I have been playing with primes and powers and came up with this equation.
(n+1)³ -n³ =P1*P2*...*Pnth
Where the result is only divisible by itself, 1 and exclusively primes and no other numbers.
I have not yet found a counter example to this relationship and am windering whether someone can either prove it wrong, tell me it's obvious or tell me i have found something. (I don't think i have but someone needs to show me why)THANKS.

I think I understand what you mean now, but I'm just checking. You want to show that A = (n+1)³ - n³ = p₁ xp₂

If A = (n+1)³ - n³ = p₁ x p₂ x p₃
this would count as a counterexample because A is then divisible by (p₁ x p₂ ), which is not prime. So we need either a proof that (n+1)³ - n³ is the product of two primes only or a proof (or counterexample) to show that (n+1)³ - n³ may be divisible by more than two primes.

This is the problem as I see it. Comments?

A counterexample
125³ - 124³ = 13 x 3577
3577 is not a prime. It is divisible by 7 and 511.

Patrick, if what Tom has written has answered your question then feel free to ignore this. I read your question as talking about square-free numbers, numbers which have no repeated prime factors, i.e. 3 x 5 x 7 x 11 is square-free whereas 3 x 5 x 7² or 7³ x 13 x 17 are not square-free as they both have 7² as a factor. If this is what you are talking about then obviously Tom's counterexample isn't a counterexample as it is square-free, I'm trying to find a non square-free number that is a counterexample but as I said above if this isn't what you're asking about then please ignore!

If n is 7, then we get 8³ -7³ =169=13² . Also for n=22 we have 1519, which is divisible by 7² .
However, this sequence does seem to have an surprising density of square-free numbers.

Two counterexamples, 72³ -71³ =7² x 313
and 23³ -22³ =7² x 31

If a number has a square factor then quite a lot of the time this will be because it has a factor of 4,9 (these are the easiest squares to be divisible by, as they are the lowest two non-unit squares). Now as it happens it is easy to see that (n+1)³ -n³ can never be divisible by 4 or 9 (because it can be written as 3n(n+1)+1, which is one more than a multiple of 6). Therefore it must divide by a larger square (25 or higher bearing in mind that 16 is also out) in order to contain a square factor. So that may be a partial explanation for why there appear so few square factors in this sequence.

Michael

Oh and it appears 25 is out too as (n+1)³ -n³ cannot be divisible by 5 ever (you can check this by substituting in n = 0,1,2,3,4). So any integer with a square factor would have to be divisible by 49 (or a higher square still). So now it is maybe not so surprising that there are so many integers in the sequence without square factors.

49 cannot be discounted in this way. If n leaves a remainder of 1 when divided by 7 then (n+1)³ -n³ will be divisible by 7. So perhaps 49 is possible.

Yours,

Michael

Ran test on sequence, got up to n = 3645 before I got bored of waiting. Found 235 numbers with square factors in sequence.

Here are the first few squares which it is possible for (n+1)³ -n³ to be divisible by (according to a simple VB program):

7²
13²
19²
31²
37²
43²
(49² )
61²
67²
73²
79²
(91² )
97²
103²
109²
127²
(133² )
139²
151²

I put the non-primes in brackets because these aren't actually building blocks (any number that is divisible by one of the squares in brackets is divisible by a smaller square in the list).

I think it is the absence of 2² , 3² , 5² which causes the shortage of numbers with square factors. (Also to a much lesser extent the absence of 11² , 17² .)

Michael

Do you think it would be possible to calculate exactly the density of square-free numbers in this sequence? It seems to approach about 0.06, from testing n up to 1000.

Dvaid

Well that would probably be very hard, but I was thinking about something you said earlier - about the overall proportion of integers without square factor being 6/pi ² . What do you think of the following?

For prime n define f(n) as overall the proportion of all integers which have a factor of 2² , 3² , 5² , 7² , ? or n² . (So they can have any prime square factor < = n² .) This would have to be done by taking the result over the first Q integers only and letting Q -> infinity. For the time being let's assume the limit exists.

Let r,s be adjacent primes (r < s and there is no prime p such that r < p < s). Now if C is large then out of the integers 1,2,3,...,C about C/s² have a factor of s² . Of these C/s² f(r) also have a prime square factor below s² . (Because all prime numbers are mutually coprime so having one prime square factor doesn't affect the likelihood of having another one below it. The factors are totally independent.)

So number of numbers out of 1,2,...,C that have s² as their lowest prime square factor is:

C/s² - f(r) C/s²

And number of numbers that have a prime square factor r² or below is:

C f(r).

