## Pi product?

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• I was just playing around with factors of F(p), and noticed a curious onvergence towards pi. 2*5*13*37*73*89*... ___________________ = pi 1*4*12*36*72*88*...
Message 1 of 16 , Oct 19, 2002
I was just playing around with factors of F(p), and noticed a curious
onvergence towards pi.

2*5*13*37*73*89*...
___________________ = pi

1*4*12*36*72*88*...

Product of F(p)
___________________________________________ = pi

Product of, one less than the factors of F(p)

I have checked this to F(1000) ofcourse, to get pi~ 3.103...
and it's slowing down fast. Could this be?
• Factors of F(p) In linear order
Message 2 of 16 , Oct 19, 2002
Factors of F(p)

In linear order<2000:
2,5,13,37,73,89,113,149,157,193,233,269,277,313,353,389,397,457,557,61
3,673,677,733,757,877,953,977,997,1069,1093,1153,1213,1237,1453,1597,1
657,1753,1873,1877,1933,1949,1993,...

Notice that there are only 5 ending in a 9:
89,269,389,1069,1949.
There are none ending in a 1, yet!

Why is there a discrimination,in linear order?
• ... No. If you keep computing it, it passes pi when you multiply by 11617/11616. 11617 is the 153rd prime which is a factor of a Fibonacci number with prime
Message 3 of 16 , Oct 19, 2002
Shane wrote:
>
> I was just playing around with factors of F(p), and noticed a curious
> onvergence towards pi.
>
> 2*5*13*37*73*89*...
> ___________________ = pi
>
> 1*4*12*36*72*88*...
>
> Product of F(p)
> ___________________________________________ = pi
>
> Product of, one less than the factors of F(p)
>
> I have checked this to F(1000) ofcourse, to get pi~ 3.103...
> and it's slowing down fast. Could this be?

No. If you keep computing it, it passes pi when you multiply by
11617/11616. 11617 is the 153rd prime which is a factor of a
Fibonacci number with prime index.

But you do raise an interesting question... Does this product
converge? Computing it for all primes < 10^6, of which 4878 meet
the criterion, it only reaches a product of approx. 3.228752.

The answer depends on how sparse such primes are, and how fast
their density drops off (assuming that it does).
• ... Just past your horizon: 2221. And more: 5381, 8861, 10061, 15761, 21481, 23021, 26141, 27941, ... ... Another good question. Of the first 1200 such
Message 4 of 16 , Oct 19, 2002
Shane wrote:
>
> Factors of F(p)
>
> In linear order<2000:
> 2,5,13,37,73,89,113,149,157,193,233,269,277,313,353,389,397,457,557,61
> 3,673,677,733,757,877,953,977,997,1069,1093,1153,1213,1237,1453,1597,1
> 657,1753,1873,1877,1933,1949,1993,...
>
> Notice that there are only 5 ending in a 9:
> 89,269,389,1069,1949.
> There are none ending in a 1, yet!

And more: 5381, 8861, 10061, 15761, 21481, 23021, 26141, 27941, ...

> Why is there a discrimination,in linear order?

Another good question. Of the first 1200 such primes:

Ending in 1: 40
Ending in 2: 1
Ending in 3: 506
Ending in 5: 1
Ending in 7: 488
Ending in 9: 164

Why such a heavy prevalance of numbers ending in 3 or 7? Note that
the ratios are also close to integers: Ratio of xxx1:xxx9 is about
1:4, ratio of xxx9:xxx7 is about 1:3, and ratio of xxx7:xxx3 is about
1:1... Interesting...
• I will just list a few of L(p), because it is just to tedious. 3,7,11,29,59,139,179,199,359,461,509,521,619,709,809,1031,1049,1511,.. . With an abundance,
Message 5 of 16 , Oct 19, 2002
I will just list a few of L(p), because it is just to tedious.

3,7,11,29,59,139,179,199,359,461,509,521,619,709,809,1031,1049,1511,..
.

With an abundance, approx 1/2 ratio between 9, and 1.
• This could be of use. We would need to know if this trend continues or reverses itself, where the primes ending in 1,3,7,9 have equal probability, over time.
Message 6 of 16 , Oct 19, 2002
This could be of use.

We would need to know if this trend continues or reverses itself,
where the primes ending in 1,3,7,9 have equal probability, over time.

Cycles maybe?
• In a message dated 20/10/02 05:27:34 GMT Daylight Time, TTcreation@aol.com ... Relevant to this, and to your Fib observations, is p. 67 of Ribenboim s NBPNR,
Message 7 of 16 , Oct 20, 2002
In a message dated 20/10/02 05:27:34 GMT Daylight Time, TTcreation@...
writes:

> I will just list a few of L(p), because it is just to tedious.
>
> 3,7,11,29,59,139,179,199,359,461,509,521,619,709,809,1031,1049,1511,..
> .
>
> With an abundance, approx 1/2 ratio between 9, and 1.
>

Relevant to this, and to your Fib observations, is p. 67 of Ribenboim's
NBPNR, where he records that Jarden (1958), Ward (1961) and others have
characterised the set of primes which divide Lucas numbers.
In particular, primes = 3, 7, 11, 19 mod 20 divide some L(n), primes = 13 or
17 mod 20 don't divide any L(n), and for p = 1, 9 mod 20 "there is no finite
set of divisibility conditions to describe completely" [whether they do or
don't]. Isn't that remarkable?

