[seqfan] Re: A000134

[David W. Cantrell]

FWIW, here are two more exceptions:

n = 10028055784215584668175827612  and n =

4770405441857974569822434308153988920977979936890

David W. Cantrell

----- Original Message ----- 
From: "David W. Cantrell" <DWCantrell at sigmaxi.net>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Sent: Friday, August 28, 2009 21:00
Subject: [seqfan] Re: A000134


> 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.
>>>>>
>>>>>
>>>>>
>>>>> _______________________________________________
>>>>>
>>>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>>
>>>>
>>>> _______________________________________________
>>>>
>>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>>
>>>
>>>
>>> _______________________________________________
>>>
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>>
>>
>>
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/