Add these together to find the number of numbers with a prime square factor s² or below (which is C f(s)):

C f(s) = C f(r) + C/s² - f(r) C/s²

f(s) = f(r) + 1/s² - f(r) /s²

Now let's define g(x) = 1 - f(x) which is what we are actually interested in (no. of square free numbers of particular factors).

1 -g(s) = 1 -g(r) + g(r)/s²

So:

g(s) = g(r) [1-1/s² ]

Now g(2) = 3/4 obviously (1/4 of all numbers have a factor of 4). Remembering r,s are adjacent primes:

g(3) = [1-1/9]g(2)

g(5) = [1-1/25][1-1/9]g(2)

In general it is easy to see by induction that:

g(p) = [1-1/2² ][1-1/3² ][1-1/5² ][1-1/7² ][1-1/11² ]...[1-1/p² ]

Now lim(p-> infinity ) g(p) = overall proportion of square free integers (as we are gradually removing all the factors).

And David says this could well be 6/pi ² . If so we just have to show:

6/pi ² = [1-1/2² ][1-1/3² ][1-1/5² ][1-1/7² ][1-1/11² ]...

But that looks like quite a daunting infinite product on the right. Help, anyone?

Yours,

Michael

Actually now I think about it that result isn't nearly as hard as it looks. First of all a simple induction proof reveals:

where $A(p)$ is the set of primes $r$ such that $2\le r\le p$, and $B(p)$ is the set of positive integers which are not multiples of any prime in $A(p)$.

Simply take the limit as $p$ tends to infinity. $B(p)$ ends up virtually $\{1\}$ and you get:

$1=\sum _{r=1}^{\infty }1/r^2\times \mathrm{proportion}\mathrm{of}\mathrm{square}\mathrm{free}\mathrm{integers}$

So proportion of square free integers $=1/\sum _{r=1}^{\infty }1/r^2=6/{\pi }^2$ by Euler's result. (See here and here .)

I've just noticed this does generalise upwards for cube free integers, fourth power free integers etc (and generalises downwards for factor free integers (i.e. prime numbers) -apparently the proportion of prime numbers is 0 which is quite intuitive.) In general the proportion of numbers with no factor of the form Xⁿ (X = positive integer > 1) is:

$1/\sum _{r=1}^{\infty }1/r^n$
This looks like the reciprocal of the zeta function (if I remember correctly a book I read).

However I should stress that all these results are dependent on the fact that the proportion of ''Xⁿ '' free numbers is indeed convergent. The above argument simply shows that if it does converge it must be $1/\zeta (n)$. I need someone's help in showing that it does converge as the approaches I have tried thus far have failed.
Yours,

Michael

If you consider;
[ 1 + 1/(2^s ) + 1/(2^2s ) + 1/(2^3s ) + ...] x [ 1 + 1/(3^s ) + 1/(3^2s ) + 1/(3^3s ) + ...] x[ 1 + 1/(5^s ) + 1/(5^2s ) + 1/(5^3s ) + ...] x [ 1 + 1/(7^s ) + 1/(7^2s ) + 1/(7^3s ) + ...] x.... where each square bracket has a prime series in it, then when this is expanded it equals 1 + 1/(2^s ) + 1/(3^s ) + 1/(4^s ) + 1/(5^s ) + .... because of unique factorisation of integers.

As 1 + 1/(n^s ) + 1/(n^2s ) + 1/(n^3s ) + ... = 1/[1 - 1/(n^s )] then this gives the infinite product form of the zeta function as Michael has written above (this was how Euler discovered the relation).


In regard to the density of squarefree numbers in the sequence (n + 1)³ - n³ = 3n² + 3n + 1, I think it equals;

[1 - (2/7² )] x [1 - (2/13² )] x [1 - (2/19² )] x [1 - (2/31² )] x [1 - (2/37² )] x .... where each of the brackets is equal to [1 - (2/p² )] where p is a prime of the form 6a + 1. I'm not certain of it but I can show that 3n² + 3n + 1 is divisible by no prime of the form 6a - 1 so obviously cannot be divisible by a prime of the form 6a - 1 squared. I'm on my way to proving that the proportion of numbers of the form 3n² + 3n + 1 that are divisible by p² is 2/p² where p is a prime of the form 6a + 1. The above formula comes from a similar idea as the infinite product for the general proportion of squarefree numbers. I'll explain this further if I manage to actually get an answer.

