Asymptote of A003242

[Christian G. Bower]

sigma(z) is the g.f. of A048272 (inverse invert transform of A003242)
From: hv at crypt.org
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Subject: Asymptote of A003242
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Subject: Re: Asymptote of A003242
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Alec Mihailovs wrote:
> From: <hv at crypt.org>
>
>> By my calculation, A003242(n) should tend to k^n, where k is given by:
>>  1 = sum_i=1^inf{ 1/(k^i + 1) }
>
> I'm not sure that it is helpful, but I noticed the following (easy to 
> prove) identity,
>
> sum(1/(k^i + 1), i=1..infinity)=-sum((-1)^i/(k^i - 1), i=1..infinity)
>
> Alec Mihailovs
>
I think this is equivalent to

1 = sum_i=1^inf{A048272(i) / k^i}

We might hope beyond hope that this latter series inverts nicely