2*(A056855(n))/(phi(n)*n)
Benoit Cloitre
abcloitre at wanadoo.fr
Sat Nov 13 16:55:19 CET 2004
It is true from Leudesdorf theorem page 101 of Hardy & Wright fifht
edition (the last line) . If I'm not wrong more is true :
12 *A056855(n)/(phi(n)*core(n)*n) is an integer where
core(n)=squarefree part of n.
12 *A056855(n)/n^2 is an integer
It seems interesting to have more properties of those numbers
S(n)=sum(1/k, 1<=k<=n, (k,n)=1)
writing :
P(n)=prod(k, 1<=k<=n, (k,n)=1) (A001783)
(i) one has denominator of S(n) divides P(n) for any n and denominator
of S(n)=P(n) iff n is a very round number (A048597)
(ii) S(n) and Harmonic numbers H(n)=sum(1/k,1<=k<=n) are related by
this property :
n doesn't divide the {denominator of H(n)-Q(n)} if and only if
{numerator of n*(H(n)-1)} - {numerator of H(n)} + {denominator of
H(n)} >0
Those n's are given by A074791 : 6,18,20,21,33,42,54,63,66,77,...
Benoit Cloitre
>> 2 * (A056855(n)) /(phi(n)*n)
>> ...
>> 2, 1, 1, 1, 5, 1, 84, 11, ...
>> (And this sequence is not in the EIS, but I am submitting the few
>> terms
>> above.)
>
> Here's more.
>
> %S Axxxxxx 2,1,1,1,5,1,84,11,184,15,193248,23,19056960,833,33740,64035,
> %T Axxxxxx 520105017600,2473,130859579289600,203685,963513600,23748417,
> %U Axxxxxx 16397141420298240000,645119,555804546402631680,8527366575
>
>
>> I asked this question years ago, but is
>> 2 * (A056855(n)) /(phi(n)*n)
>> always an integer?
>>
>> I conjecture it is.
>
> Well, the first 5000 terms are integers.
> And the conjecture is true whenever N and phi(N)/2 are coprime.
> But that leaves much in doubt.
>
> --
> Don Reble djr at nk.ca
>
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