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A179481
a(n) = 2*t(n)-1 where t(n) is the sequence of records positions of A179480.
5
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
2,1
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
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Jul 16 2010
EXTENSIONS
Edited by N. J. A. Sloane, Jul 18 2010
More terms from R. J. Mathar, Jul 18 2010
STATUS
approved