20181101, 14:07  #1 
Jan 2018
43_{10} Posts 
Revisited "Primorial as Product of Consective Number"
We all know:
2 # = 2 = 1 X 2 3 # = 6 = 2 X 3 5 # = 30 = 5 X 6 7 # = 210 = 14 X 15 and then 17 # = 510510 = 714 X 715 missing 11#,13# and so on... infact no primorial is product of consecutive numbers upto 104729# except these known ones. I found( may be refound) an interesting feature for 11#, 13#, 19# and 23#: 11 # = 2310 = 48 X 49  6 X 7 13 # = 30030 = 173 X 174  8 X 9 19 # = 9699690 = 3114 X 3115  20 X 21 23 # = 223092870 = 14936 X 14937  78 X 79 but again 29# can not be represented this way.. Can someone help me (using MATLAB for big primorials) whether 11,13, 19 and 23 are only primorial with this feature.. 
20181101, 14:57  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,061 Posts 
Quote:
Split arbitrarily the prime factors of p# into two composite factors x and y (with x>y+1). then take a=(x+y1)/2, b=(xy1)/2 and you will have p# = a*(a+1)  b*(b+1). Examples: 11# = 55 * 42 ==> a = 48 and b = 6 ==> 11# = 48 * 49  6 * 7 29# = 29#/2 * 2 ==> a = 1617423308 and b = 1617423306 ==> 29# = 1617423308*1617423309  1617423306*1617423307 (and many other ways) 

20181101, 15:23  #3  
Jan 2018
43 Posts 
Quote:
19 # = 9699690 = 3114 X 3115  20 X 21, here 3114 is int(9699690^0.5) 23 # = 223092870 = 14936 X 14937  78 X 79, here 14936 is int(223092870^0.5) thanks again.. 

20181101, 15:37  #4  
Feb 2017
Nowhere
7473_{8} Posts 
Quote:
I wrote a script that tested m = 4*p# + 1 for squareness by computing kronecker(m,q) for the next 20 primes after p. If any of them was 1, I went onto the next p without further ado. (If kronecker(m,q) = 1 for any q, then m is not a square.) If m "passed" that test, I had Pari check whether it was indeed the square of an integer (which entails extracting the integer square root, 2*k + 1), and then, if it was a square, exhibiting k and the values of 2*k + 1 (mod q) for the primes q up to p. The results were 2 1 [1] 3 2 [1, 1] 5 5 [1, 1, 1] 7 14 [1, 1, 1, 1] 17 714 [1, 1, 1, 1, 1, 1, 1] This method was quick enough that I was able to exclude the possibility of any other examples of p# = k*(k+1) for p up to 2^20 = 1048576 in seconds, not minutes. As to the other part of the question, I'm not exactly sure what the question is yet. EDIT: BTW, I got the quadratic character idea from a paper I read ages ago, ON THE BROCARDâ€“RAMANUJAN DIOPHANTINE EQUATION n! + 1 = m^{2} Last fiddled with by Dr Sardonicus on 20181101 at 16:27 Reason: Redid the paragraphing; added more info 

20181102, 00:14  #6  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3·3,061 Posts 
Quote:
43# = 114379899 * 114379900  8829 * 8830, and 114379899 = floor(sqrt(43#)) 

20181102, 01:20  #7 
Jan 2018
43 Posts 

20181102, 01:24  #8 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×3,061 Posts 
This is going to be a very long sequence.

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Sieving freakishly big MMs (was "World record" phone number?)  davieddy  Operazione Doppi Mersennes  282  20190624 07:57 
Problem E7 of Richard Guy's "Unsolved problems in number theory"  Batalov  Computer Science & Computational Number Theory  40  20130316 09:19 
Modular Subset "Product" Problem  vector  Math  15  20080322 19:52 
Would Minimizing "iterations between results file" may reveal "is not prime" earlier?  nitai1999  Software  7  20040826 18:12 
Search for a number theoretic function related to "prime divisor sums"  juergen  Math  2  20040710 23:01 