A179481 a(n) = 2*t(n)-1 where t(n) is the sequence of records positions of A179480.

3, 7, 11, 19, 23, 29, 37, 47, 53, 59, 67, 71, 79, 83, 101, 103, 107, 131, 139, 149, 163, 167, 173, 179, 191, 197, 199, 211, 227, 239, 263, 269, 271, 293, 311, 317, 347, 359, 367, 373, 379, 383, 389, 419, 443, 461, 463, 467, 479, 487, 491, 503, 509, 523, 541

Offset: 2

Author

Vladimir Shevelev, Jul 16 2010

Keywords

• nonn

Comments

Question. Is every term of this sequence prime?

From Gary W. Adamson, Sep 04 2012: (Start)

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:

13: [1, 3, 5]

......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:

11: [1, 3, 3]

......1, 1, 3; k = (b-1)/2 with r = 3 and c = 1. Therefore, 11 is in A179481 but not 13. (End)

References

• P. Hilton and J. Pedersen, A Mathematics Tapestry, Demonstrating the Beautiful Unity of Mathematics, 2010, Cambridge University Press, pp. 260-264.

Crossrefs

Cf. A179480, A179460, A179382, A179383.

Cf. A003558, A000040.

Extensions

Edited by N. J. A. Sloane, Jul 18 2010

More terms from R. J. Mathar, Jul 18 2010