Triangular_and_Fibonacci_Numbers
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Relation of Square Triangular Numbers To Mersenne Primes

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ramsey2879     Message 1 of 4  May 15, 2006
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Let M(p) = 2^(p)-1. These are known as Mersenne Numbers. I checked the 
following conjecture for values of p between 2 and 27. If and only if M
(p) is prime then the 2^(p-1)th square triangular number is divisible 
by M(p). Example the fourth square (2^{3-1})th triangular number, 1225 
is divisible by M(3).
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ramsey2879     Message 2 of 4  May 18, 2006
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--- In Triangular_Numbers@yahoogroups.com, "ramsey2879" 
<ramseykk2@...> wrote:
>
> Let M(p) = 2^(p)-1. These are known as Mersenne Numbers. I checked 
the 
> following conjecture for values of p between 2 and 27. If and only 
if M
> (p) is prime then the 2^(p-1)th square triangular number is 
divisible 
> by M(p). Example the fourth square (2^{3-1})th triangular number, 
1225 
> is divisible by M(3).
>
Thanks 
After more consideration, I realized that it is not necessary to 
store more than one term and that my test can be refined so that 
there are only p-1 terms that need to be calculated.
Here is my test 

S_0 = 1
C = p
Do 
S_0 = (4*S_0)*(8*S_0 + 1) Mod M(p)
C = C - 1
Loop until C = 1
If S_0 = 1 Then
Print "M(" & p & ") is prime"
Else
Print "M(" & p & ") is not prime"
End If.

Although this test involves more steps, it basically involves almost 
the same complexity as the Lucus-Lehmer Test since the additional 
steps are register shifts and the addition of 1
It is based upon the fact that 1 is the 2nd square triangular number 
and that The S_0's follow the sequence ST(2),ST(3),ST(5),ST(9),ST
(17).... i.e. ST(2^n+1) in the sequence of square triangular numbers 
(modulus M(p) of course). I checked the test for p = 3 to 28. It 
doesn't work for p = 2 since 3 is a factor of 6. 

Any comments?
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erszegi_andras     Message 3 of 4  May 31, 2006
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> > Let M(p) = 2^(p)-1. These are known as Mersenne Numbers. I 
checked 
> the 
> > following conjecture for values of p between 2 and 27. If and 
only 
> if M
> > (p) is prime then the 2^(p-1)th square triangular number is 
> divisible 
> > by M(p). Example the fourth square (2^{3-1})th triangular 
number, 
> 1225 
> > is divisible by M(3).

I have not found a counterexample, neither is my comment below 
necessarily helpful, but

I think it is better to choose 0 as the first index for the ST 
(square triangular) series, 
ie

ST(0)=0
ST(1)=1
ST(2)=36
ST(3)=1225, etc.

Then the explicit formula ST(n)=(((1 + sqr(2))^2n - (1 - sqr(2))^2n)/
(4* sqr(2)))^2 will also work, and then we can say that, for 
instance,

M(3)=7 divides every ST(3k) (k integer >=0), ST(0), ST(3), ST(6), 
etc.
M(5)=31 divides every ST(15k),
M(7)=127 divides every ST(63k),
M(13)=8191 divides every ST(315k), etc.

In general, it appears, prime M(p) divides ST(nk), where k is any 
integer 
>=0 and n divides M(p)-1. We can possibly also say, to get closer to 
your conjecture, that n (as in ST(nk) ) divides 2^(p-1)-1. Could we 
then rephrase your conjecture like this:

iff 2^p-1 is prime then 2^p-1 divides ST(nk) where k is any integer 
>=0 and n divides 2^(p-1)-1 ?

Regards
Andras Erszegi
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ramsey2879     Message 4 of 4  Jun 28, 2006
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--- In Triangular_Numbers@yahoogroups.com, "erszegi_andras" 
<erszegi_andras@...> wrote:
>
> > > Let M(p) = 2^(p)-1. These are known as Mersenne Numbers. I 
> checked 
> > the 
> > > following conjecture for values of p between 2 and 27. If and 
> only 
> > if M
> > > (p) is prime then the 2^(p-1)th square triangular number is 
> > divisible 
> > > by M(p). Example the fourth square (2^{3-1})th triangular 
> number, 
> > 1225 
> > > is divisible by M(3).
> 
> 
> I have not found a counterexample, neither is my comment below 
> necessarily helpful, but
> 
> I think it is better to choose 0 as the first index for the ST 
> (square triangular) series, 
> ie
> 
> ST(0)=0
> ST(1)=1
> ST(2)=36
> ST(3)=1225, etc.
> 
> Then the explicit formula ST(n)=(((1 + sqr(2))^2n - (1 - sqr(2))
^2n)/
> (4* sqr(2)))^2 will also work, and then we can say that, for 
> instance,
> 
> M(3)=7 divides every ST(3k) (k integer >=0), ST(0), ST(3), ST(6), 
> etc.
> M(5)=31 divides every ST(15k),
> M(7)=127 divides every ST(63k),
> M(13)=8191 divides every ST(315k), etc.
> 
> In general, it appears, prime M(p) divides ST(nk), where k is any 
> integer 
> >=0 and n divides M(p)-1. We can possibly also say, to get closer 
to 
> your conjecture, that n (as in ST(nk) ) divides 2^(p-1)-1. Could we 
> then rephrase your conjecture like this:
> 
> iff 2^p-1 is prime then 2^p-1 divides ST(nk) where k is any integer 
> >=0 and n divides 2^(p-1)-1 ?
> 
> Regards
> Andras Erszegi
>
Actually, it is more complicated than that, but you have the right 
idea. Note that ST(9-1) mod 9 = 0 but 9 is not prime.