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## RE: [PrimeNumbers] Consecutive primes

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• This thread reminds me of an ancient conjecture by the great (and late) Indian mathematian, Shawtoff, prove that any polynomial f(x)=sigma(i=0 to oo;a_i.x^i)
Message 1 of 16 , Apr 19, 2003
This thread reminds me of an ancient conjecture by the great (and late)
Indian mathematian, Shawtoff, prove that any polynomial f(x)=sigma(i=0 to
oo;a_i.x^i) with integer coefficients a_i, a_i !=0 for at least one i!=0,
always contains at least 1 composite.

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• Usual disclaimer: HS-dropout, no college, self-taught I ve no idea whether this little equation is already known, or whether it would even be of any interest.
Message 2 of 16 , Apr 19, 2003
Usual disclaimer: HS-dropout, no college, self-taught

I've no idea whether this little equation is already known, or
whether it would even be of any interest. But I was tinkering with
some numbers and discovered something that is rather interesting to
me.

(3+6*N)^2+2

This simple equation seems to yield a rather large number of primes.
Enough so to where I've seen some rather large runs of consecutive
primes (as many as 8 so far).
• In reponse to Mr. Berg post, ... Well, one reason might be that: (3+6*N)^2+2 == 2 ~== 0 (mod 3) and, (3+6*N)^2+2 == 9*(1+2*N)^2+2 == 9*(1+2*N)^((5-1)/2)+ 2 ==
Message 3 of 16 , Apr 19, 2003
In reponse to Mr. Berg post,

> I've no idea whether this little equation is already known, or
> whether it would even be of any interest. But I was tinkering with
> some numbers and discovered something that is rather interesting
> to me.

> (3+6*N)^2+2

> This simple equation seems to yield a rather large number of
> primes. Enough so to where I've seen some rather large runs of
> consecutive primes (as many as 8 so far).

Well, one reason might be that:

(3+6*N)^2+2 == 2 ~== 0 (mod 3)

and,

(3+6*N)^2+2 == 9*(1+2*N)^2+2 == 9*(1+2*N)^((5-1)/2)+ 2
== -1*(+-1) + 2 == +-1 + 2 ~== 0 (mod 5)

so it's not divisble by 3 or 5. That and a little luck would
probably help. I could analyze this more but I'll leave that up to
you. But if you really want to help yourself, go back to school.

Ehren
• ... Thank you for your analysis and suggestion. But at age of 50, I think I ll continue dabbling and not spend the next half dozen years book-learning - as
Message 4 of 16 , Apr 19, 2003
--- In primenumbers@yahoogroups.com, "physiguy" <physiguy@h...> wrote:

> you. But if you really want to help yourself, go back to school.

Thank you for your analysis and suggestion.

But at age of 50, I think I'll continue dabbling and not spend the
next half dozen years book-learning - as long as that's okay with
you, of course.
• ... It also isn t divisible by 13, 23, 29, 31, 37, 47.
Message 5 of 16 , Apr 19, 2003
At 09:43 PM 4/19/2003, physiguy wrote:

>Well, one reason might be that:

>so it's not divisble by 3 or 5.

It also isn't divisible by 13, 23, 29, 31, 37, 47.
• ... I could see originally from my own observation that it produced N s that were not divisible by small divisors, but I missed this point. Thank you for
Message 6 of 16 , Apr 19, 2003
--- In primenumbers@yahoogroups.com, Jud McCranie <judmccr@b...>
wrote:
> At 09:43 PM 4/19/2003, physiguy wrote:
>
> >Well, one reason might be that:
>
>
> >so it's not divisble by 3 or 5.
>
> It also isn't divisible by 13, 23, 29, 31, 37, 47.

I could see originally from my own observation that it produced N's
that were not divisible by small divisors, but I missed this point.
Thank you for bringing it out.
• There are, implementing your formula in Pari-GP: 8 primes up to n=10 47 primes up to n=100 283 primes up to n=1000 2159 primes up to n=10^4 17476 primes up to
Message 7 of 16 , Apr 20, 2003
There are, implementing your formula in Pari-GP:

8 primes up to n=10
47 primes up to n=100
283 primes up to n=1000
2159 primes up to n=10^4
17476 primes up to n=10^5
147150 primes up to n=10^6

A curious: if n is negative, say n=-k, then (3+6*n)^2 +2 = [3+6*(k-1)]^2 +2 (cuadratic <--> symmetric)

You can rewrite your formula like this:
(3+6n)^2 + 2 = 36n(n-1)+11

Taking k*n(n-1)+11, with the coefficients k=3,13,31 we have
308,405,370 primes up to 1000.

