OFFSET
1,2
COMMENTS
Let p = prime number, n = (p^2-7)/2 (mod p).
Comment: All numbers of the form 1+3k (k=0,1,2,...) are in this sequence, since 2(3k+1)+7 = 6k+9 is divisible by 3. Moreover, each of these numbers can be extended to an equidistant sequence of length k+1 and step 2k+3: This leads to the triangle T[k,m] = (3k+1)+(2k+3)*m, m=0,...,k, of elements of this sequence, because T[k,m]*2+7 = (2k+3)(2m+3) is never prime. The lines of the triangle end with m=k since the next term T[k,k+1] would be the same as the term in the following line, T[k+1,k]. (The formula T[k,m]=((2k+3)(2m+3)-7)/2 might also explain the comment involving "n=(p^2-7)/2".) [M. F. Hasler, Jun 16 2010]
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..1000
MATHEMATICA
Select[Range[200], !PrimeQ[2# + 7] &] (* Vincenzo Librandi, Nov 21 2012 *)
PROG
(PARI) for(n=1, 200, isprime(2*n+7)||print1(n", ")) \\ M. F. Hasler, Jun 16 2010
(Magma) [n: n in [1..120] | not IsPrime(2*n + 7)]; // Vincenzo Librandi, Nov 21 2012
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Vincenzo Librandi, Dec 17 2008
EXTENSIONS
Checked and extended by M. F. Hasler, Jun 16 2010
STATUS
approved