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%I #18 Feb 14 2015 10:55:02
%S 2,2,3,7,11,43,149,1013,8069,0,0,39916801,43545611,566092811,
%T 7925299211,118879488011,1609445376013,32335220736011,44771844096143,
%U 582033973248209,221172909834240011,3930072474746880013
%N a(n) = the maximum prime S possible, if S = product of b(k)'s + product of c(k)'s, where the distinct positive integers <= n are partitioned into the two sets {b(k)} and {c(k)}. a(n) = 0 if no prime S exists for that n.
%C a(0)=a(1)=2 because the product over the empty set is defined here as 1. For S to be a prime, the positive integers <= n, except 1 and the primes > n/2, must all be together in either {b(k)} or {c(k)}. If p is a prime where n/2 < p <= n, then it is possible that p is in either product of the S sum, as can 1. Terms calculated by _W. Edwin Clark_.
%H Ray Chandler, <a href="/A127165/b127165.txt">Table of n, a(n) for n=0..100</a>
%e For n = 6 we have the only prime S (and so the maximum prime S) with S = 1*2*3*4*6 + 5 = 149.
%t f[n_] := Block[{d = Divisors[Times @@ Select[Range[n], PrimeQ[ # ] && 2# > n &]]},Select[Union[d + n!/d], PrimeQ]];If[ # == {}, 0, Last[ # ]] & /@ Array[f, 30, 0] (* _Ray Chandler_, Feb 14 2007 *)
%Y Cf. A127166, A128199.
%K nonn
%O 0,1
%A _Leroy Quet_, Jan 06 2007
%E a(21)-a(35) from _Ray Chandler_, Feb 14 2007
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