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a(n) = A082893(n)-A000040(n), that is difference of n-th prime and number closest to it and divisible by n.
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%I #4 Oct 15 2013 22:31:57

%S 0,1,1,1,-1,-1,-3,-3,4,1,2,-1,-2,-1,-2,-5,-8,-7,9,9,-10,9,9,7,3,3,5,5,

%T 7,7,-3,-3,-5,-3,-9,-7,-9,-11,-11,-13,-15,-13,-19,-17,-17,-15,-23,17,

%U 18,21,22,21,24,19,18,17,16,19,18,19,22,17,8,9,12,13,4,3,-2,1,2,1,-2,-3,-4,-3,-4,-7,-6,-9,-14,-11,-16,-13,-14,-13,-14

%N a(n) = A082893(n)-A000040(n), that is difference of n-th prime and number closest to it and divisible by n.

%C n=9:p(9)=23 is between 18 and 27, closer to 27, so a(9)=27-23=+4.

%F a(n)=p[n]-n*floor[(floor(n/2)+p[n])/n], where p[n] is the n-th prime.

%t Table[Prime[n]-n*Floor[(Floor[n/2]+Prime[n])/n], {n, 1, 100}]

%Y Cf. A082893-A082901.

%K sign

%O 1,7

%A _Labos Elemer_, Apr 22 2003