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%I #7 Jun 30 2013 14:44:08
%S 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,
%T 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,
%U 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
%N Number of ways to write n as a + b/2 with a and b terms of the sequence A008407.
%C Conjecture: We have a(n) > 0 for all n > 4.
%C 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.
%H Zhi-Wei Sun, <a href="/A227083/b227083.txt">Table of n, a(n) for n = 1..1166</a>
%H A. V. Sutherland, <a href="http://math.mit.edu/~primegaps">Narrow admissible k-tuples: bounds on H(k)</a>, 2013.
%H T. Tao, <a href="http://michaelnielsen.org/polymath1/index.php?title=Bounded_gaps_between_primes">Bounded gaps between primes</a>, PolyMath Wiki Project, 2013.
%e a(10) = 2 since 10 = 2 + 16/2 = 6 + 8/2;
%e a(11) = 1 since 11 = 8 + 6/2;
%e a(25) = 1 since 25 = 12 + 26/2.
%Y Cf. A008407.
%K nonn
%O 1,9
%A _Zhi-Wei Sun_, Jun 30 2013