A190894 Auxiliary c(n) sequence used to prove some properties about Rowland's sequence. c(n) has the following recursive definition: c(1) = 5, c_(n+1) = c(n) + lfp(c(n)) - 1, where lpf(.) denotes the lowest prime factor of a number.

5, 9, 11, 21, 23, 45, 47, 93, 95, 99, 101, 201, 203, 209, 219, 221, 233, 465, 467, 933, 935, 939, 941, 1881, 1883, 1889, 3777, 3779, 7557, 7559, 15117, 15119, 15131, 30261, 30263, 30315, 30317, 30323, 60645, 60647, 121293, 121295, 121299, 121301, 121401

Offset

1,1

Comments

This sequence is matched with r(n)=A190895(n). 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. For any other odd initial condition a(1) greater than 3, there is an analog c(n) sequence, with c(1) = a(1) - 2.

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

For further information, see the references.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

F. Chamizo, D. Raboso, and S. Ruiz-Cabello, On Rowland's sequence, Vol. 18(2), 2011, #P10.

E. S. Rowland, A natural prime-generating recurrence, J. Integer Seq., 11(2): Article 08.2.8, 13, 2008.

Eric Rowland, A Bizarre Way to Generate Primes, YouTube video, 2023.

Formula

c(1) = 5; c(n+1) = c(n) + lfp(c(n)) - 1.

Example

For n=2, c(n) = 5 + lpf(5) - 1 = 5 + 5 - 1 = 9
For n=3, c(n) = 9 + lfp(9) - 1 = 9 + 3 - 1 = 11

Mathematica

NestList[#+FactorInteger[#][[1, 1]]-1&, 5, 50] (* Harvey P. Dale, Jun 10 2016 *)

Crossrefs

Cf. A020639, A106108, A137613, A190895.

Sequence in context: A100141 A314604 A309137 * A214262 A288143 A120228

Adjacent sequences: A190891 A190892 A190893 * A190895 A190896 A190897

Keyword

nonn

Author

Serafín Ruiz-Cabello, May 23 2011

Status

approved