A206913 Greatest binary palindrome <= n; the binary palindrome floor function.

0, 1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 9, 9, 9, 9, 15, 15, 17, 17, 17, 17, 21, 21, 21, 21, 21, 21, 27, 27, 27, 27, 31, 31, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 33, 45, 45, 45, 45, 45, 45, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 51, 63, 63, 65, 65, 65, 65

Offset: 0

Author

Hieronymus Fischer, Feb 13 2012

Keywords

• nonn
• base

Comments

Also the greatest binary palindrome < n + 1;

For n > 0, a(n-1) is the greatest binary palindrome < n.

Examples

			a(0) = 0 since 0 is the greatest binary palindrome <= 0;
a(1) = 1 since 1 is the greatest binary palindrome <= 1;
a(2) = 1 since 1 is the greatest binary palindrome <= 2;
a(3) = 3 since 3 is the greatest binary palindrome <= 3.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A006995, A206915, A206920, A030301.

Sequences related to palindromic floor and ceiling: A175298, A206913, A206914, A261423, A262038, and the large block of consecutive sequences beginning at A265509.

Programs

Haskell

a206913 n = last $ takeWhile (<= n) a006995_list
-- Reinhard Zumkeller, Feb 27 2012

Formula

Let n > 2, p = 1 + 2*floor((n-1)/2), m = floor(log_2(p)), q = floor((m+1)/2), s = floor(log_2(p-2^q)),

F(x, r) = floor(x/2^q)*2^q + Sum_{k = 0...q - 1} (floor(x/2^(r-k)) mod 2)*2^k;

If F(p, m) <= n then a(n) = F(p, m), otherwise a(n) = F(p-2^q, s).

By definition: F(p, m) = floor(p/2^q)*2^q + A030101(p) mod 2^q; also: F(p-2^q, s) = floor((p-2^q)/2^q)*2^q + A030101(p-2^q) mod 2^q; [Edited and corrected by Hieronymus Fischer, Sep 08 2018]

a(n) = A006995(A206915(n));

a(n) = A006995(A206915(A206914(n+1))-1);

a(n) = A006995(A206916(A206914(n+1))-1).