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%I #9 Dec 17 2013 03:55:47
%S 0,0,0,1,1,2,1,2,3,1,1,3,3,3,3,2,4,5,3,4,4,4,4,4,3,5,4,5,4,5,3,4,7,4,
%T 5,6,4,8,8,4,4,4,7,5,6,5,6,8,4,6,8,6,7,6,6,5,5,9,7,9,7,6,8,7,7,8,6,9,
%U 9,6,6,12,9,6,10,8,9,12,7,7,11,5,10,9,9,10,7,11,8,9,6,8,14,10,8,8,10,12,9,6
%N a(n) = |{0 < m < 2*n: m = sigma(k) for some k > 0, and 2*n - 1 - m and 2*n - 1 + m are both prime}|, where sigma(k) is the sum of all (positive) divisors of k.
%C Conjecture: (i) a(n) > 0 for all n > 3.
%C (ii) For any even number 2*n > 0, 2*n + sigma(k) is prime for some 0 < k < 2*n.
%C See also A233793 for a related conjecture.
%C Clearly part (i) of the conjecture implies Goldbach's conjecture for even numbers 2*(2*n - 1) with n > 3; we have verified part (i) for n up to 10^8. Concerning part (ii), we remark that 1024 is the unique positive integer k < 1134 with 1134 + sigma(k) prime, and that sigma(1024) = 2047 > 1134.
%H Zhi-Wei Sun, <a href="/A233864/b233864.txt">Table of n, a(n) for n = 1..10000</a>
%e a(7) = 1 since sigma(5) = 6, and 2*7 - 1 - 6 = 7 and 2*7 - 1 + 6 = 19 are both prime.
%e a(10) = 1 since sigma(6) = sigma(11) = 12, and 2*10 - 1 - 12 = 7 and 2*10 - 1 + 12 = 31 are both prime.
%e a(11) = 1 since sigma(7) = 8, and 2*11 - 1 - 8 = 13 and 2*11 - 1 + 8 = 29 are both prime.
%t f[n_]:=Sum[If[Mod[n,d]==0,d,0],{d,1,n}]
%t S[n_]:=Union[Table[f[j],{j,1,n}]]
%t PQ[n_]:=n>0&&PrimeQ[n]
%t a[n_]:=Sum[If[PQ[2n-1-Part[S[2n-1],i]]&&PQ[2n-1+Part[S[2n-1],i]],1,0],{i,1,Length[S[2n-1]]}]
%t Table[a[n],{n,1,100}]
%Y Cf. A000040, A000203, A002372, A002375, A232270, A233544, A233654, A233793.
%K nonn
%O 1,6
%A _Zhi-Wei Sun_, Dec 16 2013
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