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Integers n such that the century defined by the interval [100n+1,100n+100] (i.e., the (n+1)-st century) contains exactly three Ormiston prime pairs.
1

%I #9 Mar 07 2016 13:09:15

%S 651507,1057749,1263729,1666162,2527374,3620083,3850824,4925140,

%T 5418973,8255451,8300166,8318772,8389776,9829099,10876176,11419227,

%U 12195316,14241925,15170550,15864637,15876420,16868593,16892233,17123505,18032409

%N Integers n such that the century defined by the interval [100n+1,100n+100] (i.e., the (n+1)-st century) contains exactly three Ormiston prime pairs.

%C Ormiston pairs must be of form (100n+13,100n+31), (100n+37,100n+73), (100n+79,100n+97).

%C There cannot be more than three Ormiston prime pairs in a century.

%o (PARI) is(n)=isprime(100*n+13) && nextprime(100*n+17)==100*n+31 && isprime(100*n+37) && nextprime(100*n+39)==100*n+73 && isprime(100*n+79) && nextprime(100*n+81)==100*n+97 \\ _Charles R Greathouse IV_, Mar 07 2016

%Y Cf. A162893.

%K base,nonn

%O 1,1

%A _Ray Chandler_ and _Ki Punches_, Jul 16 2009