A206915 The index (in A006995) of the greatest binary palindrome <= n; also the 'lower inverse' of A006995.

1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 16, 16, 16, 16, 16, 16, 16

Offset: 0

Author

Hieronymus Fischer, Feb 15 2012

Keywords

• nonn
• base

Comments

The greatest m such that A006995(m)<= n;

The number of binary palindromes <= n;

n is palindromic iff a(n)=A206916(n);

a(n) is the number of the binary palindrome A206913(n);

if n is a binary palindrome, then A006995(a(n))=n, so a(n) is 'inverse' with respect to A006995.

Partial sums of the binary palindromic characteristic function A178225.

Examples

			a(1)=2 since 2 is the index number of the greatest binary palindrome <= 1;
a(5)=4 since there are only 4 binary palindromes (namely 0,1,3 and 5) which are less than or equal to 5;
a(10)=6 since A006995(6)=9<=10, but A006995(7)=15>10, and so that, 6 is the index number of greatest binary palindrome <= 10;
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A006995, A206914, A206916, A206920.

Programs

Mathematica

A178225[n_]:=Boole[PalindromeQ[IntegerDigits[n,2]]];
Accumulate[Array[A178225,100,0]] (* Paolo Xausa, Oct 15 2023 *)

Python

def A206915(n):
    l = n.bit_length()
    k = l+1>>1
    return (n>>l-k)-(int(bin(n)[k+1:1:-1] or '0',2)>(n&(1<Chai Wah Wu, Jul 24 2024

Formula

a(n) = max(m | A006995(m) <= n);

a(A006995(n)) = n;

A006995(a(n)) <= n, equality holds true iff n is a binary palindrome;

Let p = A206913(n), m = floor(log_2(p)) and p>2, then:

a(n) = (((5-(-1)^m)/2) + sum_{k=1..floor(m/2)} (floor(p/2^k) mod 2)/2^k)) * 2^floor(m/2).

a(n) = (1/2)*((6-(-1)^m)*2^floor(m/2) - 1 - sum_{k=1..floor(m/2)} (-1)^floor(p/2^k) * 2^(floor(m/2)-k))).

a(n) = (5-(-1)^m) * 2^floor(m/2)/2 - 3*sum_{k=2..floor(m/2)} (floor(p/2^k) * 2^floor(m/2)/2^k) + (floor(p/2) * 2^floor(m/2)/2 - 2*floor((p/2) * 2^floor(m/2)) * floor((m-1)/m+1/2).

Partial sums S(n) = sum_{k=0..n} a(k):

S(n) = (n+1)*a(n) - A206920(a(n)).

G.f.: g(x) = (1+x+x^3+sum_{j>=1} x^(3*2^j)*(f_j(x)+f_j(1/x)))/(1-x), where the f_j(x) are defined as follows:

f_1(x) = x, and for j>1,

f_j(x) = x^3*product_{k=1..floor((j-1)/2)} (1+x^b(j,k)), where b(j,k)=2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k>1, and b(j,1)=(2+(-1)^j)*2^(floor((j-1)/2)+1).