Quote:
Originally Posted by Don Blazys
Wow, what a monster result! Sure beats my measly 25 digit calculation!
How much calculating time did that take?
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It took just under 6 minutes (358 sec) to calculate
for
. The speed of the calculation depends on the size (number of integer digits) of the numbers, so it gets slower as it proceeds: calculating
for
took an additional 176 sec. This is on 2004 commodity (ie: cheap) laptop, using a programming language not optimal for high-precision arithmetic (
MUMPS).
I spent a couple of hours performing a few steps with the equivalent of paper and pencil, then writing the program. The program is just a strait-forward automation of the algebra of writing
in terms of
for each
, via the formula
which always has a solution of the form
where j, k, r, and s are integers
, then solving the inequality
in terms of
My program is in a less well-known language, and calls my own unlimited precision subroutines. This is it:
Code:
K P s P=2,P(2)="",J=1,K=0,M=0,N=1 f w "(",$p(J,"/"),"*A",$p($s(K'<0:"+"_K,1:K),"/"),")/(",$p(M,"/"),"*A",$p($s(N'<0:"+"_N,1:N),"/"),")" w " P=",P," " d SRM(.j,P,N),SRS(.j,j,K),SRM(.k,P,M),SRS(.k,J,k),SRA(.m,j,N),SRS(.n,k,M) w $p(j,"/"),"/",$p(k,"/")," <= A < ",$p(m,"/"),"/",$p(n,"/") r R,! s j=J,k=K,m=M,n=N d SRM(.J,P,m),SRM(.K,P,n),SRM(.M,P,m),SRM(.M,M,-1),SRA(.M,M,j),SRM(.N,P,n),SRM(.N,N,-1),SRA(.N,N,k) s P=P'=2+1+P,I=0 F S I=$O(P(I)) S:'I P(P)=1 Q:'I S:P#I=0 P=P+2,I=0
But I’d not expect someone unfamiliar with the MUMPS language to be able to easily read it. In pseudo code, it’s:
- Initialize 4 integers j, k, m and n to: j=1; k=0; m=0; n=1
- Get the nth prime number P
- is between (P*n -k)/(P*m -j) and ((P+1)*n -k)/((P+1)*m -j). Note that the program doesn’t tell which is the lower, inclusive bound, and which the upper, exclusive one.
- Assign new values to j, k, m and n:
j= P*m; k= P*n;
m= j -P*m; n= k -P*n
(note that calculations must be simultaneous, ie: the j and k used to update m and n must be their pre-update values)
- Repeat steps 2 – 4 until desired precision reached.
There’re some interesting math questions raised here, among them “does
exist that can generate all the primes?” Note that this is a very different question than “Can
be written in terms other than all of the primes?” which is what your conjecture asserted to prove by example.
The question of the existence of appears to me to be equivalent to a restatement of the
Reimann hypothesis as it applies to the distribution of primes (its original description). In short, if there is no unexpectedly large gap in the distribution of the primes, then the RH is true, and
exists. If such a gap exists, the RH is false, and
doesn’t exists.
We can illustrate this by following the above program with the assumption that the
, rather than its actual value of 11. This gives output (note that
is written A in this output:
Code:
1. (1*A+0)/(0*A+1) P=2
2/1 <= A < 3/1
2.000000000000000000000000000000 <= A < 3.000000000000000000000000000000
2. (0*A+2)/(1*A-2) P=3
-8/-3 > A <= -10/-4
2.666666666666666666666666666666 > A <= 2.500000000000000000000000000000
3. (3*A-6)/(-3*A+8) P=5
46/18 <= A < 54/21
2.555555555555555555555555555555 <= A < 2.571428571428571428571428571428
4. (-15*A+40)/(18*A-46) P=7
-362/-141 > A <= -408/-159
2.567375886524822695035460992907 > A <= 2.566037735849056603773584905660
5. (126*A-322)/(-141*A+362) P=23
8648/3369 <= A < 9010/3510
2.566933808251706737904422677352 <= A < 2.566951566951566951566951566951
Because the ranges
and
don’t overlap,
could not exist if the first 5 primes were 2, 3, 5, 7, 23 rather than 2, 3, 5, 7, 11.
Don, you seem to having the same thought, evidenced by your question
Quote:
Originally Posted by Don Blazys
For instance, do you expect that someone will ever find a "non-trivial zero" of the Riemann zeta function with a "real part" other than (1/2)? I don't.
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which is another phrasing of the Riemann hypothesis. I share your belief that the RH is true, though, along with every known person in the world, I’ve not proven it.
That
exists is, as I note, very different than the question of it being describable other than in terms of the primes.
The
estimating technique for we’ve shown in this thread is unimpressive in the sense that, along with the generating function
all it shows is that we can generate the first n primes if we already know the first n primes.
We’ve show that
for
![\frac{\left[\left(sin(x^{(1/2)})\right)-1\right]^{-1}}{\pi^2+\left[\ln\left(\ln\left(2\left[\left[2(x^{-1}+1)\right]^{-1}+1\right]\right)\right)\right]^{-1}}-1=0](http://hypography.com/forums/latex/img/38faa33b0c09c540a50b72059a2ee470-1.gif "\frac{\left[\left(sin(x^{(1/2)})\right)-1\right]^{-1}}{\pi^2+\left[\ln\left(\ln\left(2\left[\left[2(x^{-1}+1)\right]^{-1}+1\right]\right)\right)\right]^{-1}}-1=0")
or, the prettier
, but I’m not aware of a proof that no such expression can exist, though I’ve a strong suspicion none can. If it did, its implications would be staggering. A proof that no such expression can exists would be profound.
