[seqfan] Re: a curiosity, about A017979
Robert Israel
israel at math.ubc.ca
Mon Dec 28 01:24:55 CET 2009
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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