Andrew - I'd completely missed the mod 6 connection! Yes you're right that 3n² + 3n + 1 cannot be divisible by 6k-1. But I had to use Gauss's theorem of Quadratic Reciprocity (see here for anyone who doesn't know this theorem) to prove it via (6n+3)² = -3 (mod 6k-1). (It does hold for negative primes doesn't it?) Can you do it otherwise?

In fact do you know of anywhere I could find a proof of Gauss's Theorem of Quadratic Reciprocity? I hadn't even heard of it before the Fibonacci discussion. From what I've done it appears to be unbelievably useful. I've already spent about an hour trying to prove it this morning and I've got absolutely nowhere which is not surprising if Gauss was the first one to derive it!

Anyway for prime 6k+1 we want:

3n² + 3n + 1 = 0 (mod 6k+1)

Then we know that as 6k+1 is prime it is valid to write:

n = (-3 +- sqrt(-3))/6 (mod 6k+1)

where sqrt(-3) is equivalent to an integer via Gauss's theorem of Quadratic Reciprocity, and fractions are also equivalent to an integer because of Euclid's algorithm (which is explained somewhere else on NRICH). And so we have two valid values for which differ by an integer equivalent to:

sqrt(-3)/3, and this not equivalent to 0 as this would mean -3 = 0 (mod 6k+1).

So there are indeed exactly 2 values n that satisfy:

3n² + 3n + 1 = 0 (mod 6k+1)

if 6k+1 is prime. The two values for n (a,b) which satisfy this equation are related:

a + b = 6k

But as for dividing by (6k+1)² ... I'm not sure. The solutions certainly come in pairs (this is easy to show). But how to prove there are any at all??

Just a comment on your suggested infinite product. I'm pretty sure you're right, but you are using the probability relation P(A and B) = P(A) x P(B) which only holds when A,B are independent. So we have to show that a member of a sequence being divisible by one particular prime is independent of it dividing by another given prime. But I think we're OK here. Because to mod p the sequence 3n² + 3n + 1 goes in cycles of length p (p prime). As any two given primes are coprime then the lengths of the two sequences are coprime so I think I agree it is independent, so your relation should hold. I'll need to convince myself of this properly soon.

So if that's OK then all that remains is to be proven is that for prime p = 1 (mod 6) there exists an integer a such that:

3a² + 3a + 1 = 0 (mod p² )

or equivalently there exists b such that:

12b² + 1 = 0 (mod p² ).

Yours,

Michael

Would it help if I mentioned all primes above 3 are of the form 6k +- 1, or do you know that already? Might help you later on to save time

Gauss gave eight proofs of the quadratic reciprocity law and there are apparently over 50 floating around in the literature, I have a proof in a book but its pretty long and refers to other theorems in the book so I can't write it all out here, sorry! I expect you'll be able to find one on the internet somewhere or in a number theory book.

Why is sqrt(-3) equivalent to an integer?

I agree that 3n² + 3n + 1 = 0 (mod 6k+1) (*) has two solutions and am also not sure about the case with (6k+1)² .
This may help though,
if a is a solution to (*) then (a + yp) is also a solution where p = 6k+1 and we'll assume there is a value of y = t such that f(a+tp) is divisible by p² .
If f(n) = 3n² + 3n + 1 then using the Taylor series, f(a + tp) = f(a) + tp x f'(a) + t² p² x f''(a)/2 and all further terms are equal to 0 as f''(x) is constant.

Taking this modulo p² and using f''(x)=f''(a)=6,
f(a+tp) = f(a) + tp x f'(a) + 3 x t² p² (mod p² ) but f(a+tp) is presumed to be divisible by p² and 3 x t² p² is so we are left with;

f(a) + tp x f'(a) = 0 (mod p² )
f(a) = -tp x f'(a) (mod p² )
and as f(a) is divisible by p,
t x f'(a) = -f(a)/p (mod p² )

f'(a)=3 x (2a + 1) so
3 x t x (2a + 1) = -(3n² + 3n + 1)/p(mod p² ).

This is a linear equation in t which should probably be solvable but I'm still working on it.

I know what you mean about the independent bit, in the general square free number theory above it works because 1/p² of the integers are divisible by p² where p is a prime and 1/q² of the integers are divisible by q² where q is a prime and we can prove that 1/(pq)² of the integers are divisible by (pq)² so the proof is OK. I can't prove however the equivalent for the 3n² + 3n + 1 idea. Any ideas?