Amazingly your formula gets more primes than near all the other formulas with k primes (up to k=80).

I was thinking that all the k*n(n-1)+11 must have the same ratio primes/(count_of_n) as n becomes bigger, but maybe this isn't true. Anyone knows? For example, for n=10^5, for k=41, there are 10416 primes, which are over 7000 primes less than with k=36 (the k suggested by Berg). For k=31, we have 21719 primes for n=10^5.

Let's take again the formula: k*n(n-1)+11

With the squares:
k=1, n=1000 , primes: 289, n=10^5, primes: 15662
k=4, n=1000 , primes: 177, n=10^5, primes: 9884
k=9, n=1000 , primes: 206, n=10^5, primes: 12075
k=16, n=1000 , primes: 276, n=10^5, primes: 16698
k=25, n=1000 , primes: 213, n=10^5, primes: 13031
k=49, n=1000 , primes: 231, n=10^5, primes: 14381
k=64, n=1000 , primes: 234, n=10^5, primes: 14111
k=81, n=1000 , primes: 237, n=10^5, primes: 14760

PARI code
{ k=? , j=? }
{ psum=0;for(n=0,10^j,if(isprime(k*n*(n-1)+11),psum=psum+1));psum }

Let's search now for the five optimal k for n up to 10^3, 10^4 and for k up to 1000

I define the yield of the formula as number of primes / maximun n

n up to 10^3

k= 276, primes: 345. Yield: 0,345
k= 91, primes: 354. Yield: 0,354
k= 1000, primes: 361. Yield: 0,361
k= 31, primes: 369. Yield: 0,369
k= 13, primes: 404. Yield: 0,404

The average number of primes is 153.097 (counting multiples of 11 that trivially have only 1 prime, n=1)

The average yield is 0,153

Curios: 144*n*(n-1)+11 is always composite if n>1 (surely this holds for every M^2*n*(n-1)+(M-1) )

n up to 10^4

k= 133, primes: 2665. Yield: 0, 267
k=31, primes: 2693. Yield: 0, 269
k= 91, primes: 2722. Yield: 0, 272
k= 1000, primes: 2774. Yield: 0,277
k= 13, primes: 3033. Yield: 0,303

The average number of primes is 1198.

The average yield is 0,120

We observe that the numbers k are moreless the same. The other combinations are:
n=10^3
k=133, primes: 328. Yield: 0,328

n=10^4
k= 276, primes: 2601. Yield: 0,260

If we pick the first 10^4 numbers "as are", then by the PNT we will have about 10^4/log(10^4) = 1085 prime numbers. Using any of the k*n*(n-1)+11 formulas we have 10% more primes in average, and twice the amount if we make a good choice (like k=13) - yes, the sizes of the numbers are very different in both cases, i'm only supossing that we want to find N primes in at most P trys).

This problem remembered me the NumberSpiral of our friend Robert Sacks (http://www.numberspiral.com/index.html), because this formulas look like his P curves.

Now an (important?) question: if we take A*n(n-1)+B, what is the pair of numbers (A,B) that maximizes the amount of primes we get up to a maximum n? What about other non-linear formulas, like, for example, A*n(n-1)+B*n+C or A*n(n-1)*(n-7)+B ?

Is all this studied yet?

Jose Brox
http://espanol.groups.yahoo.com/group/Telecomunicacion

----- Original Message -----
From: j_m_berg
Sent: Sunday, April 20, 2003 2:27 AM
Subject: [PrimeNumbers] Consecutive primes

Usual disclaimer: HS-dropout, no college, self-taught

I've no idea whether this little equation is already known, or
whether it would even be of any interest. But I was tinkering with
some numbers and discovered something that is rather interesting to
me.

(3+6*N)^2+2

This simple equation seems to yield a rather large number of primes.
Enough so to where I've seen some rather large runs of consecutive
primes (as many as 8 so far).

Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
The Prime Pages : http://www.primepages.org/

[Non-text portions of this message have been removed]
• ... Sort of. See A17 in Richard Guy s Unsolved Problems in Number Theory. The asymptotic density of primes given by a quadratic with integer coefficients is
Message 8 of 16 , Apr 20, 2003
> Now an (important?) question: if we take A*n(n-1)+B, what is the pair
> of numbers (A,B) that maximizes the amount of primes we get up to a
> maximum n? What about other non-linear formulas, like, for example,
> A*n(n-1)+B*n+C or A*n(n-1)*(n-7)+B ?
>
> Is all this studied yet?

