Hello, Group.

Can someone calculate the length of N= (2^1660693)*(2^1660693+21)+1?

I esimated it as [(2*1660693)*((log 2)/(log 10))] as someone suggested
and arrived at L(N)= 999,999.  How inaccurate is it?

Thanks, Bill.

>Can someone calculate the length of N= (2^1660693)*(2^1660693+21)+1?

>I esimated it as [(2*1660693)*((log 2)/(log 10))] as someone suggested
>and arrived at L(N)= 999,999. How inaccurate is it?

GMP gives : 998838

[Non-text portions of this message have been removed]

--- leavemsg1 <leavemsg1@...> wrote:
> Hello, Group.
>
> Can someone calculate the length of N= (2^1660693)*(2^1660693+21)+1?
>

? 2*1660693*log(2)/log(10)
999836.8131784078460461793726

> I esimated it as [(2*1660693)*((log 2)/(log 10))] as someone suggested
> and arrived at L(N)= 999,999.  How inaccurate is it?

My answer's very accurate,
It appears that you've mixed 1660693 and 1660963.

Phil

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--- Jacques Tramu <jacques.tramu@...> wrote:
> >Can someone calculate the length of N= (2^1660693)*(2^1660693+21)+1?
>
> >I esimated it as [(2*1660693)*((log 2)/(log 10))] as someone suggested
> >and arrived at L(N)= 999,999. How inaccurate is it?
>
> GMP gives : 998838

Maybe it does, maybe it doesn't, but it's the wrong tool for the job.

Calculating the value of a simple expression you're only going to find the size
of is lazy thinking.

The size of a^b is b times the size of a.
The size of a*b is the sum of the sizes of a and b.

Phil

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I thought I would post some results of my limited studies of
Sierpinski/ Riesel series in bases other than 2,3,4 and 5. Some of you
with efficient programming should be able to take this study further.
Here in Bangladesh I am unable to access my Maple software, which
would have speeded things up a lot.

Some of you might have seen recent Yahoo postings relating to the
hypothesis that there is a covering set for every base for both Riesel
and Sierpinski series. This is going to be difficult to prove as such,
but certainly it seems very likely. I am likely to move my research to
the special base form base=2^n-1, where I need to find a covering set
in principle for all values and these may not be easy to find.

Anyway here is information on bases 6 to 18:

Base 6:

Will need a lot of work – a covering set, repeating every 12 n is
[7,13,31,37,43]. Lowest Sierpinski and Reisel candidates not known.
But neither can be greater than 4488211.

Another possible covering set, with repeat of 12 is [7,43,37,31,97]

Base 7:

Totally horrible. Possible covering set with repeat every 24 n is
[19,5,43,1201,13,181,193,73], also 5 other sets perming 73, 193 and 409.

Sierpinski and Riesel numbers are both lower than 162643669672445

Work is needed to find a low k value which is Riesel or Sierpinski.

Base 8:

Covering set [3,5,13] covering every 4 n. The corresponding Sierpinski
number is 47, but it is not proven for the small fact that k=1 is
known not to have small primes. (Think about it: 8^n+1= 2^3n+1

For Reisel k=112 looks the most likely candidate. A prime needs to be
found however for k=14

Base 9:

Covering set every 6n for [5,7,13,73]. Alternative covering set every
8 n for [5,41,17,193]. Lowest mooted Sierpinski is 2344 (k=439 is not
Sierpinski because the k is also trivial). Lowest conjectured Riesel
is 74, so should be easy to prove, but 4,16,36,64 are proving pesky.
Note 16 and 64 are subsets of 4.

Base 10:

Covering set every 6n is [11,37,7,13]. Lowest known non trivial
Sierpinski is 9351, lowest Riesel 10176. No work done yet to prove these.

Base 11:

Covering set every 6n is [3,7,19,37]. Lowest Sierpinski k is thought
to be 1490. 3 ks need to be eliminated 416, 958, 1468

Lowest Riesel is thought to be 4624. 12k's still need to be
eliminated. 62, 682, 862, 904, 1528, 2410, 2690, 3110, 3544, 3788,
4208, 4564.

Base 12:

Covering set every 6k is [13,157,7,19]. Lowest Sierpinski mooted to be
14600 and lowest Riesel 16329. No work on this yet.

