A127165 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.

2, 2, 3, 7, 11, 43, 149, 1013, 8069, 0, 0, 39916801, 43545611, 566092811, 7925299211, 118879488011, 1609445376013, 32335220736011, 44771844096143, 582033973248209, 221172909834240011, 3930072474746880013

Offset

0,1

Comments

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.

Links

Ray Chandler, Table of n, a(n) for n=0..100

Example

For n = 6 we have the only prime S (and so the maximum prime S) with S = 1*2*3*4*6 + 5 = 149.

Mathematica

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 *)

Crossrefs

Cf. A127166, A128199.

Sequence in context: A087384 A179283 A234850 * A100683 A153940 A049905

Adjacent sequences: A127162 A127163 A127164 * A127166 A127167 A127168

Keyword

nonn

Author

Leroy Quet, Jan 06 2007

Extensions

a(21)-a(35) from Ray Chandler, Feb 14 2007

Status

approved