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%I #22 Feb 15 2017 16:56:31
%S 0,1,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,1,0,0,3,0,0,1,0,0,
%T 2,0,0,2,1,0,2,1,0,3,0,0,1,1,0,2,0,0,1,2,0,2,2,0,4,0,0,1,2,0,2,0,1,1,
%U 3,0,2,2,0,2,0,0,1,2,0,2,1,1,2,4,0,1,2
%N Number of decompositions of 2n into an unordered sum of two Ramanujan primes.
%C Suggested by _John W. Nicholson_.
%C There are 95 zeros in the first 10000 terms. Are there more? Related to Goldbach's conjecture. - _T. D. Noe_, Jan 27 2012
%C There are no other zeros in the first 10^8 terms. a(n) > 0 for n from 1313 to 10^8. - _Donovan Johnson_, Jan 27 2012
%H Donovan Johnson, <a href="/A204814/b204814.txt">Table of n, a(n) for n = 1..10000</a>
%e a(29) = 3. 2*29 = 58 = 11+47 = 17+41 = 29+29 (11, 17, 29, 41 and 47 are all Ramanujan primes). 58 is the unordered sum of two Ramanujan primes in three ways.
%Y Cf. A045917, A104272, A173634.
%K nonn
%O 1,29
%A _Donovan Johnson_, Jan 27 2012
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