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Old 2016-03-10, 23:59   #1
MattcAnderson
 
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"Matthew Anderson"
Dec 2010
Oregon, USA

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Hi Math People

I want to make you all aware of my latest math work.

https://docs.google.com/viewer?a=v&p...WM2ZjY1ZGUyNDQ

Given h(n) ~ n^2 + n + 41.
Assume n is an integer.
There seem to be patterns in the composite values of h(n) for certain n.

More soon.

Matt A
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Old 2016-03-11, 00:41   #2
science_man_88
 
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Quote:
Originally Posted by MattcAnderson View Post
Hi Math People

I want to make you all aware of my latest math work.

https://docs.google.com/viewer?a=v&p...WM2ZjY1ZGUyNDQ

Given h(n) ~ n^2 + n + 41.
Assume n is an integer.
There seem to be patterns in the composite values of h(n) for certain n.

More soon.

Matt A
is it that they are all of form n^2+n+41 ? sorry first thought was that but "We assume that n is an integer. We only

consider positive integer values for n in this progect." wouldn't that be the whole numbers or the natural numbers then also the typo progect is present. pairs is mistyped on page 5 and modular arithmetic has to repeat eventually if that's the pattern you saw.
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Old 2016-03-11, 10:42   #3
MattcAnderson
 
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Quote:
Originally Posted by MattcAnderson http://www.mersenneforum.org/images/...s/viewpost.gif
Hi Math People

I want to make you all aware of my latest math work.

https://docs.google.com/viewer?a=v&p...WM2ZjY1ZGUyNDQ

Given h(n) ~ n^2 + n + 41.
Assume n is an integer.
There seem to be patterns in the composite values of h(n) for certain n.

More soon.

Matt A


is it that they are all of form n^2+n+41 ? sorry first thought was that but "We assume that n is an integer. We only

consider positive integer values for n in this progect." wouldn't that be the whole numbers or the natural numbers then also the typo progect is present. pairs is mistyped on page 5 and modular arithmetic has to repeat eventually if that's the pattern you saw.



Hi Science_Man_88

Thank you for your question and comment. I understand that you question is "Are all the composite integers of the form n^2 + n + 41?" The answer to this question is yes.

I appreciate the note about the typo on page 5.

Admittedly, the word document for the above file cannot be found tonight. I have made a new .pdf file with .doc file on USB drive.

Pease see matt's project 5th cut
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Old 2016-03-11, 12:57   #4
CRGreathouse
 
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Quote:
Originally Posted by MattcAnderson View Post
"Are all the composite integers of the form n^2 + n + 41?" The answer to this question is yes.
I take it you intend to allow n to take on (at least) arbitrary real values, rather than just integers?
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Old 2016-03-13, 04:06   #5
MattcAnderson
 
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Quote:
Originally Posted by MattcAnderson http://www.mersenneforum.org/images/...s/viewpost.gif
"Are all the composite integers of the form n^2 + n + 41?" The answer to this question is yes.

I take it you intend to allow n to take on (at least) arbitrary real values, rather than just integers?

Hi Math People ,

I was incorrect in the quote above. Numbers like 4 and 6 are not of the form n^2 + n + 41 assuming n is an integer.

@Greathouse You are correct. It would require n to be (at least) an arbitrary real value in order for all the composite integers to be of the form n^2 + n + 41.

I restate a statement. It is my conjecture that all the composite positive integers of the form n^2 + n + 41 ( to wit 41^2 , 41*43 , 43*47 , 47*53) form an interesting pattern. The first bifurcation happens at an n value of 163. So 163^2 + 163 +41 is 26773.

Regards ,
Matt
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Old 2016-03-16, 00:19   #6
bhelmes
 
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A beautiful evening,

the quadratic irreducible polynomial f(n)=n^2+n+41 is one of quadratic irreducible polynomials which can be used for generating infinite prime sequence.

A nice implementation of a basic algorithm and some result for n<2^30 can be found under http://devalco.de/basic_polynomials/...p?a=1&b=1&c=41

If you like the prime generators for quadratic irreducible polynomial
have a look for the polynomial f(n)=n^2+1 resp. :
http://devalco.de/quadr_Sieb_x%5E2+1.php

One of the properties of the quadratic polynomial is the discriminant
D:=b^2-4ac for ax^2+bx+c

I am looking for some more properties in order to find a better order or system to classify the polynomials.

There are some good tries, see at http://devalco.de/#106

If you have special mathematical question, i will try to give an answer.

There was another thread before which dealed with this topic,
resp. with the polynomial f(n)=2n^2-1 and i was a bit displeased about how some people discussed the topic.

Nevertheless the quadratic polynomial and the prime generating sequences
are a nice subject. You will even find out that the sieve of Eratosthenes is
a special case of prime generating of quadratic polynomials.

Greetings from the primes
Bernhard
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Old 2016-03-19, 00:37   #7
a1call
 
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Well here is my take on the form:


x^2+1x+41= x(x+1)+41
the two addends are coprime for all values except for multiples of (x= 41m definitely not coprime) and probably not coprime for multiples (x=3m and x=7m)
being coprime means the sum will not divide 41.
I can not see why the smallest factor is 41. Could you get smaller factors for larger x?
if not why not?

Last fiddled with by a1call on 2016-03-19 at 00:52
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Old 2016-03-19, 01:45   #8
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Quote:
Originally Posted by a1call View Post
Well here is my take on the form:


x^2+1x+41= x(x+1)+41
the two addends are coprime for all values except for multiples of (x= 41m definitely not coprime) and probably not coprime for multiples (x=3m and x=7m)
being coprime means the sum will not divide 41.
I can not see why the smallest factor is 41. Could you get smaller factors for larger x?
if not why not?
well by the pigeonhole principle if it hasn't divided by smaller primes by x=41 it never will because all modular remainder holes there can be will have had to have been used up, because at most p modular classes exist mod p once the values of x less than p have been tested if none divide by p then the "pattern" has had to start repeating or reversing mod p. either way even if it's not a pattern by x=p if p hasn't divided then p never will. given that all entries up to x=41 are prime nothing but themselves divides them so no smaller divisor can exist.

Last fiddled with by science_man_88 on 2016-03-19 at 01:47
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Old 2016-03-19, 12:18   #9
a1call
 
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My point is that if you choose some other prime than 41 such as 7, you could get a divisor less than that prime:
4x5+7=27
But a mod analysis of all the primes less than 41 for the form n(n+1)+41 shows/proves that indeed there can be no divisor less than 41. It is just a coincidence that no complement to the reminder of 41/p for p<41 happens to be a product of 2 consecutive integers. All the compliments happen to be primes or squares or numbers like 10 which can not be a product of 2 consecutive integers.

Last fiddled with by a1call on 2016-03-19 at 12:21
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Old 2016-03-19, 21:40   #10
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Quote:
Originally Posted by a1call View Post
It is just a coincidence that ...
Algebraic Number Theory can provide some insight here.
There is a summary on Wikipedia: https://en.wikipedia.org/wiki/Heegner_number
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Old 2016-03-21, 17:25   #11
MattcAnderson
 
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Dec 2010
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Thank you for your comments.

You can view some

Power point slides

They are regarding the expression n^2 + n + 41.

Regards,
Matt
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