A240301 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).

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, 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, 8, 10, 8, 6, 6, 10, 9, 9, 7, 7, 6, 9, 10, 9, 8, 8, 12, 13

Offset: 2

Author

Lei Zhou, Apr 03 2014

Keywords

• nonn
• easy

Comments

a(n)>0 for n > 1.

It is conjectured that every positive integer appears a positive finite number of times in this sequence.

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

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.

Examples

			For n=2, p_3=prime(2)=3, 2*2=1+3. One instance found, so a(2)=1;
...
For n=8, p_3=prime(8)=19, 2*11=3+19, 2*13=7+19. Two instances found, so a(8)=2;
...
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.
		

Links

• Lei Zhou, Table of n, a(n) for n = 2..10001

Crossrefs

Cf. A000040, A240232.

Programs

Mathematica

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}]