A227083 Number of ways to write n as a + b/2 with a and b terms of the sequence A008407.

0, 0, 1, 0, 1, 1, 1, 1, 2, 2, 1, 3, 1, 2, 2, 3, 2, 3, 2, 3, 3, 3, 3, 4, 1, 4, 4, 2, 3, 5, 3, 4, 5, 4, 3, 7, 4, 4, 3, 6, 5, 5, 3, 6, 5, 6, 4, 6, 4, 6, 7, 5, 5, 7, 4, 6, 6, 7, 4, 7, 6, 5, 8, 5, 6, 9, 6, 5, 6, 7, 8, 8, 6, 7, 7, 9, 7, 7, 5, 9, 10, 6, 8, 9, 8, 10, 7, 8, 7, 11, 8, 7, 9, 9, 10, 10, 8, 9, 8, 13

Offset: 1

Author

Zhi-Wei Sun, Jun 30 2013

Keywords

• nonn

Comments

Conjecture: We have a(n) > 0 for all n > 4.

For every k = 2, ..., 342, the value of A008407(k) has been determined by T. J. Engelsma. Since A008407(343)/2 > A008407(342)/2 = 2328/2 = 1164, if n <= 1166 can be written as A008407(j) + A008407(k)/2 with j > 1 and k > 1 then neither j nor k exceeds 342. Based on this we are able to compute a(n) for n = 1, ..., 1166.

Examples

			a(10) = 2 since 10 = 2 + 16/2 = 6 + 8/2;
a(11) = 1 since 11 = 8 + 6/2;
a(25) = 1 since 25 = 12 + 26/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..1166
• A. V. Sutherland, Narrow admissible k-tuples: bounds on H(k), 2013.
• T. Tao, Bounded gaps between primes, PolyMath Wiki Project, 2013.

Crossrefs

Cf. A008407.