DOH!!! Yes of course you're correct. I tried a similar thing but made a horrendous algebraic blunder, so that I got a quadratic relationship with t not a linear one which made it unworkable. I think you can avoid Taylor's series though:

If there exists a so that 0 < = a < = p-1and

f(a) = 3a² + 3a + 1 = 0 (mod p)

then we can write:

3a² + 3a + 1 = kp (mod p² )

Now we know that:

f(a+tp) = 3(a+tp)² + 3(a+tp) + 1 = 3a² + 3a + 1 + 3t(2a+1)p = p[k + t x 3(2a+1)] (mod p² )

So we require:

k + t x 3p(2a+1) = 0 (mod p)

Now 2a+1 is coprime with p if 0 < = a < = p-1 so by Euclid's algorithm there is a solution for t in:

t x 3p(2a+1) = 1 (mod p) [if you don't know what Euclid?s algorithm is, I can direct you to the discussion where I learnt about it first].

And multiplying both sides of that equation by q reveals that there is a solution for t in:

t x 3p(2a+1) = q (mod p)

as well for all q. Therefore as the left hand side is periodic (with sequence length = p) we can see that there is exactly ONE solution in t for:

k + t x 3p(2a+1) = 0 (mod p)

for a given a.

But there are two possible values for a as we have agreed. They will each have an associated value for t.

In fact there can never be one solution because f(n) = f(p² -1-n) (mod p² ) and if you set n = (p² -1)/2 you find that 1 = 0 (mod p² ) which is a contradiction. So there are indeed two solutions for

3n² + 3n + 1 = 0 (mod p² ) where prime p = 1 (mod 6) as we were hoping.

Just one minor correction to what you wrote. You said:

3 x t x (2a+1) = -(3n² + 3n + 1)/p (mod p² )

when I think you meant:

3 x t x (2a+1) = -(3n² + 3n + 1)/p (mod p). Otherwise I agree with everything you've done.

Yours,

Michael

Sorry - I appear to have gained a p in the equations mod p. They should have read:

k + t x (2a+1) = 0 (mod p)

then

t x 3(2a+1) = 1 (mod p)

etc.

Please try to ignore all the ps in the mod p equations. If someone out there has the power to remove them then I'd be extremely grateful.

Very sorry about that,

Michael

PS Neil - yeah thanks, I was already aware of that. Andrew's point is that the only primes that do occur in this sequence are 6k+1 never 6k-1...

Right so I think where we're at is that the equation 3n² + 3n + 1 has two distinct solutions modulo p² where p is of the form 6x + 1and no solutions where p is of the form 6x -1, as there are no solutions where p = 2 or 3 this covers all the primes.

Yes, the p² should have been a p, I used the Taylor expansion because it comes from a more general theory for finding solutions to f(x) = 0 (mod p^j ) where you find solutions mod p then mod p² then mod p³ etc.

As I see the problem the final step is proving that if p₁ , p₂ , p₃ ,... are distinct primes of the form 6x + 1, the equation 3n² + 3n + 1 = 0 (mod (p₁ xp₂ x ... x p_n )² ) has 2ⁿ distinct solutions.

Here follows a proof of the quadratic reciprocity law, I tried scanning the book I have and it came out much better than I thought it would so didn't take too long to do. I don't think we need some of what I called theorem 2 but didn't want to risk removing bits we do need by mistake.

Theorem 1;

Lemma of Gauss. For any odd prime p let (a,p) = 1. Consider the integers a, 2a, 3a, ..., {(p - 1)/2}a and their least positive residues modulo p. If n denotes the number of these residues that exceed p/2, then (a/p) = (-1)ⁿ .
(a/p) is the Legendre symbol, see the last thing I wrote in Fibonacci in clock arithmetic for a definition.

Proof;

If $a$ is odd, this implies $n\equiv \sum _{j=1}^{(p-1)/2}[ja/p]$ (mod 2). If $a=2$ it implies $n\equiv (p^2-1)/8$ (mod 2) since $[2j/p]=0$ for $1\le j\le (p-1)/2$. The theorem now follows from Theorem 1.
Theorem 3;

The Gaussian reciprocity law. If p and q are distinct odd primes, then

(p/q)(q/p) = (-1)^{((p -1)/2)((q -1)/2)}

Another way to state this is: if p and q are distinct odd primes of the form 4k + 3, then one of the congruences x² = p (mod q) and x² = q (mod p) is solvable and the other is not; but if at least one of the primes is of the form 4k + 1, then both congruences are solvable or both are not.

