# [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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>
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>

```