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Number of primes of the form: |P_i +- P_n / P_i| or |P_i/2 +- 2P_n / P_i|, where 2 <= i <= n, P_i = the i-th primorial number (A002110).
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%I #36 Mar 28 2015 17:18:12

%S 3,7,10,11,16,13,13,15,10,12,14,16,11,13,7,21,14,25,13,10,12,9,19,14,

%T 15,11,6,15,7,13,12,9,13,9,15,17,12,12,9,7,7,8,11,14,11,14,19,15,11,9,

%U 11,11,11,15,15,17,7,15,11,16,11,18,8,10,13,9,10,13,7,17

%N Number of primes of the form: |P_i +- P_n / P_i| or |P_i/2 +- 2P_n / P_i|, where 2 <= i <= n, P_i = the i-th primorial number (A002110).

%C a(n) is defined for n > 1.

%C Equal primes obtained in |P_i +- P_n / P_i| and in |P_i/2 +- 2P_n / P_i| are counted as different cases, as demonstrated in the examples.

%C It is conjectured that all terms of this sequence are greater than zero, but this conjecture might have exceptions.

%H Lei Zhou, <a href="/A238506/b238506.txt">Table of n, a(n) for n = 2..500</a>

%e -------------------- For n = 2, a(2) = 3: --------------------

%e P_2 = 2*3 = 6.

%e P_2 + P_2 / P_2 = 6 + 1 = 7 is prime (case 1);

%e |P_2 - P_2 / P_2| = |6 - 1| = 5 is prime (case 2);

%e P_2 / 2 + 2P_2 / P_2 = 6/2 + 2 = 5 is prime (case 3);

%e |P_2 / 2 - 2P_2 / P_2| = |6/2 - 2| = 1 is not prime;

%e three ways found, so a(2) = 3.

%e -------------------- For n = 3, a(3) = 7: --------------------

%e P_2 = 2 * 3 = 6; P_3 = 2 * 3 * 5 = 30.

%e P_2 + P_3 / P_2 = 6 + 5 = 11 is prime (case 1);

%e |P_2 - P_3 / P_2| = |6 - 5| = 1 is not prime;

%e P_2 / 2 + 2P_3 / P_2 = 6/2 + 2 * 30/6 = 3 + 10 = 13 is prime (case 2);

%e |P_2 / 2 - 2P_3 / P_2| = |6/2 - 2 * 30/6| = |3 - 10| = 7 is prime (case 3);

%e P_3 + P_3 / P_3 = 30 + 1=31 is prime (case 4);

%e |P_3 - P_3 / P_3| = |30 - 1| = 29 is prime (case 5);

%e P_3 / 2 + 2P_3 / P_3 = 30/2 + 2 * 30/30 = 15 + 2 = 17 is prime (case 6);

%e |P_3 / 2 - 2P_3 / P_3| = |30/2 - 2 * 30/30| = |15 - 2| = 13 is prime (case 7);

%e seven ways found, so a(3) = 7.

%t Table[ct = 0; p1 = 1; Do[p1 = p1*Prime[i], {i, 2, n}]; p2 = p1; Do[p2 = p2/Prime[i]; p3 = p1/p2; If[PrimeQ[2*p2 + p3], ct++]; If[PrimeQ[Abs[2*p2 - p3]], ct++]; If[PrimeQ[p2 + 2*p3], ct++]; If[PrimeQ[Abs[p2 - 2*p3]], ct++], {i, 2, n}]; ct, {n, 2, 78}]

%Y Cf. A000040, A002110.

%K nonn,hard

%O 2,1

%A _Lei Zhou_, Feb 27 2014