A043261 Sum of the binary digits of the n-th base-2 palindrome.

0, 1, 2, 2, 3, 2, 4, 2, 3, 4, 5, 2, 4, 4, 6, 2, 3, 4, 5, 4, 5, 6, 7, 2, 4, 4, 6, 4, 6, 6, 8, 2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 6, 7, 6, 7, 8, 9, 2, 4, 4, 6, 4, 6, 6, 8, 4, 6, 6, 8, 6, 8, 8, 10, 2, 3, 4, 5, 4, 5, 6, 7, 4, 5, 6, 7, 6, 7, 8, 9, 4, 5, 6, 7, 6, 7, 8, 9, 6, 7, 8

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• base

Examples

			The fourth base-2 palindrome is 5 = 101_2, so a(4) = 1+0+1 = 2.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006995 (base-2 palindromes), A057148.

Programs

Maple

b:= proc(n) option remember;
  procname(floor(n/2)) end proc;
b(1):= 0; b(2):= 1;
c:= proc(n) option remember;
  procname(floor(n/2)) + (n mod 2) end proc;
c(1):= 0; c(2):= 1;
A043261:= n -> 2*c(n) - (n mod 2)*b(n);
A043261(2):= 1;# Robert Israel, Apr 06 2014

Python

def A043261(n):
    if n == 1: return 0
    a = 1<<(l:=n.bit_length()-2)
    m = a|(n&a-1)
    return (m.bit_count()<<1) - (0 if a&n else m&1) # Chai Wah Wu, Jun 13 2024

Formula

Let b(1) = 0, b(2) = 1, otherwise b(2*n-1) = b(n-1) and b(2*n) = b(n).

Let c(1) = 0, c(2) = 1, otherwise c(2*n-1) = c(n-1)+1 and c(2*n) = c(n).

Then for n >= 2, a(2*n-1) = 2*c(2*n-1) - b(2*n-1) and a(2*n) = 2*c(2*n).

Extensions

edited by Robert Israel, Apr 06 2014