http://research.att.com/~njas/sequences/?q=id:A118960|id:A001750
In particular the following questions could be asked:
- how can 2+1/2 (a fraction) be partitioned into primes?
- Doesn't the formula indicate this is a duplicate of  A006254,
- Is this an erroneous version of A001750 with a typo at a(0), 1 for 10?

Richard