[seqfan] Re: a curiosity, about A017979

[Maximilian Hasler]

1e-35 corresponds to about 13 terms
(exp(2 pi * 13)  > 10^35).

The 16 sequences starting with
1,1,1,2,2,3,4,5,6,8,10,12,15
are even closer to the same value, 2^(-3/8) exp(-23 pi/12)

So IMO this is just a coincidence.

Happy holidays & best wishes for the next year, nevertheless!

Maximilian

On Sun, Dec 27, 2009 at 8:24 PM, Robert Israel <israel at math.ubc.ca> wrote:
> Just to clarify, Simon is using an offset of 1 rather than 0 here, i.e.
>
>> Digits:= 50:
>   Q1:=evalf(Sum(floor(2^((n-1)/3))/exp(2*Pi*n),n=1..infinity));
>
>       Q1 := .18709366108449968820887622505118307225836270463014e-2
>
>> Q2:= evalf(1/2*2^(5/8)/exp(23*Pi/12));
>
>       Q2 := .18709366108449968820887622505118273509656597828342e-2
>
>> Q1-Q2;
>
>       .33716179672634672e-35
>
>
>
> Cheers,
> Robert Israel
>
>
> On Sun, 27 Dec 2009, Simon Plouffe wrote:
>
>>
>> hello seqfan,
>>
>>  I have this curiosity,
>>
>>  The sequence A017979 when evaluated at exp(-2*Pi)
>> is equal to
>>
>>            5/8
>>          2
>>   1/2 ----------
>>           23 Pi
>>       exp(-----)
>>            12
>>
>> to a precision of 35 decimal digits.
>>
>> That is : sum(a(n)/exp(2*Pi*n),n=1..infinity), where a(n) = A017979(n).
>>
>>  Can anyone find why ?
>>
>> recall : A017979, is :
>>
>> %S A017979 1,1,1,2,2,3,4,5,6,8,10,12,16,20,25,32,40,50,64,80,101,
>> %T A017979 128,161,203,256,322,406,512,645,812,1024,1290,1625,2048,
>> %U A017979 2580,3250,4096,5160,6501,8192,10321,13003,16384,20642
>> %N A017979 Powers of cube root of 2 rounded down.
>>
>> Curious isn't ?
>>
>> Cheers and happy new year,
>>
>> simon plouffe
>>
>>
>>
>>
>>
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>>
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>>
>
>
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