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

2,7

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.

Links

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

Example

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.

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}]

Crossrefs

Cf. A000040, A240232.

Sequence in context: A094701 A210452 A347037 * A289641 A209312 A054715

Adjacent sequences: A240298 A240299 A240300 * A240302 A240303 A240304

Keyword

nonn,easy

Author

Lei Zhou, Apr 03 2014

Status

approved