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

1,2

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.

Links

Table of n, a(n) for n=1..51.

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.

Example

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.

Crossrefs

Cf. A106108, A190894.

Sequence in context: A369273 A277095 A332482 * A129286 A334813 A083450

Adjacent sequences: A190892 A190893 A190894 * A190896 A190897 A190898

Keyword

nonn

Author

Serafín Ruiz-Cabello, May 23 2011

Status

approved