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%I #17 Aug 12 2022 15:13:29
%S 3,7,11,19,23,29,37,47,53,59,67,71,79,83,101,103,107,131,139,149,163,
%T 167,173,179,191,197,199,211,227,239,263,269,271,293,311,317,347,359,
%U 367,373,379,383,389,419,443,461,463,467,479,487,491,503,509,523,541
%N a(n) = 2*t(n)-1 where t(n) is the sequence of records positions of A179480.
%C Question. Is every term of this sequence prime?
%C From _Gary W. Adamson_, Sep 04 2012: (Start)
%C In answer to the primality question and pursuant to the Coach Theorem of Hilton and Pedersen: phi(b) = 2 * k * c, with b an odd integer and k in A003558, and c (the numbers of coaches) in A135303; iff phi(b) = (b-1) then b = p, prime. This implies that if b has one coach and k = (b-1)/2, b must be prime since phi(b) = 2 * k * c = 2 * (b-1)/2 * 1 = (b-1). Conjectures: all terms in A179481 have one coach with k = (b-1)/2 and are therefore primes. Next, if A179480(n) is a new record high value, then so is A003558(n-1); but not necessarily the converse (e.g. 13), and the corresponding value of k for b is (b-1)/2. Examples: b = 13 has one coach with k (sum of bottom row terms ) = 6 = A003558(6); and r (number of entries in each row) = 3:
%C 13: [1, 3, 5]
%C ......2, 1, 3. This example satisfies the primality requirements since phi(13) = 12 = 2 * k * c = 2 * 6 * 1; but not the new record requirement for r = 3 since A179480(6) = 3, corresponding to 11, not 13. As shown in the coach for 11:
%C 11: [1, 3, 3]
%C ......1, 1, 3; k = (b-1)/2 with r = 3 and c = 1. Therefore, 11 is in A179481 but not 13. (End)
%D P. Hilton and J. Pedersen, A Mathematics Tapestry, Demonstrating the Beautiful Unity of Mathematics, 2010, Cambridge University Press, pp. 260-264.
%Y Cf. A179480, A179460, A179382, A179383.
%Y Cf. A003558, A000040.
%K nonn
%O 2,1
%A _Vladimir Shevelev_, Jul 16 2010
%E Edited by _N. J. A. Sloane_, Jul 18 2010
%E More terms from _R. J. Mathar_, Jul 18 2010
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