I know you are restricting your attention to L(p) for p _prime_, but with
these previous results in mind, it would seem interesting to do your analyses
w.r.t. mod 20, not mod 10.

Mike

[Non-text portions of this message have been removed]
• Subject: Re: Linear factors of F(p), L(p) ADVERTISEMENT ... I had just noticed that again. P = 1 mod 20 ratio ~ 1/1 P = 11 mod 20 ratio ~ 1/2 P = 9 mod 20
Message 8 of 16 , Oct 20, 2002
Subject: Re: Linear factors of F(p), L(p)

> w.r.t. mod 20, not mod 10.
>
> Mike

I had just noticed that again.

P = 1 mod 20
ratio ~ 1/1
P = 11 mod 20

ratio ~ 1/2

P = 9 mod 20
ratio ~ 1/1
P = 19 mod 20

Does this continue to hold though?
• Mike Oakes wrote, For p = 1, 9 mod 20 there is no finite ... they do or don t]. Isn t that remarkable? There seems to be a certain probability that mirrors
Message 9 of 16 , Oct 20, 2002
Mike Oakes wrote,
For p = 1, 9 mod 20 "there is no finite
> set of divisibility conditions to describe completely" [whether
they do or don't]. Isn't that remarkable?

There seems to be a certain probability that mirrors p=11,19 mod 20.
Remarks?

1<=11
9<=19 ?
• Jack Brennen wrote ... For primes
Message 10 of 16 , Oct 20, 2002
Jack Brennen wrote

> Computing it for all primes < 10^6,
> of which 4878 meet the criterion,
> it only reaches a product of approx.
> 3.228752.

For primes < 10^7 it attains the value

3.25453335...

I'll try letting the code run up to 10^8.

David
• It looks as if the product converges, about as slowly as Brun s twin-prime constant. Here, at random, is a screenful of prime {q,p} with q|F(p), ordered by the
Message 11 of 16 , Oct 20, 2002
It looks as if the product converges, about
as slowly as Brun's twin-prime constant.

Here, at random, is a screenful
of prime {q,p} with q|F(p),
ordered by the divisor q:

[58470317, 3248351],
[58470793, 29235397],
[58470869, 14617717],
[58471069, 4872589],
[58471213, 29235607],
[58471513, 29235757],
[58473397, 29236699],
[58473433, 29236717],
[58473977, 9745663],
[58474193, 9745699],
[58474573, 29237287],
[58474777, 29237389],
[58475069, 14618767],
[58475617, 29237809],
[58476521, 1461913],
[58477813, 29238907],
[58478593, 29239297],
[58478677, 29239339],
[58479437, 9746573],
[58479697, 29239849],

At this size, slightly more than a half
are Drobot-Cunningham pairs:
q=2*p-1 with p=13 or 17 mod 20.

If drobots remained a finite fraction, the log of
the product would converge in a manner similar
to Brun's constant, since
integral dx/(x*log(x)^2)
=1/1og(x)+const
is kosher at large x.

David
• David: With these linear,primes mod 20, Are the remainder ratios, Jack and I have observed correct? Or is this just a local phenomenon? Shane F.
Message 12 of 16 , Oct 20, 2002
David:
With these linear,primes mod 20,
Are the remainder ratios, Jack and I have observed correct?
Or is this just a local phenomenon?

Shane F.
• Primes q
Message 13 of 16 , Oct 20, 2002
Primes q < 10^8 dividing some F(p) with prime p;
final digit pattern for the divisor, q:

Ending in 1....6835
Ending in 2.......1
Ending in 3..104149
Ending in 5.......1
Ending in 7..104217
Ending in 9...36781

Total....... 251984

of which

Drobots......152904 (60.7%)

prod_{q<10^8} 1/(1-1/q)
=3.27365650101...
• Since gcd(F(n),F(m))=F(gcd(m,n)), we may associate to each prime q at most one prime p such that q|F(p). For 10^8 q|F(p), the final-digit p-pattern is
Message 14 of 16 , Oct 20, 2002
Since gcd(F(n),F(m))=F(gcd(m,n)),
we may associate to each prime q
at most one prime p such that q|F(p).

For 10^8>q|F(p), the
final-digit p-pattern is
............!

Ending in 1...14378
Ending in 2........
Ending in 3...27047
Ending in 5.......1
Ending in 7..105132
Ending in 9..105426

PS: The largest F(p) with
prime index p and a prime divisor
q<10^8 has more than 10 million decimal digits.
• Would it not be odd if it reached 3.359885666....? Although, I must confess, it is a long shot. The above is the sum of the reciprocals of the Fibonacci.
Message 15 of 16 , Oct 20, 2002
Would it not be odd if it reached 3.359885666....?

Although, I must confess, it is a long shot.

The above is the sum of the reciprocals of the Fibonacci.
• ... This number is [veta 1]+1 The answer is no, when [veta s] is small, these two numbers are farthest apart. Conversely when s is large they converge, at a
Message 16 of 16 , Oct 23, 2002
>"mistermac39" wrote:
> Would it not be odd if it reached 3.359885666....?>
> Although, I must confess, it is a long shot.
> The above is the sum of the reciprocals of the Fibonacci.

This number is [veta 1]+1

The answer is no, when [veta s] is small, these two numbers are
farthest apart. Conversely when s is large they converge, at a
wonderful rate.

I believe you were off by 1.

See veta message,
Shane F.
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