Base 13:

Covering set every 3 n, [7,17,5]. Lowest Sierpinski is proven to be
132. Lowest non trivial Riesel is thought to be 302, but need to prove
k=288 is not.

Base 14:

With a covering set every 2n [3,5] this proves to be easy. Proven
Sierpinski and Riesel values are both proven at 4.

Base 15:

Horrible. A covering set is [241,113,211,17,1489,13,3877], and
Sierpinski and Riesel values are therefore less than 7330957703181619.
As bad as the base 3 problem.

Base 16:

Sierpinski number not known, quite complex to calculate. Covering set
[7,13,19,37,73] have 36 combinations to check Sadly the simple
covering set 7,13,17 only appears in trivial solutions. On the Riesel
side not so sad, conjectured to be 120. However n=9 is a problem, this
number is well studied as 9.16^n-1 = 9.2^(4n)-1, and no values for n
less than 207177

Base 17:

Covering set every 4 n, [3,5,29]. Sierpinski value of 278, the
following candidates need a prime 88,92,160,244,262. On the Riesel
side, the lowest is conjectured to be 86, with only k=44 needing to
find a prime.

Base 18:

Covering set every 6n is [19,13,5]. Sierpinski value is 398. 4 k
candidates seek primes 18,324,122,381. Riesel is proven to be 246

Any help anyone can give in taking this to higher bases, or proving
the mooted Sierpinskis and Riesels is welcome. Postings please to
http://www.mersenneforum.org/showthread.php?t=6895

Regards

Robert Smith

1,2,1,3,0,1,4,2,0,1,5,0,0,0,1,6,3,2,0,0,1,7,0,0,0,0,0,1,8,4,0,2,0,0,0,
%T A126988
1,9,0,3,0,0,0,0,0,1,10,5,0,0,2,0,0,0,0,1,11,0,0,0,0,0,0,0,0,0,1,12,6,4
,
%U A126988 3,0,2,0,0,0,0,0,0
%N A126988 Triangle, row sums = Sigma(n).
%C A126988 Row sums = A000203, (Sigma(n)): 1, 3, 4, 7, 6, 12, 8,
15,... Sigma(n) is the sum of the divisors of the integer n.
Conjecture: The sequence of parsed terms in Sigma(n) is the reversal
of non-zero row terms in the triangle A126988.
%D A126988 David Wells, "Prime Numbers, the Most Mysterious Figures
in Math", John Wiley & Sons, Inc, 2005, Appendix B.
%F A126988 Row sums of a triangle in which k-th column (k=0,1,2...)
is (1,2,3,...) interspersed with n consecutive zeros, starting after
the "1".
%e A126988 First few rows of the triangle are:
%e A126988 1;
%e A126988 2, 1;
%e A126988 3, 0, 1;
%e A126988 4, 2, 0, 1;
%e A126988 5, 0, 0, 0, 1;
%e A126988 6, 3, 2, 0, 0, 1;
%e A126988 7, 0, 0, 0, 0, 0, 1;
%e A126988 8, 4, 0, 2, 0, 0, 0, 1;
%e A126988 9, 0, 3, 0, 0, 0, 0, 0, 1;
%e A126988 10, 5, 0, 0, 2, 0, 0, 0, 0, 1;
%e A126988 ..
%e A126988 Sigma(12) = 28 = (from tables): (1 + 2 + 3 + 4 + 6 + 12).
%e A126988 Sigma(12) = 28, from 12-th row of A126988 = (12 + 6 + 4 +
3 + 2 + 1), deleting the zeros, from left to right.
%Y A126988 Cf. A000203.
%K A126988 nonn,tabl,uned
%O A126988 1,2
%A A126988 Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 31 2006

I get about 6.5*10^999836.  This would indicate a length of 999837
digits.

