%I
%S 1,1,1,1,1,1,2,2,2,2,1,2,3,1,3,3,2,2,4,3,4,4,4,2,3,4,3,3,5,5,5,4,5,4,
%T 5,4,5,5,6,4,4,7,4,6,7,6,7,5,4,5,4,6,8,7,7,7,7,4,8,9,8,5,9,6,7,8,4,8,
%U 8,10,8,6,6,10,9,9,7,7,6,9,10,9,8,8,12,13
%N a(n) is the number of sets of three positive integers p_1 < p_2 < p_3 such that 2*p_2 = p_1 + p_3, where p_i (i=1,2,3) is either 1 or a prime number and p_3 = prime(n).
%C a(n)>0 for n > 1.
%C It is conjectured that every positive integer appears a positive finite number of times in this sequence.
%C The sequence of records is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 18, 20, 21, 22, 23, 24, 25, 27, 28, 31, 33, 34, 35, 36, 39, 40, 41, 42, 43, 46, 47, 48, 50, 51, 53, 55, 56, 58, 61, 62, 64, 65, 66, 70, 71, 72, 74, 76, 78,...  _R. J. Mathar_, May 02 2014
%C Alternative definition for p_i is p_1 is either 1 or an odd prime, p_2 is an odd prime after a(2) and p_3 is prime(n).  _Jon Perry_, Apr 17 2014.
%H Lei Zhou, <a href="/A240301/b240301.txt">Table of n, a(n) for n = 2..10001</a>
%e For n=2, p_3=prime(2)=3, 2*2=1+3. One instance found, so a(2)=1;
%e ...
%e For n=8, p_3=prime(8)=19, 2*11=3+19, 2*13=7+19. Two instances found, so a(8)=2;
%e ...
%e For n=30, p_3=prime(30)=113, 2*59=5+113, 2*71=29+113, 2*83=53+113, 2*101=89+113, 2*107=101+113. Five instances found, so a(30)=5.
%t Table[p = Prime[n]; ct = 0; pp = p; While[pp = NextPrime[pp, 1]; diff = p  pp; diff < pp, cp = pp  diff; If[(PrimeQ[cp])  (cp == 1), ct++]]; ct, {n, 2, 87}]
%Y Cf. A000040, A240232.
%K nonn,easy
%O 2,7
%A _Lei Zhou_, Apr 03 2014
