|
|
A257933
|
|
Prime p such that sqrt(p+2) is semiprime (A001358).
|
|
1
|
|
|
79, 223, 439, 1087, 1223, 2399, 3023, 4759, 5927, 8647, 14159, 14639, 21023, 24023, 25919, 28559, 31327, 33487, 42023, 47087, 56167, 61007, 64007, 67079, 70223, 71287, 89399, 90599, 91807, 95479, 104327, 112223, 116279, 126023, 137639, 152879, 172223, 199807
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
The terms are not congruent to 1 (mod 10).
The sequence contains no Mersenne prime p=2^t-1. Since p > 79, t is an odd prime and p+2 = 2^t+1 is divisible by 3. So, since 2^t+1 should be square, 2^t+1 is divisible by 9, i.e., (2^t+1)/3 == 0 (mod 3). (1)
Note that either t=6k+1 or t=6m+5. In each case, (1) is impossible.
Indeed, if t=6k+1, then (2^t+1)/3 = (2*(4^k)^3+1)/3 = (2*(3+1)^(3*k)+1)/3 == (2*binomial(3*k,1)*3+2+1)/3 == 1(mod 3), and analogously in case t=6*m+5, (2^t+1)/3 == 2 (mod 3): a contradiction.
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
Prime 79 is in the sequence because sqrt(79+2) = 9 = 3*3 which is semiprime.
Prime 1223 is in the sequence because sqrt(1223+2) = 35 = 5*7 which is semiprime.
|
|
MATHEMATICA
|
Select[Prime@Range@18000, PrimeOmega[Sqrt[#+2]]==2&]//Quiet (* Ivan N. Ianakiev, May 13 2015 *)
|
|
PROG
|
(PARI) issemi(n)=bigomega(n)==2
(PARI) list(lim)=my(v=List(), k=sqrt(lim+2), t); forprime(p=2, sqrt(k), forprime(q=p, k\p, if(isprime(t=(p*q)^2-2), listput(v, t)))); Set(v) \\ Charles R Greathouse IV, May 13 2015
(Perl) use ntheory ":all"; forprimes { say if is_power($_+2, 2) && scalar(factor(sqrtint($_+2)))==2 } 1e7; # Dana Jacobsen, May 13 2015
(Perl) use ntheory ":all"; sub list { my($lim, $k, $t, $p, %v)=shift; $k=sqrt($lim+2); forprimes { $p=$_; forprimes { $t=($p*$_)**2-2; $v{$t}++ if is_prime($t); } $p, int($k/$p); } int(sqrt($k)); my @v=sort{$a<=>$b} keys %v; @v; } say for list(1e10); # Translation of PARI, Dana Jacobsen, May 13 2015
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|