- David
--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
wrote:
>
> --- Jacques Tramu <jacques.tramu@...> wrote:
> > >Can someone calculate the length of N= (2^1660693)*(2^1660693+21)
+1?
> >
> > >I esimated it as [(2*1660693)*((log 2)/(log 10))] as someone
suggested
> > >and arrived at L(N)= 999,999. How inaccurate is it?
> >
> > GMP gives : 998838
>
> Maybe it does, maybe it doesn't, but it's the wrong tool for the
job.
>
> Calculating the value of a simple expression you're only going to
find the size
> of is lazy thinking.
>
> The size of a^b is b times the size of a.
> The size of a*b is the sum of the sizes of a and b.
>
> Phil
>
> ()  ASCII ribbon campaign      ()    Hopeless ribbon campaign
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>
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>

--- matrixmonitor <matrixmonitor@...> wrote:
> 1,2,1,3,0,1,4,2,0,1,5,0,0,0,1,6,3,2,0,0,1,7,0,0,0,0,0,1,8,4,0,2,0,0,0,
> %T A126988
> 1,9,0,3,0,0,0,0,0,1,10,5,0,0,2,0,0,0,0,1,11,0,0,0,0,0,0,0,0,0,1,12,6,4
> ,
> %U A126988 3,0,2,0,0,0,0,0,0
> %N A126988 Triangle, row sums = Sigma(n).

That much seems true.

> %C A126988 Row sums = A000203, (Sigma(n)): 1, 3, 4, 7, 6, 12, 8,
> 15,... Sigma(n) is the sum of the divisors of the integer n.
> Conjecture: The sequence of parsed terms in Sigma(n) is the reversal

What the blazes are 'parsed' terms?

> of non-zero row terms in the triangle A126988.
> %D A126988 David Wells, "Prime Numbers, the Most Mysterious Figures
> in Math", John Wiley & Sons, Inc, 2005, Appendix B.
> %F A126988 Row sums of a triangle in which k-th column (k=0,1,2...)
> is (1,2,3,...) interspersed with n consecutive zeros, starting after
> the "1".

This sequence isn't the row sums of any triangle, it is a triangle.

The descriptions are garbled, just as would be expected from:

> %A A126988 Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 31 2006

So, matrixmonitor, I don't remember seeing you post before. You're not just
another sock-puppet for that previous known spewer of
/stuff-of-questionable-quality/ are you?

Phil

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On 1/9/07, Raman (Tuch) * Tay <vr_aman@...> wrote:
> satisfaction satisfactio
> satisfactin satisfacti

Unlike other people, he _can_ get satisfaction!

Christ van Willegen

Raman is an asshole who doesn't even know how to mount a denial of
service attack properly.

I, together with many other people in the CNT community and denizens of
www.mersenneforum.com, have been seeing this sort of stuff since last
year.

A half-way decent spam-removal mechanism deals with him without any
fuss.


Paul

On Tue, 2007-01-09 at 10:34, Christ van Willegen wrote:
> On 1/9/07, Raman (Tuch) * Tay <vr_aman@...> wrote:
> > satisfaction satisfactio
> > satisfactin satisfacti
>
> Unlike other people, he _can_ get satisfaction!
>
> Christ van Willegen
>
>
>
>


[Non-text portions of this message have been removed]

This is an interesting example of how small numbers don't always
behave the same as big numbers, and it's somewhat related to prime
numbers (well, factorization anyway)...


Start enumerating those values of sigma(N) which are divisible by 17
(sigma function here being the sum of positive integer divisors).

The first few are of course:

   sigma(67) == 68
   sigma(101) == 102
   sigma(128) == 255
   sigma(134) == 204

There seems to be no scarcity of such numbers -- for N < 10^6, we
have sigma(N) divisible by 17 a grand total of 80608 times,
a hit rate of just over 8%.


Now also enumerate those values of sigma(N) which are congruent
to 5 (modulo 6):

   sigma(2401) == 2801
   sigma(9604) == 19607
   sigma(21609) == 36413
   sigma(28561) == 30941

Okay, these are a little bit scarce, but still, there are 24
of them for N < 10^6, and 1644 of them for N < 10^10.  There
are clearly an infinite number of them -- for instance, given
any prime p of the form 6*a+1, N = p^4 will work.


The big question is, how far do you have to go to find a number
which is in both lists?

Or for those who like their problem stated succinctly:
What is the smallest N such that sigma(N) is congruent to
17 (modulo 102)?


   Jack

Apologies if people get 2 copies - yahoo seems flakey, so I'm trying
again.

--- In primenumbers@yahoogroups.com, "jbrennen" <jb@...> wrote:
> What is the smallest N such that sigma(N) is congruent to
> 17 (modulo 102)?