Sort of. See A17 in Richard Guy's Unsolved Problems in Number Theory.

The asymptotic density of primes given by a quadratic with integer
coefficients is believed to be c*sqrt(n)/log(n).

Of course, c can be as low as 0. An easy example is x^2+x+4, which
is never prime.

The value of c can be made high as well. Guy gives the "best"
example with the function:

x^2+x+132874279528931

for which he gives c = 5.0870883.

This polynomial is never divisible by any prime smaller than 181,
and asymptotically should yield many primes.

For x in the range [0 ... 10000], this polynomial yields 3141 primes,
which compares quite favorably with all of your examples.

Euler's well known function x^2+x+41 has corresponding c = 3.3197732,
and in the range [0 ... 10000] yields 4149 primes, quite a good count.
But the polynomial above should overtake it eventually.

Another function is x^2+x+27941, with c = 3.6319998. This one gives
an even more impressive 4466 primes in the range [0 ... 10000]. But
it too should eventually be surpassed by the one above.
• To address the mathematics, as long as p is a prime for which -2 is a quadratic residue and relatively prime to 6, p will divide (3+6*n) ^2+2 twice in a string
Message 9 of 16 , Apr 20, 2003
To address the mathematics, as long as p is a prime for which -2 is a
quadratic residue and relatively prime to 6, p will divide (3+6*n)
^2+2 twice in a string of p consecutive values for n. Once you solve
x^2+2==0 mod p, you can then solve 3+6*n==x mod p. For this n value,
and every p thereafter (n+p, n+2p, n+3p...) the value will always be
divisible by p. Such a prime is p=11, for which 3^2=9==-2 mod 11.
Of course, the other root is (-3)^2==-2 mod 11. So, if n is 0 or 10
mod 11, (3+6*n)^2+2 is divisible by 11 (and if n is large enough,
this guarantees it is not prime). The longest possible string of
primes is therefore n==1,2,3,4,5,6,7,8,9 mod 11, and, with finding a
string of 8, you are almost to the predicted maximum.

Other such primes are 17, 19, 41, 43, 59, 67, 73, 83, 89, 97.
Perhaps the Chinese remainder theorem could be employed to limit the
length of possible prime sequences further, using this info.

To address the other topics, I support your curiosities, and urge you
to continue having fun with math topics. I would suggest that your
individual "educational status" is irrelevant, and you don't need to
post that at all. It is hard to ignore insensitive people, but it is
also a lot more wonderful world to live in when you remove their
influence from your life (as much as is possible).

--- In primenumbers@yahoogroups.com, "j_m_berg" <jberg@e...> wrote:
> Usual disclaimer: HS-dropout, no college, self-taught
>
> I've no idea whether this little equation is already known, or
> whether it would even be of any interest. But I was tinkering with
> some numbers and discovered something that is rather interesting to
> me.
>
> (3+6*N)^2+2
>
> This simple equation seems to yield a rather large number of
primes.
> Enough so to where I've seen some rather large runs of consecutive
> primes (as many as 8 so far).
• ... I only count probable primes in the following and may be one off because some results are old and started at x=1 - too lazy to check x=0 now.
Message 10 of 16 , Apr 20, 2003
Jack Brennen wrote:
> Sort of. See A17 in Richard Guy's Unsolved Problems in Number Theory.
>
> The asymptotic density of primes given by a quadratic with integer
> coefficients is believed to be c*sqrt(n)/log(n).
>
> Of course, c can be as low as 0. An easy example is x^2+x+4, which
> is never prime.
>
> The value of c can be made high as well. Guy gives the "best"
> example with the function:
>
> x^2+x+132874279528931
>
> for which he gives c = 5.0870883.
>
> This polynomial is never divisible by any prime smaller than 181,
> and asymptotically should yield many primes.
>
> For x in the range [0 ... 10000], this polynomial yields 3141 primes,
> which compares quite favorably with all of your examples.
>
> Euler's well known function x^2+x+41 has corresponding c = 3.3197732,
> and in the range [0 ... 10000] yields 4149 primes, quite a good count.
> But the polynomial above should overtake it eventually.
>
> Another function is x^2+x+27941, with c = 3.6319998. This one gives
> an even more impressive 4466 primes in the range [0 ... 10000]. But
> it too should eventually be surpassed by the one above.

I only count probable primes in the following and may be one off because some
results are old and started at x=1 - too lazy to check x=0 now.
x^2+16161x+6067 gives 4761 primes to x=10000.

