%I #18 Nov 14 2018 20:30:06
%S 2,1,1,1,2,1,3,1,2,1,1,1,1,1,4,1,2,1,1,1,3,1,1,1,1,1,2,1,1,1,2,1,2,1,
%T 1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,5,1,2,1,1,1,
%U 1,1,1,1,3,1,1,1,1,1,1,1,1,1,1,1,2,1,1
%N Rank of the n-th binary palindrome. The minimal number of iterations A206915(A206915(...A206915(A006995(n))...) such that the result is not a binary palindrome, a(3)=1.
%C The number of iterations such that A006995(n)=A006995(A006995(A006995(…(A206922(n)…) [For n<>3] .
%H Antti Karttunen, <a href="/A206921/b206921.txt">Table of n, a(n) for n = 1..65537</a>
%F a(n)=k, where k can be determined by the following iteration: set k=0, p(0)=A006995(n). Repeat while A178225(p(k))==1, set k=k+1, p(k)=A206915(p(k-1)) end repeat [for n<>3].
%F Recursion for n<>3:
%F Case 1: a(n)=1, if n is not a binary palindrome;
%F Case 2: a(n)=a(A206915(n))+1, else.
%F Formally: a(n)=if (A178225(n)==0) then 1 else a(A206915(n))+1
%e a(1)=2, since A006995(1)=0=A006995(A006995(2)) [==> 2 iterations; 2 is not a binary palindrome];
%e a(3)=1 by definition;
%e a(4)=1, since A006995(4)=5=A006995(4) [==> 1 iteration; 4 is not a binary palindrome];
%e a(7)=3, since A006995(7)=15=A006995(A006995(A006995(4))) [==> 3 iterations; 4 is not a binary palindrome];
%o /* C program fragment, omitting formal details, n!=3 */
%o k=0;
%o p=A006995(n);
%o while A178225(p)==1
%o {
%o k++;
%o p=A206915(p);
%o }
%o return k;
%o (PARI)
%o up_to = 65537;
%o A178225(n) = (Vecrev(n=binary(n))==n);
%o A206915list(up_to) = { my(v=vector(up_to+1), s=0); for(n=1,up_to+1,s += A178225(n-1); v[n] = s); (v); };
%o v206915 = A206915list(up_to);
%o A206915(n) = v206915[1+n];
%o A206921(n) = if((3==n)||!A178225(n),1,1+A206921(A206915(n))); \\ _Antti Karttunen_, Nov 14 2018
%Y Cf. A006995, A206922, A178225, A206915, A154809.
%K nonn,base
%O 1,1
%A _Hieronymus Fischer_, Mar 12 2012