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First of four consecutive primes p,q,r,s such that p+q+r+s is divisible by A001414(p+q).
3

%I #13 Mar 20 2022 04:32:16

%S 7,53,151,263,269,311,317,457,587,761,883,919,1049,1303,1423,1579,

%T 1637,1759,1993,2063,2111,2789,2851,2903,3137,3533,3593,3797,3889,

%U 3967,4153,4219,4463,4507,4663,5179,5843,6353,6737,6829,7937,8563,8999,9059,9241,9437,9439,9629,10039,10831,10847

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

%C If x is a prime and 2*x-y, 2*x+y, 2*x+z and 2*x+32-z are consecutive primes where 0 < y < z < 16, then 2*x-y is a term. Thus Dickson's conjecture implies there are infinitely many terms.

%H Robert Israel, <a href="/A352459/b352459.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3) = 151 is a term because 151,157,163,167 are consecutive primes, and A001414(151+157) = A001414(2^2*7*11) = 2+2+7+11 = 22 divides 151+157+163+167 = 638.

%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(p+q)[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, A352480, A352534.

%K nonn

%O 1,1

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