2*k+3 = p

"An odd number d1 is composite (not prime)
if and only if is product of 2 generic odd numbers d2 and d3 different from 1" It follows that if d1 is composite then there exist m,n>=1 such that
d1=(2m+1)(2n+1) = 4mn+2m+2n+1.
Let d1= 2k+3 (with k>=0) so we have that 2k=4mn+2m+2n-2;
k=2mn+m+n-1 (m,n >=1). We have so built the triangular matrix: (TdL)

3
6 11
9 16 23
12 21 30 39
15 26 37 48 59
= = = = = = = = = = =

defined, as written, by A[m,n] = k = 2mn+m+n-1
(with m, n integers and >=1; k>=0 and m>=n).
Substituting in the expression p=2k+3 all values of k which don't belong to TdL we can have the series of odd prime numbers:
For k = (0, 1, 2, 4, 5, 7, 8, 10, 13, 14, 17, 19, ..)
we have p =(3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,.).
If a odd number is prime or not only verifying if k belongs or not to TdL.

Example.
if N=2k+3=47, then k = 22, that doesn't belong
to TdL, therefore 47 is prime.

if N = 77, then k = 37, belonging to TdL. So 77 isn't prime. Actually, k=37 element is in the position A(5,3)]; 77=(2*5+1)(2*3+1) = 11*7.

After that, we have now the problem to establish, less or more fast,if a given k belongs or not to TdL.

We have seen (generally speaking), that:
"if p is prime, an integer k >= (p^2 -3)/2 belongs to
TdL if k = (p^2 - 3) /2 (mod p)".

Si chiede di verificare.
Thanks for help.

Vincenzo Librandi
vincenzo.librandweoz.alice.it
.

Relevant Pages

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