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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))...).

%I A206922 #25 Jan 13 2025 05:43:01
%S A206922 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 A206922 28,29,30,11,32,12,34,35,36,37,38,39,40,41,42,43,44,13,46,47,48,49,50,
%U A206922 14,52,53,54,55,56,57,58,59,60,61,62,4,64,16,66,67,68
%N 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))...).
%C A206922 If n is not a binary palindrome, then a(n)=n.
%C A206922 For n>3: a(n)<n, iff n is a binary palindrome.
%C A206922 For n<>3: The number of iterations such that A006995(n)= A006995(A006995(A006995(...(...(r))...) is given by A206921(n).
%F A206922 a(n) <= n for n > 1.
%F A206922 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 A206922 Recursion for n<>3:
%F A206922   Case 1: a(n)=n, if n is not a binary palindrome;
%F A206922   Case 2: a(n)=a(A206915(n)), else.
%F A206922 Formally: a(n)=if (A178225(n)==0) then n else a(A206915(n)).
%e A206922 a(1)=2, since A006995(1) = 0 = A006995(A006995(2)).
%e A206922 a(3)=3, since A006995(3) = 3 = A006995(A006995(A006995(...(3)...).
%e A206922 a(7)=4, since A006995(7) = 15 = A006995(A006995(A006995(4)).
%e A206922 a(9)=6, since A006995(9) = 21 = A006995(A006995(6)).
%o A206922 (C) /* C program fragment, omitting formal details, n!=3 */
%o A206922 k=0;
%o A206922 p=A006995(n);
%o A206922 while A178225(p)==1
%o A206922 {
%o A206922   k++;
%o A206922   p=A206915(p);
%o A206922 }
%o A206922 return p;
%Y A206922 Cf. A006995, A206921, A178225, A206915, A154809.
%K A206922 nonn,base
%O A206922 1,1
%A A206922 _Hieronymus Fischer_, Mar 12 2012