Only going to 10000 gives small coefficients a big advantage, because the
candidates simply become smaller:
x^2+x+27941 gives 4466 primes to 10000 but only 35150 to 100000.
x^2+x+854032997 gives 4366 primes to 10000 and 41320 to 100000.

Try to beat my record in CYF NO. 13 at www.shyamsundergupta.com/canyoufind.htm
This allows negative and repeated primes and I got 44484 primes to 100000 in
x^2-100001x+2498695637
The puzzle asks for 50000 but that is beyond me.
According to my computations, 92.4% of the candidates will not have a prime
factor below 250 for x^2+x-7731189253. It only gave 41503 primes to 100000.

There are also results and discussions in this german paper:
www.birkhauser.ch/journals/1700/papers/9054002/90540064.pdf
Even if you don't know german, you may get something out of it.

--
Jens Kruse Andersen
• ... Adam, That is an absolutely horrible thing to say, remove their influence from your life. Whether you realize it or not, everybody influences everybody
Message 11 of 16 , Apr 20, 2003
> It is hard to ignore insensitive people, but it is also a lot more
> wonderful world to live in when you remove their influence from
> your life (as much as is possible).

That is an absolutely horrible thing to say, "remove their influence
from your life." Whether you realize it or not, everybody influences
everybody else in this world and that's what life is all about. And
you don't have any idea how sensitive I am. I am a math tutor, and I
haven't had a student yet that hasn't gotten an A. I was also a
cancer researcher at NCI and developed a computer program that
discovers new cancer treatment drugs. The one and only program that
is able to do this. So, if you get cancer you'll be glad you didn't
remove my influence from your life. I care about people
tremendously, and I've made it my personal and professional goal in
life to help as many people as I can. When I see somebody who looks
like they need help I'll do whatever I can to help them. Even if
that means rebuke. You know, rebuke is not a bad thing. What would
happen if you never scolded a child. I'll tell you what would
happen, they would never learn, they would never accomplish
anything, they would never have a happy and productive life, and the
list goes on. I certainly respect Mr. Berk and only want the best
for him. And I do not mean that in any way other than the way I said
it. When he refered to "book-learning" as if it was something he was
terrified of, I felt a need to help, whether or not he wanted my
advice. I saw a way I could help him and I and tried to do so. That
is all there was to it. There was no insensitivity ever directed
towards him or you. It would only be insensitive if I ignored his
situation. I hope this has cleared up my intentions. So, encourage
Mr. Berk to indepently study, but I also want to clarify that "book-
learning" is not a bad thing, no matter how old you are.

Ehren
• At 07:24 PM 4/20/2003, physiguy wrote: like they need help I ll do whatever I can to help them. Even if ... We have to remember, though, that there are many
Message 12 of 16 , Apr 20, 2003
At 07:24 PM 4/20/2003, physiguy wrote:

like they need help I'll do whatever I can to help them. Even if
>that means rebuke. You know, rebuke is not a bad thing.

We have to remember, though, that there are many types of people on this
list. Some are students, some have an interest in prime numbers but aren't
mathematically sophisticated, etc. For instance, a bright 12 year old
might make some observation about primes, and we wouldn't want to rebuke
them, would we?
• ... I ve spent three decades as a systems design engineer in Silicon Valley and chances are I ve helped design much of the systems you ve programmed. ... Thank
Message 13 of 16 , Apr 20, 2003
> haven't had a student yet that hasn't gotten an A. I was also a
> cancer researcher at NCI and developed a computer program that
> discovers new cancer treatment drugs. The one and only program that
> is able to do this.

I've spent three decades as a systems design engineer in Silicon
Valley and chances are I've helped design much of the systems you've
programmed.

> like they need help I'll do whatever I can to help them. Even if
> that means rebuke.

Thank you for rebuking me. However who asked you to do so? I was
presenting something of interest to myself (and others, judging by
the thread). As I said before, it was for you to decide whether to
investigate it or ignore it.

> ... What would
> happen if you never scolded a child. I'll tell you what would
> happen, they would never learn, they would never accomplish
> anything, they would never have a happy and productive life, and
the
> list goes on.

At 50, I doubt I am a child. And I seem to have a happy and
productive life after three decades as a self-taught systems
engineer. However it now becomes obvious why you feel that you're
qualified to rebuke me - given that you view me as a child.

> it. When he refered to "book-learning" as if it was something he
was
> terrified of, I felt a need to help, whether or not he wanted my
> advice. I saw a way I could help him and I and tried to do so.

