[seqfan] Re: A000134

[David Wilson]

Almost suggesting a sequence.

----- 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 6:07 PM
Subject: [seqfan] Re: A000134


> 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.
>>>>>>
>>>>>>
>>>>>>
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