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%I #6 Jul 21 2017 13:37:54
%S 3,4,3,4,2,5,3,7,4,1,6,4,6,3,5,2,8,3,1,17,16,2,3,8,4,2,1,18,13,4,29,
%T 10,2,1,15,1,6,3,46,12,33,1,4,19,52,2,12,53,31,2,12,1,14,23,3,13,15,5,
%U 2,24,19,9,15,67,46,122,10,62,43,1,18,3,5,257,130,297,1,577,402,218,173,2
%N a(n) is the multiplier of the primorial which furnishes the "one-line proof" for primality of prime(n) associated with A079900(n); i.e., k*(p+k) = primorial(q)*a(n), where k = A079900(n), p = prime(n) and q = sqrt(p).
%C A one-line proof looks like this: 149 = 5*5*11 - 2*3*3*7. For each prime Q up to the square root of p(n), either the left product or the right product is divisible by Q, but not both. It follows that the difference is not divisible by any such Q and so is prime. Multiply the left and right products together, then divide out each Q once, the sequence gives the remaining product.
%C The generating program listed for A079900 tries every k from 1 up until a solution is found. A study of this sequence was used to design a generating program that skips k values where i*Q# < k*(p+k) << (i+1)*Q#.
%F a(n) = A079900(n)*(prime(n) + A079900(n))/primorial(sqrt(prime(n))).
%e a(28) = 18 because prime(28) = 107, A079900(28) = 28 and 28*(107 + 28) = 3780 = (2*3*5*7)*18.
%t a[p_] := Module[{prod, k}, prod=Times@@Prime/@Range[PrimePi[Sqrt[p]]]; For[k=1, True, k++, If[GCD[p, k]==1&&Mod[k*(p+k), prod]==0, Return[a[p]=k*(p+k)/prod]]]]; a/@Prime/@Range[70]
%Y Cf. A079900.
%K nonn
%O 1,1
%A Bruce A. Leenstra (leenstrb(AT)comcast.net), Aug 24 2006
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