A190895 Auxiliary r(n) sequence used to prove some properties about Rowland's sequence: r(1) = 1, and r(n) = 1/2*(c(n)+1), where c(n) is A190894, for n>1.

1, 5, 6, 11, 12, 23, 24, 47, 48, 50, 51, 101, 102, 105, 110, 111, 117, 233, 234, 467, 468, 470, 471, 941, 942, 945, 1889, 1890, 3779, 3780, 7559, 7560, 7566, 15131, 15132, 15158, 15159, 15162, 30323, 30324, 60647, 60648, 60650, 60651, 60701, 60702, 121403, 121404, 242807, 242808, 242810

Offset: 1

Author

Serafín Ruiz-Cabello, May 23 2011

Keywords

• nonn

Comments

This sequence is matched with another auxiliary sequence called c(n) (A190894). Rowland's sequence (A106108) can be easily described in terms of c(n) and r(n). Also, they can be used to prove easily that the difference between two consecutive terms is always 1 or a prime.

This sequence is related to Rowland's sequence (A106108) with initial condition a(1)=7.

Sequence r(n) satisfies 2r(n) - 1 = c(n), for any n>1.

For further information, see the references.

Examples

			For n = 2, r(2) = 1/2 * (c(2) + 1) = 1/2 * (9 + 1) = 5.
For n = 3, r(3) = 1/2 * (c(3) + 1) = 1/2 * (11 + 1) = 6.
		

Links

• F. Chamizo, D. Raboso, and S. Ruiz-Cabello, On Rowland's sequence, Electronic J. Combin., Vol. 18(2), 2011, #P10.
• E. S. Rowland, A natural prime-generating recurrence, J. Integer Seq., 11(2): Article 08.2.8, 13, 2008.

Crossrefs

Cf. A106108, A190894.