Proof;

Let S be the set of all pairs of integers (x,y) satisfying 1 < = x < = (p - 1)/2, 1 < = y < = (q - 1)/2. The set S has (p - 1)(q - 1)/4 members. Separate this set into two mutually exclusive subsets S₁ and S₂ according as qx > py or qx < py. Note that there are no pairs (x,y) in S such that qx = py. The set S₁ can be described as the set of all pairs (x,y) such that 1 < = x < = (p - 1)/2, 1 < = y < qx/p. The number of pairs in $S_1$ is then seen to be $\sum _{x=1}^{(p-1)/2}[qx/p]$. Similarly $S_2$ consists of the pairs $(x,y)$ such that $1\le y\le (q-1)/2$, $1\le x<px/q$, and the number of pairs in $S_2$ is $\sum _{y=1}^{(q-1)/2}[py/q]$. Thus we have

$\sum _{j=1}^{(p-1)/2}[qj/p]+\sum _{j=1}^{(q-1)/2}[pj/q]=(p-1)/2\times (q-1)/2$
and hence (p/q)(q/p) = (-1)^{((p -1)/2)((q -1)/2)} by theorem 2.
Q.E.D. (at last)

This turned into something of an epic and I'd be very surprised if there aren't a few little mistakes, if anything doesn't seem to make sense say where it is and I'll check the book.

Also can you explain why sqrt(-3) is equivalent to an integer.

Andrew -many many thanks for including that derivation. You said it didn't take you long but you must have spent ages putting in all those sums and subscripts. I'm extremely grateful.

I'm no longer ashamed of not being able to prove the theorem when I had a go a couple of days ago. It is amazing! But equally amazing is the fact that I think I understand it. I think the absolute key behind the whole thing is that r₁ , r₂ ..., p-s₁ , p-s₂ , ...are just the integers 1,2,3,...,(p-1)/2 in a different order. This is an amazing discovery.

Once again many, many thanks.

Sqrt(-3) -I've got no idea on whether it is standard to write it like this, but I think I've proved that there aren't any contradictions if the base we're working to is prime .

What sqrt(-3) means (to me) is that if x = sqrt(-3) then x² = -3. So if x = sqrt(-3) (mod p) then x² = -3 (mod p). An example to show the kind of thing I meant.

Let x = sqrt(-3) (mod 7).

Now -3 = 4 (mod 7)

So x = sqrt(4) (mod 7) = ±2 (I'm not sure whether it is a good idea to take one over the other by convention. On the whole I think not.)

In general there can't be more than 2 values because if

a² = b² (mod p)

then

(a-b)(a+b) = 0 (mod p)

So either a = b (mod p) or a = -b (mod p), so there are 2 solutions.

Sometimes there isn't a value for sqrt(x) under a prime base. I think you've been calling these ?quadratic non-residues?.

Now you may ask why is this useful. Well maybe it isn't. But I was using it as follows.

We want to solve 3n² + 3n + 1 = 0 (mod p) [this is from my last but one message].

Quadratically:

n = (-3±sqrt(9-12))/6 = (-3±sqrt(-3))/6 (mod p)

Now when does it have a solution? Well as p is prime, if there is a solution for sqrt(-3) then there is also a solution for n as you're allowed to do things like dividing through by 6 with prime bases (I think).

So we need sqrt(-3) to have a value. By Gauss's reciprocity law (as -3 =/= 3 (mod 4)):

x² = -3 (mod p)

has a solution if and only if:

x² = p (mod -3) is soluble.

Now this is true if p = 1 (mod 3) but not true if p = -1 (mod 3). Therefore as you noticed the original equation does have solutions if p = 1 (mod 6) but not if p = -1 (mod 6). And we can also see if the solutions you get for p = 1 (mod 6) are identical. (i.e. if the quadratic has a repeated solution). If it did then:

(-3+sqrt(-3))/6 = (-3-sqrt(-3))/6 (mod p)

So:

sqrt(-3) = -sqrt(-3) (mod p)

2sqrt(-3) = 0 (mod p)

Squaring:

4(-3) = (mod p)

So p has to be 3 (we are not bothering about 2 in this as Gauss's law doesn?t even work for it). And indeed p = 3 is the only other prime to have one solution. Anything higher doesn't satisfy 12 = 0 (mod p).

So there are indeed two distinct solutions to the equation.

You could say it just amounts to a different way of writing things, perhaps it isn't done by anybody but me! It just seemed sort of natural.

