A352480 - OEIS

%I #12 Mar 20 2022 04:29:32

%S 11,19,23,37,97,109,137,263,277,307,401,409,617,757,877,1039,1187,

%T 1321,1481,1567,1871,2179,2333,2543,2861,3371,3617,3697,4649,4783,

%U 5639,5651,6547,6689,6779,6883,7687,8807,9371,9437,9767,10331,11317,11777,11927,13523,14503,15683,16921,17291,19073,19553

%N First of four consecutive primes p,q,r,s such that p+q+r+s is divisible by A001414(q+r).

%C If x is a prime and 2*x-y,2*x-z,2*x+z,2*x+y+32 are consecutive primes with 0 < z < y, then 2*x-y is a term.  Thus Dickson's conjecture implies there are infinitely many terms.

%H Robert Israel, <a href="/A352480/b352480.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3) = 23 is a term because 23, 29, 31, 37 are consecutive primes, and A001414(29+31) = A001414(2^2*3*5) = 12 divides 23+29+31+37 = 120.

%p q:= 2: r:= 3: s:= 5:

%p R:= NULL: count:= 0:

%p while count < 50 do

%p   p:= q; q:= r; r:= s; s:= nextprime(s);

%p   a:= add(t[1]*t[2], t = ifactors(q+r)[2]);

%p   if (p+q+r+s) mod a = 0 then count:= count+1; R:= R, p; fi

%p od:

%p R;

%Y Cf. A001414, A352459, A352534.

%K nonn

%O 1,1

%A _J. M. Bergot_ and _Robert Israel_, Mar 17 2022