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%I #9 Feb 05 2017 00:42:48
%S 3,13,19,73,131,181,197,193,379,521,397,601,1093,1093,1231,1153,1871,
%T 1297,2243,1801,2269,2861,3037,3313,4001,4993,3673,5209,5743,4621,
%U 5333,4481,6733,8161,9241,9001,9029,7069,8737,9601
%N a(n) is the smallest prime p such that there is a multiplicative subgroup H of Z/pZ, of even size and of index n, such that for any two cosets H1 and H2 of H, H1 + H2 contains all of (Z/pZ)\0.
%C It must be that p = a(n) is equivalent to 1 mod 2n. The fact that H is of even size means H = -H. The finite integral relation algebra with n symmetric flexible diversity atoms is representable over Z/pZ, where p = a(n).
%H Jeremy F. Alm, <a href="/A281998/b281998.txt">Table of n, a(n) for n = 1..1000</a>
%H Jeremy F. Alm, <a href="/A281998/a281998.pdf">Plot of a(n) on n</a>
%Y Cf. A282001
%K nonn
%O 1,1
%A _Jeremy F. Alm_, Feb 04 2017
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