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%I #9 Feb 11 2024 13:41:04
%S 5,11,13,97,101,109,277,293,389,653,739,863,1019,1297,1319,1321,1481,
%T 1759,1979,2293,2557,2719,3209,3511,3889,3947,5419,5701,5987,6079,
%U 6529,7127,7639,7753,7853,9343,9433,10499,10781,10831,11131,11489,12619,13523,14083,15227,15937,19501,22247,22511
%N First of four consecutive primes p,q,r,s such that p+q+r+s is divisible by A001414(r+s).
%H Robert Israel, <a href="/A352534/b352534.txt">Table of n, a(n) for n = 1..10000</a>
%e a(3) = 13 is a term because 13, 17, 19, 23 are consecutive primes and A001414(19+23) = A001414(2*3*7) = 12 divides 13+17+19+23 = 72.
%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(r+s)[2]);
%p if (p+q+r+s) mod a = 0 then count:= count+1; R:= R, p fi
%p od:
%p R;
%t Select[Partition[Prime[Range[2600]],4,1],Divisible[Total[#],Total[Times@@@FactorInteger[ Total[ Take[#,-2]]]]]&][[;;,1]] (* _Harvey P. Dale_, Feb 11 2024 *)
%Y Cf. A001414, A352459, A352480.
%K nonn
%O 1,1
%A _J. M. Bergot_ and _Robert Israel_, Mar 20 2022
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