I've spent my life "learning" since I'm a self-taught systems
engineer. The fact that I choose not to spend the next half decade
getting a degree in math, is simply that I see no need to do so for a
hobby! Thank you for your "help", but your entire theory was
incorrect given that I obviously have zero fear of "learning". This
includes the dozens of courses I've taken on subjects that I have
choosen to pursue.

> is all there was to it. There was no insensitivity ever directed
> towards him or you. It would only be insensitive if I ignored his
> situation.

Personally I would have preferred you to ignore my "situation" if it
means you weren't so damned condescending! People don't like having
someone tell them that their hobby is equivilent to cloud watching!
While I obviously won't create any breakthroughs, it is my right to
enjoy my life as I wish!
• After being on this list for several months, I begin to understand others frustration with your post, Jon. Polynomials are not composite, rather, they are
Message 14 of 16 , Apr 22, 2003
After being on this list for several months, I begin to understand
others' frustration with your post, Jon. Polynomials are not
composite, rather, they are reducible or irreducible, and you really
need to also mention the ring that is attached to that status,
reducible over... or irreducible over.... Since you connected your
post to the thread, I would assume that you meant "a polynomial that
has a function value which is composite."

As to it being a conjecture, one of my previous post sketched a proof
that for infinity many primes p there are infinitely many values x
for which f(x) is divisible by p^2. So, to answer the conjecture,
yes, function values are composite, not just once but, infinitely
often.

See: Message 11921 of 12227

--- In primenumbers@yahoogroups.com, "Jon Perry" <perry@g...> wrote:
> This thread reminds me of an ancient conjecture by the great (and
late)
> Indian mathematian, Shawtoff, prove that any polynomial f(x)=sigma
(i=0 to
> oo;a_i.x^i) with integer coefficients a_i, a_i !=0 for at least one
i!=0,
> always contains at least 1 composite.
>
> Jon Perry
> perry@g...
> http://www.users.globalnet.co.uk/~perry/maths/
> BrainBench MVP for HTML and JavaScript
> http://www.brainbench.com
• A slight modification in this polynomial yields a slightly higher count of primes. In searching for primes of the form ax^2+bx+c, it is good to put the axis
Message 15 of 16 , May 4, 2003
A slight modification in this polynomial yields a slightly higher
count of primes. In searching for primes of the form ax^2+bx+c, it
is good to put the axis of symmetery at -b/2a=50000.5 so that when x
yields a prime value then 100001-x also yields a prime value (what we
Americans call a "two-fer," for two for one sales). The general
principle is sound but one should realize that, once a candidate
formula is found, the value at x=1 might be composite while the value
at x=100001 might be prime, so that shifting x by 1 would up the
prime count. In this fashion I identified a shift of 5357 giving 16
more primes. The polynomial x^2-89287*x+1991687729 gives 44500 Prp
on the range x=1..100000.

snip>>>>

Try to beat my record in CYF NO. 13 at
www.shyamsundergupta.com/canyoufind.htm
> This allows negative and repeated primes and I got 44484 primes to
100000 in
> x^2-100001x+2498695637
> The puzzle asks for 50000 but that is beyond me.
> According to my computations, 92.4% of the candidates will not have
a prime
> factor below 250 for x^2+x-7731189253. It only gave 41503 primes to
100000.
<<<snip
• ... Nice work. I also considered trying to shift when I found the symmetric quadratic. I knew the requested 50000 primes was far out of reach and was too lazy
Message 16 of 16 , May 5, 2003
> A slight modification in this polynomial yields a slightly higher
> count of primes. In searching for primes of the form ax^2+bx+c, it
> is good to put the axis of symmetery at -b/2a= so that when x
> yields a prime value then 100001-x also yields a prime value (what we
> Americans call a "two-fer," for two for one sales). The general
> principle is sound but one should realize that, once a candidate
> formula is found, the value at x=1 might be composite while the value
> at x=100001 might be prime, so that shifting x by 1 would up the
> prime count. In this fashion I identified a shift of 5357 giving 16
> more primes. The polynomial x^2-89287*x+1991687729 gives 44500 Prp
> on the range x=1..100000.
>

Nice work. I also considered trying to shift when I found the symmetric
quadratic. I knew the requested 50000 primes was far out of reach and was too
lazy to try the shift. I did no heuristics but did not expect as many as 16
extra primes.
Brian Trial's best attempt at www.shyamsundergupta.com/canyoufind.htm is a
shift of my second best, but he found it independently. It only has one extra
prime and only shifts the symmetri axis from 50000.5 to 49999.5. Maybe he did
not shift but started with a slightly asymmetric quadratic on the interval
1..100000.

--
Jens Kruse Andersen
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