A136138 a(n+1)=sopfr(4a(n)+1), with sopfr=A001414. Finishes with the cycle (34, 137, 67, 269, 362, 36).

1, 5, 10, 41, 19, 18, 73, 293, 43, 173, 24, 97, 389, 179, 242, 39, 157, 54, 38, 23, 34, 137, 67, 269, 362, 36, 34

Offset

0,2

Comments

In the class of recurrence sequences a(n+1)=sopfr(C*a(n)+D), with C=4, D=1. It is a simple example where it finishes with a nontrivial cycle.

Links

Table of n, a(n) for n=0..26.

Formula

a(n+1) = A001414(4*a(n)+1).

Mathematica

sopfr = Function[x, Plus @@ Map[Times @@ # &, FactorInteger[x]]]; NestList[sopfr[4# + 1] &, 1, 40]
NestList[Total[Times@@@FactorInteger[4*#+1]]&, 1, 30] (* Harvey P. Dale, Sep 23 2019 *)

Crossrefs

Cf. A136136, A136137.

Sequence in context: A001692 A038070 A364738 * A321307 A270288 A271805

Adjacent sequences: A136135 A136136 A136137 * A136139 A136140 A136141

Keyword

fini,nonn

Author

Carlos Alves, Dec 16 2007

Status

approved