An upper bound is 160470643909878751793805444097921

Phil

--- jbrennen <jb@...> wrote:
> What is the smallest N such that sigma(N) is congruent to
> 17 (modulo 102)?

Well, an upper bound is 160470643909878751793805444097921

Phil



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I think you can prove that Phil's example 103^16 is the smallest
such number.  Sigma(n) has factors (1+p+p^2+...+p^k) for p^k||n  (<-
largest power of p that divides n).  For sigma(n) to be 5 mod 6, all
the factors have to be either 1 or 5 mod 6 (i.e., no factors that
are 2,3,4,0 mod 6).  By examination, for p=2 and p=3,
(1+p+p^2+...+p^k) is never 0 mod 17 and 1 or 5 mod 6.

For p=3, we get 1+3+3^2+...+3^k odd when k is even but then
1+3+3^2+...+3^k=(3^(k+1)-1)/2 while the order of 3 mod 17 is 16 so 3^
(k+1)-1 is divisible by 17 only when 16 divides k+1.  It can not be
true that k is even and 16 divides k+1.

For p=2, we get 1+2+2^2+...+2^k is 1 mod 3 when k is even.  But the
order of 2 mod 17 is 8, so 2^(k+1)-1 == 0 mod 17 means (k+1) is even.

The only other available primes are either 1 or 5 mod 6.  If p is 5
mod 6 and 1+p+...+p^k is either 1 or 5 mod 6, then k is even.  If p
is 1 mod 6 and 1+p+...+p^k is either 1 or 5 mod 6, then k is either
4 or 6 mod 6.

For k even, solve 1+x+...+x^k==0 mod 17 and find the first solution
happens when k=16 and x==1 mod 17.  The first prime that is 1 mod 17
is 103.  103^16 is the first number that solve sigma(n)==17 mod 102.

--- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@...> wrote:
>
> I think you can prove that Phil's example 103^16 is the smallest
> such number.

Yes, and you seemed to give a convincing argument.  :)


Did anyone notice that I basically gave away the key to finding
Phil's example?  I stated that solving sigma(N) == 5 (mod 6)
was easily done by picking a prime p of the form 6*x+1 and
then taking p^4, or p^(5-1).

This is of course easily generalized...

So to solve sigma(N) == 17 (mod 102), find a prime of the
form 102*x+1.  The first such is of course 103.  And then
set N = 103^(17-1) = 103^16.



The tricky part is proving that 103^16 is in fact the minimal
solution.

My method for doing so:

Note that N must be divisible by a prime power p^a which
has sigma(p^a) divisible by 17 but not divisible by 2 or 3.

Note also that sigma(p^a) = (p^(a+1)-1)/(p-1).

In order for sigma(p^a) to be an odd number (not divisible
by 2), we need to have either p or a be even.

If p is even, it must be 2, and we would have:
p^(a+1) == 1 (mod 17).  It turns out that this is true
when a == 7 (mod 8); however, when a == 7 (mod 8), we
also have sigma(2^a) divisible by 3.  So that doesn't
work.

So p must be odd and a must be even.  In order for
p^(a+1)-1 (the numerator of the expression above) to
be divisible by 17, (a+1) must be divisible by the
order of p modulo 17.  Note that the order of p
modulo 17 must divide evenly into 16 (a consequence
of Fermat's Little Theorem).  So we have an
odd number (a+1) which must be divisible by a
divisor of 16.  Clearly the divisor (the order of
p modulo 17) must be 1.  This implies that the
prime p is in fact of the form 17*x+1.

And once we know that p == 1 (mod 17), we can see
easily that a must be of the form 17*y-1.

The smallest prime p of the form 17*x+1 is 103,
and the smallest exponent a of the form 17*y-1 is 16.

So the smallest prime power which has sigma(p^a)
divisible by 17, but not by 2 or 3, is 103^16.

So any solution N to the original puzzle must be
divisible by a prime power >= 103^16.



There are not a lot of ordered pairs (A,B) such
that A > 2 and such that N = (B+1)^(A-1) is the
smallest solution to sigma(N) == A (modulo B).

I know of these three:  (5,6), (5,30), and (17,102).

There could be others, if anyone wants to try
to find them.  It's interesting that for those
three I know of, A is always a Fermat prime.

The PRP's 10^19000+79387 and 10^19000-38663 produce a prime number GAP of
118,050

Milton L. Brown

[Moderators note: Expected gap is ~43750, so this is ~2.7x that.]

