20160310, 23:59  #1 
"Matthew Anderson"
Dec 2010
Oregon, USA
2×3×5×29 Posts 
prime producing polynomial 2
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 
20160311, 00:41  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
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. 

20160311, 10:42  #3 
"Matthew Anderson"
Dec 2010
Oregon, USA
2·3·5·29 Posts 
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 
20160311, 12:57  #4 
Aug 2006
3·1,993 Posts 

20160313, 04:06  #5 
"Matthew Anderson"
Dec 2010
Oregon, USA
2×3×5×29 Posts 
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 
20160316, 00:19  #6 
Mar 2016
2·3·59 Posts 
prime generator for quadratic polynomials
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^24ac 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^21 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 
20160319, 00:37  #7 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2088_{10} Posts 
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 20160319 at 00:52 
20160319, 01:45  #8  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Last fiddled with by science_man_88 on 20160319 at 01:47 

20160319, 12:18  #9 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4050_{8} Posts 
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 20160319 at 12:21 
20160319, 21:40  #10 
Dec 2012
The Netherlands
2^{2}×431 Posts 
Algebraic Number Theory can provide some insight here.
There is a summary on Wikipedia: https://en.wikipedia.org/wiki/Heegner_number 
20160321, 17:25  #11 
"Matthew Anderson"
Dec 2010
Oregon, USA
2·3·5·29 Posts 
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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