I think I've proved your infinite product formula. But it is a bit fiddly (not hard) and I would like to make sure I don't make as many mistakes as usual. Unfortunately I won't be able to write for two weeks (starting tomorrow 7 am). I'll try to post my proof this evening or early tomorrow for you to comment on. It is really only a matter of organising easy ideas.

I look forward to seeing how the discussion has developed over the next two weeks. (I'm going to try to get hold of and read a copy of the book I started at school: ''From Polynomials to Sums of Squares ''as this looked like it had a lot of good stuff that might help us here.)

See you,

Michael

Michael, have a nice holiday (I assume!).

The formatting didn't take too long as you get into a rhythm and also I'd done the first half without too much formatting before noticing the numerous sigmas.

I understand how you're using the square roots now, I also think its probably OK but I think its not recognised because of the fact that as in your example sqrt(-3) is equivalent to 2 different integers mod 7 whereas a constant is assumed to be equivalent to only one integer in modular arithmetic. I think that if we use it with care then we're probably OK.

The way that I solved the quadratic was;
in general if we want to solve ax² + bx + c = 0 (mod p) assuming that a isn't a multiple of p(odd prime) then multiply the equation by 4a and complete the square to get
(2ax + b)² = b² - 4ac (mod p)

so in our case we get (3 x (2x + 1))² = -3 (mod p) and as 3 x (2x+1) for x from 1 to p are just a rearrangement of the integers 1 to p (mod p) the equation reduces to v² = -3 (mod p) which is what you had.

Yes, the last part of the proof isn't too difficult, I thought it was going to be really bad because I thought that we would need to find the solutions for a particular prime 6x+1 (mod (6x+1)² ) and then show that they intersected with the other primes at convenient places but then found a way to avoid this, I'll write it out in due course.

Any idea what the infinite product actually equals in a closed form?

Hi Andrew,

Yeah, you're right about square roots, and it being a bit confusing as it's multivalued. And completing the square was the obvious alternative to the way I did it. (I did that too as a check.)

Just to quickly outline the proof. Ithink I've simplified it a bit.

The sequences a,a+p² ,a+2p² ,... and b, b+q² ,b+2q² ,...

intersect at numbers at an interval of p² q² as can be seen by Euclid's algorithm (and then comparing prime factors of the relavant equation).

But concerning our sequence f(n) 3n² + 3n + 1 we have to duplicate this four times as f(p² -1-n)=f(n) (mod p). So in an interval of p² q² you get a single intersection of:

a,a+p² ,a+2p² and b, b+p² ,...

and then one when you replace a with p² -1-a, another when you replace b with p² -1-b, and another when you replace both.

So the chance of a prime in the sequence dividing p² and q² is 4/p² q² . This means that the probability relation:

P(A)P(B) = P(A and B)

is satisfied for A = divides by p² , B = divides by q² . Now do the same for showing that events dividing by p² q² and r² are also independent, and all other combinations. It is very easy to see all events are independent.

Therefore not dividing by p² is independent of not dividing by q² . And so on. Therefore the probability relation holds.

I'm sorry if that was a bit rushed...

Anyway, we can compare notes on the infinite product in two weeks time.

Michael

Andrew - any luck with the infinite product? I had a go at it about a week ago and got an interesting intermediate result but I can't do the problem at the moment. Of course it is tempting to suspect that there is no closed form solution...

Thanks,

Michael

No luck I'm afraid, I can't even find a solution by trial and error using pi and e etc, maybe there isn't a solution.

Hi Andrew,

Yes, I also suspect there isn't a solution in closed form. (I don't know how you'd go about proving things like that though.) I really thought there was one because I had a method using periodicity properties of complex numbers which looked quite promising. But it simply doesn't work so there's not much point in even writing about it. I'll have another go tomorrow though. If there is a way of doing it I think we're going to have to be a lot more creative than my attempts so far!

There's a web page you might find useful, called the "Inverse Symbolic Calculator" (a quick search on Yahoo or something should bring this page up). You type in a number, and it suggests possible closed forms for it, by comparing with a huge database of possibilities. You might like to give that a go for your infinite product.

Thanks Dan. I tried to find it numerically using VB and if you consider primes less than 3.5 x 10⁶ then I think the number comes out as:

0.934842033

but I'm not condident after the 6th decimal place. Plugging this into the website you mentioned (which looks incredibly useful generally) it suggested:

1) A root of a quintic equation with integral coefficients.

2) An infinite product which was a lot more complicated than the one we're considering.

I suspect neither of these are correct. I'll see if I can find a few more decimal places and try again.

Thanks,

Michael

Thanks Michael, that whole discussion was pretty interesting.