I created a program that is testing for potential primes at 10^1,000,000,000+.
   My test yesterday was from 10^1,000,000,000 to 10^1,000,000,000+1,000,000
   it took 9 minutes and 5 seconds to run the test. I was able to eliminate all
but 9112 as being prime. That is using a small prime library of primes up to
1,000,000 approx.
   Just to make sure it is clear that is 10 to the 1 billion. Got a lot of
questions about that on another forum.


---------------------------------
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[Non-text portions of this message have been removed]

--- george hayes <gr.hayes@...> wrote:
> I created a program that is testing for potential primes at
> 10^1,000,000,000+.
>   My test yesterday was from 10^1,000,000,000 to 10^1,000,000,000+1,000,000
>   it took 9 minutes and 5 seconds to run the test. I was able to eliminate
> all but 9112 as being prime. That is using a small prime library of primes up
> to 1,000,000 approx.
>   Just to make sure it is clear that is 10 to the 1 billion. Got a lot of
> questions about that on another forum.

From the speed, I can only assume you're running on a programmable calculator,
or a PDA. NewPGen, which is notoriously inefficient for the initial ramp-up of
tiny primes can get to about 2 million in only a few seconds.

And from your sieving depth, I'd have thought that about 72000 candidates
should remain, so it sounds as if you have a bug.

Phil
Phil

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At 02:36 PM 1/13/2007, Phil Carmody wrote:

>--- george hayes <<mailto:gr.hayes%40yahoo.com>gr.hayes@...> wrote:
> > I created a program that is testing for potential primes at
> > 10^1,000,000,000+.
> > My test yesterday was from 10^1,000,000,000 to 10^1,000,000,000+1,000,000
> > it took 9 minutes and 5 seconds to run the test. I was able to eliminate
> > all but 9112 as being prime. That is using a small prime library
> of primes up
> > to 1,000,000 approx.
> > Just to make sure it is clear that is 10 to the 1 billion. Got a lot of
> > questions about that on another forum.
>
> From the speed, I can only assume you're running on a programmable
> calculator,

He says that he is doing billion-digit primes.

--- Jud McCranie <j.mccranie@...> wrote:
> At 02:36 PM 1/13/2007, Phil Carmody wrote:
> >--- george hayes <<mailto:gr.hayes%40yahoo.com>gr.hayes@...> wrote:
> > > I created a program that is testing for potential primes at
> > > 10^1,000,000,000+.
> > > My test yesterday was from 10^1,000,000,000 to 10^1,000,000,000+1,000,000
> > > it took 9 minutes and 5 seconds to run the test. I was able to eliminate
> > > all but 9112 as being prime. That is using a small prime library
> > of primes up
> > > to 1,000,000 approx.
> > > Just to make sure it is clear that is 10 to the 1 billion. Got a lot of
> > > questions about that on another forum.
> >
> > From the speed, I can only assume you're running on a programmable
> > calculator,
>
> He says that he is doing billion-digit primes.

No, 'potential primes'. I can do that instantly by doing nothing.

However, if I wanted to make use of a "small prime library of primes up to
1,000,000 approx.", then I'd probably sieve the range with that the, and I'd
take something more comparable to a second than ten minutes.

There's less than one thousandth of a prime in that range, and certainly zero
provable ones even if there's one probable-prime.

Phil


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Is there a simple answer for the reason that

((2*T)^721) mod 115921 = 1 for all primes T except 13, 37, and 241

And, 721 =7 * 103 is the smallest such number.

Note that 115921 = 13 * 37 * 241 is a Carmichael Number.

How does on obtain the the number 721?

Milton L. Brown

--- miltbrown@... wrote:
> Is there a simple answer for the reason that
>
> ((2*T)^721) mod 115921 = 1 for all primes T except 13, 37, and 241

? Mod(2*2,115921)^721
Mod(4, 115921)
? Mod(2*3,115921)^721
Mod(6, 115921)
? Mod(2*5,115921)^721
Mod(10, 115921)
...

This is worse than your normal hit-rate, Milton.

> How does on obtain the the number 721?

Hard to say without being crude.

