# [seqfan] Re: A000134

David W. Cantrell DWCantrell at sigmaxi.net
Fri Aug 28 23:00:24 CEST 2009

```Heuristics would, I think, suggest that your conjecture is correct,
and that there are infinitely many exceptions. Another exception is
n = 427422694871; whether that's the smallest exception > 2,
I don't know.

If you want a simple formula for which heuristics would, I think,
suggest that there should be no exceptions, try

a(n) = round( pi (n - 1/4) + 1/(8 pi (n - 1/4)) )

David W. Cantrell

----- Original Message -----
From: "David Wilson" <dwilson at gambitcomm.com>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Sent: Friday, August 28, 2009 17:37
Subject: [seqfan] Re: A000134

>I would conjecture that there are other (very sparse) exceptions. If
>we
> knew the exceptions very far out, it would be an easy compute:
>
> a(n) =
> round(pi*(n-1/4))+1 if n is an exception
> round(pi*(n-1/4) otherwise.
>
> Robert Israel wrote:
>> Maple confirms this, and will give you as many additional terms as
>> you
>> wish:
>>
>>> B:= BesselJ(0,n-1/2)*BesselJ(0,n+1/2);
>>
>>> map(combine, asympt(B,n,4));
>>
>> (cos(1)+sin(2*n))/Pi/n-1/4*cos(2*n)/Pi/n^2+1/32*(-sin(2*n)-4*sin(1))/Pi/n^3+O(1/(n^4))
>>
>> Cheers,
>> Robert Israel
>>
>> On Thu, 27 Aug 2009, Gerald McGarvey wrote:
>>
>>> This inspired me to look at the shape of the Bessel function of
>>> order 0,
>>> in PARI:    b(n) = besselj(0,n-1/2)*besselj(0,n+1/2)
>>>
>>> The function b(n) appears to asymptotically approach the following
>>> function:
>>>
>>> f(n) = 1/(n*Pi) * (sin(2*n) + cos(1))
>>>
>>> The difference between these functions also looks like a damped
>>> sinusoid,
>>> around   1/(4*n^2*Pi)*cos(2*n)
>>>
>>> Regards,
>>> Gerald McGarvey
>>>
>>> At 10:03 PM 8/26/2009, David Wilson wrote:
>>>> For most n, A000134(n) = round(pi*(n-1/4)). For 1 <= n <= 1000,
>>>> the only
>>>> exception is n = 2.
>>>>
>>>>
>>>>
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```