A179921 a(n) = prime(n) if n<=3; for n>3, a(n) is the smallest prime >a(n-1), such that the denominator of fraction (a(n-1)-a(n-2))/(a(n)-a(n-1)) did not appear earlier.

2, 3, 5, 7, 13, 23, 31, 53, 67, 79, 113, 131, 151, 193, 233, 271, 307, 353, 379, 409, 457, 557, 613, 691, 761, 809, 883, 907, 1013, 1069, 1123, 1181, 1213, 1279, 1361, 1423, 1483, 1571, 1657, 1709, 1811, 1933, 1997, 2087, 2179, 2273, 2341, 2459

Offset: 1

Author

Vladimir Shevelev, Jan 12 2011

Keywords

• nonn

Comments

Using Dirichlet's theorem on arithmetic progressions, it is easy to prove that the sequence is infinite. The sequence of the corresponding denominators begins with 2,1,3,5,4,11,7,6,17, ...

Examples

			The first four terms 2,3,5,13 give three denominators: 2,1,3. Then a(5) is not in {17, 19}, since (13-5)/(17-13) = 2/1, (13-5)/(19-13) = 4/3 and denominators 1 and 3 already appeared earlier. Since (13-5)/(23-13) = 4/5 and 5 is not yet in the denominator sequence, a(5) = 23.
		

Crossrefs

Cf. A168253, A178942, A179210, A179234, A179256, A179328.

Extensions

Edited by Alois P. Heinz, Jan 12 2011