cp's OEIS Frontend

A256705 Numbers n such that if we define f(n,m) by the recursion f(n,1)=n, f(n,k+1)= A007672(n,k), the subsequence f(n,k), with n constant, is eventually periodic with period two.

%I A256705 #14 Jul 18 2016 12:36:44
%S A256705 9,16,25,45,49,63,75,80,81,99,112,117,121,125,128,147,153,169,171,175,
%T A256705 176,207,208,225,243,245,250,256,261,275,279,289,304,315,325,333,343,
%U A256705 361,363,368,369,375,387,405,423,425,441,464,475,477,486,495,496,500,507,512,525,529,531,539,549,560
%N A256705 Numbers n such that if we define f(n,m) by the recursion f(n,1)=n, f(n,k+1)= A007672(n,k), the subsequence f(n,k), with n constant, is eventually periodic with period two.
%C A256705 If n is in the sequence, there is an index k such that, for m>=k, S(f(n,m))=S(f(n,m+1)), where S(n) is Kempner function A002034(n).
%C A256705 If n is not in the sequence, there is an index k such that S(f(n,m))>S(f(n,m+1)) for m<k, and f(n,m)=1 for m>=k.
%e A256705 16 is in the sequence because f(16,1) = 16, f(16,2) = A007672(16) = 45, f(16,3) = A007672(45) = 16, f(16,4) = A007672(16) = 45, ..., a sequence with period 2.
%e A256705 15 is not in the sequence, then f(15,1) = 15, f(15,2) = A007672(15) = 8, f(15,3) = A007672(8) = 3, f(15,4) = A007672(3) = 2, f(15,5) = A007672(2) = 1, f(15,6) = A007672(1) = 1, ..., a sequence with period 1.
%o A256705 (PARI) b(n)=my(m=1,x=n,as=1,p); while(x>1, m++; p=gcd(x,m); x/=p; as*=m/p); as /* A007672(n) */
%o A256705 for(i=1,10^3, m=i; v=1; while(m>1&&v, n=b(m); if(m==b(n), v=0; print1(i,", ")); m=n))
%Y A256705 Cf. A000142, A002034, A007672.
%K A256705 nonn
%O A256705 1,1
%A A256705 _Antonio Roldán_, Apr 08 2015