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A206922 Root of the n-th binary palindrome. Least number r > 1 such that A006995(n) can be represented by a finite or infinite number of iterations A006995(A006995(A006995(...(...(r))...). 1

%I #23 Feb 20 2023 18:32:16

%S 2,2,3,4,4,6,4,8,6,10,11,12,13,14,4,16,8,18,19,20,6,22,23,24,25,26,10,

%T 28,29,30,11,32,12,34,35,36,37,38,39,40,41,42,43,44,13,46,47,48,49,50,

%U 14,52,53,54,55,56,57,58,59,60,61,62,4,64,16,66,67,68

%N Root of the n-th binary palindrome. Least number r > 1 such that A006995(n) can be represented by a finite or infinite number of iterations A006995(A006995(A006995(...(...(r))...).

%C If n is not a binary palindrome, then a(n)=n.

%C For n>3: a(n)<n, iff n is a binary palindrome.

%C For n<>3: The number of iterations such that A006995(n)= A006995(A006995(A006995(...(...(r))...) is given by A206921(n).

%F a(n) <= n for n > 1.

%F a(n)=p(k), where p(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)=n, if n is not a binary palindrome;

%F Case 2: a(n)=a(A206915(n)), else.

%F Formally: a(n)=if (A178225(n)==0) then n else a(A206915(n)).

%e a(1)=2, since A006995(1) = 0 = A006995(A006995(2)).

%e a(3)=3, since A006995(3) = 3 = A006995(A006995(A006995(...(3)...).

%e a(7)=4, since A006995(7) = 15 = A006995(A006995(A006995(4)).

%e a(9)=6, since A006995(9) = 21 = A006995(A006995(6)).

%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 p;

%Y Cf. A006995, A206921, A178225, A206915, A154809.

%K nonn,base

%O 1,1

%A _Hieronymus Fischer_, Mar 12 2012

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