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A127166 a(n) = the minimum 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. 3

%I #18 Feb 14 2015 10:56:45

%S 2,2,3,5,11,23,149,179,1187,0,0,3628811,43545611,43545743,7925299211,

%T 9144576143,1609445376013,32335220736011,44771844096143,

%U 582033973248209,52672757806189,18804174520322717,267682954936324199

%N a(n) = the minimum 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="/A127166/b127166.txt">Table of n, a(n) for n=0..100</a>

%e For n = 6 we have the only prime S (and so the minimum 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, First[ # ]] & /@ Array[f, 30, 0] (* _Ray Chandler_, Feb 14 2007 *)

%Y Cf. A127165, 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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Last modified April 19 18:05 EDT 2024. Contains 371798 sequences. (Running on oeis4.)