[seqfan] RE : Uphi(m)=UPhi(n)=3/4*(m*n)^(1/2)
koh
zbi74583.boat at orange.zero.jp
Fri Nov 7 02:04:57 CET 2008
yk> %N A000001 Numbers m,n such that UnitaryPhi(m)=UnitaryPhi(m)=3/4*(m*n)^(1/2), n<=m
Ricard wrote :
>There is an obvious typo in this definition (?!). I am also strongly in favor of including explicit OEIS A-number references to rare functions such as UnitaryPhi(.) into submissions of that kind (see the discussion in the seqfan list of Oct 2006).
I rewrote the definition.
%N A000001 Number m such that UnitaryPhi(m)=UnitaryPhi(n)=3/4*(m*n)^(1/2), n<=m
Is it better?
>I ran the following Maple prog and confirm that this list is complete in the range y<=x<=21760. There are no further values <= 21760 in the sequence. A Java version is available on request.
A047994 := proc(n) option remember ;
local ifacts;
ifacts := ifactors(n)[2] ;
mul( op(1,op(i,ifacts))^op(2,op(i,ifacts))-1,i=1..nops(ifacts)) ;
end:
x := 1:
while true do
uphix := A047994(x) ;
if uphix mod 3 = 0 then
for y from 1 to x do
uphiy := A047994(y) ;
if uphix^2 = uphiy^2 and uphix^2 = 9/16*x*y then
print(x,ifactors(x)[2],y,ifactors(y)[2]) ;
fi ;
od :
fi:
x := x+1 :
od :
I think if you use f( )=(4/3*UnitaryPhi(m))^2/m then the prog will run much faster.
If n=f(f(n)) then n is a member of a(n).
Maximilian wrote :
>I found this a bit arbitrary. why 3/4 ?
The original problem is the following.
UnitaryPhi(m)=UnitaryPhi(m)=u/v*(m*n)^(1/2), n<=m
Where u/v<=1
u=v=1
Only one example exists m=n=1
u=1,v=2
A030163 and no example exists for m,n are not the same
u=2,v=3
It looks like the case of u=1,v=2
u=3,v=4
So, it is the smallest u,v of which equation has examples for different m,n
>is it for amicable pairs allowed to have x=y ?
I allow it.
>btw, I just realized that almost all sequences found at
http://www.research.att.com/~njas/sequences/?q=Amicable%20Number
>do lack a self-contained definition. E.g.
http://www.research.att.com/~njas/sequences/A005276
>- the only definition is via the link to MathWorld. I think we should make OEIS self-contained ; who knows when MathWorld will become restricted to people who buy a Mathematica license ... ;-)
I agree with you.
I know that once Eric Weisstein fought against CRC to defend the freedom of MAthWorld.
So I suppose that such a thing will never become.
Yasutoshi
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