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))...).

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, 28, 29, 30, 11, 32, 12, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 13, 46, 47, 48, 49, 50, 14, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 4, 64, 16, 66, 67, 68

Offset: 1

Author

Hieronymus Fischer, Mar 12 2012

Keywords

• nonn
• base

Comments

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

For n>3: a(n)

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

Examples

			a(1)=2, since A006995(1) = 0 = A006995(A006995(2)).
a(3)=3, since A006995(3) = 3 = A006995(A006995(A006995(...(3)...).
a(7)=4, since A006995(7) = 15 = A006995(A006995(A006995(4)).
a(9)=6, since A006995(9) = 21 = A006995(A006995(6)).
		

Crossrefs

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

Programs

C

/* C program fragment, omitting formal details, n!=3 */
k=0;
p=A006995(n);
while A178225(p)==1
{
  k++;
  p=A206915(p);
}
return p;

Formula

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

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].

Recursion for n<>3:

Case 1: a(n)=n, if n is not a binary palindrome;

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

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

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