Phil

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I am not sure what you are saying, but

((2*2)^721) mod 115921 = 1

no? Also,

((2*3)^721) mod 115921 = 1

but not

((2*13)^721) mod 115921 =/= 1

--- miltbrown@... wrote:
> I am not sure what you are saying,

You're playig this gaem again, I see:
http://www.politicsforum.org/images/flame_warriors/flame_46.php

I'm talking correct maths. This could be why you don't recognise it.

Please don't top post. Responses have the potential, not always realised alas,
of making more sense when they follow the thing to which they are a response.

> but
>
> ((2*2)^721) mod 115921 = 1
>
> no?

What part of my

? Mod(2*2,115921)^721
Mod(4, 115921)

did you not understand?

> Also,
>
> ((2*3)^721) mod 115921 = 1

? Mod(2*3,115921)^721
Mod(6, 115921)

Please don't make me beat you round the head with all 103680 totatives.

> but not
>
> ((2*13)^721) mod 115921 =/= 1

2*13 is not a totative, as 13|115921

This is beginners' stuff, Milton.

I remember in the past that because of your abuses, the moderators agreed to
not only moderate for _relevance_, but also for _correctness_. Alas I forgot
that earlier, but have remembered it now. If it's wrong or misleading, I will
not approve it.

To help you get your maths straightened out, the keywords you should be
googling for are "carmichael", "lambda", and "function"; preferably in very
close proximity.

Phil

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A new twin prime record from "Twin Prime Search" with 58711 digits has
beaten the previous record of 51780 digits.

http://primes.utm.edu/top20/page.php?id=1

old: http://primes.utm.edu/primes/page.php?id=77996

new: http://www.mersenneforum.org/showthread.php?t=6964

Congrats to TPS

Paul

Congratulations to Eric_V and TPS !!!! This is a big
one !


	 primenumbers@yahoogroups.com
--- Paul Underwood <paulunderwood@...>
schrieb:

> A new twin prime record from "Twin Prime Search"
> with 58711 digits has
> beaten the previous record of 51780 digits.
>
> http://primes.utm.edu/top20/page.php?id=1
>
> old: http://primes.utm.edu/primes/page.php?id=77996
>
> new:
> http://www.mersenneforum.org/showthread.php?t=6964
>
> Congrats to TPS
>
> Paul
>
>







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Hi,

Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2
and 3 where p is prime???

Bill

--- leavemsg1 <leavemsg1@...> wrote:
> Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2
> and 3 where p is prime???

Humans and mail are slow. Computers and Pari/GP are fast.

? for(i=2,12,print(i" : "2^(i-1)%6" "3^(i-2)%6" "(2^(i-1)+3^(i-2))%6))
2 : 2 1 3
3 : 4 3 1
4 : 2 3 5
5 : 4 3 1
6 : 2 3 5
7 : 4 3 1
8 : 2 3 5
9 : 4 3 1
10 : 2 3 5
11 : 4 3 1
12 : 2 3 5

So it's not just true when p's an (odd) prime, but true whenever p's any odd
number.

Phil


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On 1/16/07, leavemsg1 <leavemsg1@...> wrote:
> Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2
> and 3 where p is prime???

Yes.

2^whatever is even, 3^whatever is odd, so 2^x + 3^y - 1 is even.

2^even = 1 (mod 3), 3^whatever = 0 (mod 3), so
2^even + 3^whatever - 1 is divisible by 3.

So if p is odd (not necessarily prime, and not 2), then the
divisibility will happen as you have said.

--Joshua ucker

--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...>
wrote:

[Stuff - just see: http://tech.groups.yahoo.com/group/primenumbers/
message/18206?threaded=1&var=1&l=1 ]

On sci.math just now, Gerry Myerson has chipped in the following:

<<<
It's shown in Niven, Zuckerman, and Montgomery that if p is 1 mod 4
then ( (p - 1) / 2 ) factorial squared is -1 mod p (p. 54), and
it's an exercise to show that if p is 3 mod 4 then the quantity is
plus-or-minus 1 (exercise 2.1.18).

Maybe something interesting happens with ( (p - 1) / 3 ) factorial,
for those primes congruent 1 mod 3.
>>>

I instantaniously bought Niven, Zuckerman, and Montgomery (huge, not
petit), and unless anyone else wants to chip in with a review right
now, I shall do so when it arrives in a month or so.

G^HMerry unorthodox Christmas to me!

Phil