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Minimum numbers k such that (k^2*2^n + 1) is prime.

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`%I #2 Mar 31 2012 13:20:28
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`%S 1,1,3,1,6,2,3,1,6,5,3,4,12,2,6,1,3,10,15,5,9,5,18,25,9,13,9,14,12,7,
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`%T 6,9,3,17,9,9,15,12,9,6,6,3,3,11,42,18,21,9,66,10,33,5,27,7,48,80,24,
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`%U 40,12,20,6,10,3,5,3,7,3,79,75,63,96,40,48,20,24,10,12,5,6,15,3,22,72,11
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`%N Minimum numbers k such that (k^2*2^n + 1) is prime.
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`%C 3 divides a(2k+1) for k>0. Corresponding primes of the form (k^2*2^n + 1) are listed in A122912[n] = {3,5,73,17,1153,257,1153,257,18433,25601,18433,65537,1179649,65537,1179649,65537,1179649,26214401,117964801,...}. There are repeating patterns in a(n) such that for many n a(n) = 2*a(n+2) and a(n+1) = 2*a(n+3). For example, {6,2,3,1}, {12,2,6,1}, {42,18,21,9}, {96,40,48,20,24,10,12,5,6}, {66,10,33,5}, {48,80,24,40,12,20,6,10,3}, {366,38,183,19}. These patterns correspond to identical twin runs in A122912[n] such that A122912[n] = A122912[n+2] and A122912[n+1] = A122912[n+3]. The final index of many such twin runs is perfect power such as {8,16,25,64,81,100,...}.
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`%F a(n) = Sqrt[ (A122912[n] - 1) / 2^n ].
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`%Y Cf. A112912.
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`%K nonn
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`%O 1,3
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`%A _Alexander Adamchuk_, Sep 18 2006
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