Revisited "Primorial as Product of Consective Number"

[MARTHA]

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..

[Batalov]

Originally Posted by MARTHA

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..

Every primorial can be represented this way.
Split arbitrarily the prime factors of p# into two composite factors x and y (with x>y+1).
then take a=(x+y-1)/2, b=(x-y-1)/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)

[MARTHA]

Originally Posted by Batalov

Every primorial can be represented this way.
Split arbitrarily the prime factors of p# into two composite factors x and y (with x>y+1).
then take a=(x+y-1)/2, b=(x-y-1)/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)

Thanks for quick reply.. though you are technically correct, my observation was:
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..

[Dr Sardonicus]

Originally Posted by MARTHA

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. <snip>

I note that p# = k*(k+1) when 4*p# + 1 = (2*k+1)^2, and that, therefore, (2*k + 1)^2 == 1 (mod q), so that 2*k + 1 == 1 or -1 (mod q), for every q <= p. Now 2*k + 1 is about twice the square root of p#, which seems small to fulfill all the congruence conditions.

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 BROCARD–RAMANUJAN DIOPHANTINE EQUATION n! + 1 = m²

[Batalov]

But then you have A192579.
Related?

[Batalov]

Originally Posted by MARTHA

Thanks for quick reply.. though you are technically correct, my observation was:
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..

Well, then,
43# = 114379899 * 114379900 - 8829 * 8830, and 114379899 = floor(sqrt(43#))

[MARTHA]

Originally Posted by Batalov

Well, then,
43# = 114379899 * 114379900 - 8829 * 8830, and 114379899 = floor(sqrt(43#))

Wow.. Thanks a lot indeed..

Now we can proceed to make an interesting oeis Integer sequence starting: 11, 13, 19, 23 and newly found 43... with this property

[Batalov]

This is going to be a very long sequence.