- Jack wrote:

Because sigma(x) is odd if and only if x is a square or

twice a square.

The density of x such that sigma(x) is odd == 0.

[jp : http://mathworld.wolfram.com/DivisorFunction.html (eqn28)

Try finding n such that n divides sigma(n-1)+sigma(n)+sigma(n+1).

Heuristics would indicate an infinite number of such n...

2, 5, 7, 33, 18336, 19262, 38184, 54722, ...

I've just submitted this to the EIS.

---

From the link, the equations can be morphed into:

(n-1).sigma(n-1) = 2 + i.phi(n-1)

n.sigma(n) = 2 + j.phi(n)

(n+1).sigma(n+1) = 2 + k.phi(n+1)

n(n^2-1).sigma(n(n^2-1)) = [2 + i.phi(n-1)].[2 + j.phi(n)].[2 + k.phi(n+1)]

so the answers occur whenever the RHS contains n^2 as a factor.

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths

BrainBench MVP for HTML and JavaScript

http://www.brainbench.com

-----Original Message-----

From: Phil Carmody [mailto:thefatphil@...]

Sent: 02 July 2002 13:35

To: primenumbers

Subject: RE: [PrimeNumbers] Factor patterns

--- Jon Perry <perry@...> wrote:> A swerve.

Look at sigma(n)%2.

>

> Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always

> even?

>

> (PARI/GP)

>

> for

> (n=2,1000,write("sigmax.txt",(sigma(n-1)+sigma(n)+sigma(n+1))%2))

Mostly 0.

Each of the many remotely isolated 1s will cause 3 consecutive 1s in

your expression. 101 causes a 11011 in yours, and 11 causes 1001 in

yours.

No magic.

Phil

=====

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Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/ > Are there any more numbers sharing this property?

I believe not; you can do it mod 6, mod 12, mod 24

and then there are exceptions.

Easy to see why: we require a modulus M such that

n^2=1 mod M for all n with gcd(n,M)=1

M=24 is easy.

We cannot have any larger M with gcd(M,5)=1,since

5^2-1=24.

So try M=12*5.

Now we have problems with

7^2-1=48

unless M=12*5*7.

Now we have problems with

11^2-1=120

It's clear enough that we are are losing out,

though I do not have a proof..

David- (sorry about the complete tripe in previous post - the n.sigma(n) eqv. 2 mod

phi(n) is only true for n=p, and n=4,6,22).

--

Continuing Jack's work:

for (n=2,100000,if ((sigma(n-1)+sigma(n)+sigma(n+1))%(n-1)==0,

write("sigmaconn.txt",n,":",sigma(n-1)%(n-1),":",+sigma(n)%(n-1),":",sigma(n

+1)%(n-1),":",(sigma(n-1)+sigma(n)+sigma(n+1)))))

2,3,24,64,227,291,784,1883,7731,18547,25723,30397,94358

for (n=2,100000,if ((sigma(n-1)+sigma(n)+sigma(n+1))%(n+1)==0,

write("sigmaconp.txt",n,":",sigma(n-1)%(n+1),":",+sigma(n)%(n+1),":",sigma(n

+1)%(n+1),":",(sigma(n-1)+sigma(n)+sigma(n+1)))))

8,21,22,23,57,157,505,1053,2147,2273,3311,4679,5931,7898,22682

--

A related question - when does sigma(n,2)%sigma(n,1)==0?

If n is a square, and also:

20,50,117,180,200,242,325,450,468,500,578,605,650,800,968,980

for (n=2,10000, if (sigma(n,2)%sigma(n,1)==0 &&

!issquare(n),write("sigmasigmasq.txt",n)))

Anyone spot the missing link?

Jon Perry

perry@...

http://www.users.globalnet.co.uk/~perry/maths

BrainBench MVP for HTML and JavaScript

http://www.brainbench.com