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a(n+1)=sopfr(2a(n)+1), with sopfr=A001414. Finishes with 10 (fixed point).
5

%I #4 Sep 24 2013 00:41:43

%S 1,3,7,8,17,12,10

%N a(n+1)=sopfr(2a(n)+1), with sopfr=A001414. Finishes with 10 (fixed point).

%C The class of recurrence sequences a(n+1)=sopfr(C*a(n)+D) may finish with a fixed point or with a cycle.

%F a(n+1)=A001414(2*a(n)+1)

%t sopfr = Function[x, Plus @@ Map[Times @@ # &, FactorInteger[x]]]; NestList[sopfr[2# + 1] &, 1, 10]

%Y Cf. A001414.

%K fini,full,nonn

%O 1,2

%A _Carlos Alves_, Dec 16 2007