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a(n) is the least number j such that j*T_k +/- 1 is n-th prime for some k-th triangular number.
6

%I #6 Mar 30 2012 18:40:42

%S 1,2,1,1,1,2,3,2,4,1,2,1,2,2,8,9,4,4,1,2,2,1,3,2,16,10,17,3,2,4,6,2,1,

%T 5,10,10,2,27,6,29,4,2,1,32,3,3,1,8,38,23,3,2,2,7,43,4,6,2,1,10,47,14,

%U 2,4,4,53,5,12,58,35,59,3,61,62,1,64,5,6,40,3,2,2,12,12,8,74,10,76,2,2,6,4

%N a(n) is the least number j such that j*T_k +/- 1 is n-th prime for some k-th triangular number.

%C Eventually all primes p appear since (p +/-1) times 1(1+1)/2 equals (p +/- 1).

%C If we asked for the least number k then k always equals 1 since all primes p appear since (p +/-1) times 1(1+1)/2 equals (p +/- 1).

%C The k's for the corresponding j's are: round(sqrt(2p/j)).

%C First occurrence of i is A125770: 1, 2, 7, 9, 34, 31, 54, 15, 16, 26, 148, 68, 398, 62, 193, 25, 27, 140, 550, 397, 107, 113, ...,.

%e a(1) = 1 because 1*1+1 = 2 which is the first prime,

%e a(2) = 2 because 2*1+1 = 3 which is the second prime,

%e a(3) = 4 because 1*6-1 = 5 which is the third prime,

%e a(8) = 3 because 2*10-1 = 19 which is the eighth prime, ...

%t triQ[n_] := IntegerQ@ Sqrt[8n + 1]; f[n_] := Block[{j = 1, p = Prime@n}, While[ !triQ[(p - 1)/j] && !triQ[(p + 1)/j], j++ ]; j]; Array[f, 92]

%Y Cf. A000217, A125765, A125766, A125767, A125768, A125770.

%K nonn

%O 1,2

%A _Jonathan Vos Post_ & _Robert G. Wilson v_, Dec 01 2006