A145037 Number of 1's minus number of 0's in the binary representation of n.

0, 1, 0, 2, -1, 1, 1, 3, -2, 0, 0, 2, 0, 2, 2, 4, -3, -1, -1, 1, -1, 1, 1, 3, -1, 1, 1, 3, 1, 3, 3, 5, -4, -2, -2, 0, -2, 0, 0, 2, -2, 0, 0, 2, 0, 2, 2, 4, -2, 0, 0, 2, 0, 2, 2, 4, 0, 2, 2, 4, 2, 4, 4, 6, -5, -3, -3, -1, -3, -1, -1, 1, -3, -1, -1, 1, -1, 1, 1, 3, -3, -1, -1, 1, -1, 1, 1, 3, -1, 1, 1

Offset: 0

Author

Reikku Kulon, Sep 30 2008

Keywords

• sign
• base
• easy

Comments

Column 2 of A144912 (which begins at n = 2).

Zeros in that column correspond to A031443.

Examples

			From _Michel Marcus_, Feb 12 2022: (Start)
Viewed as an irregular triangle:
   0;
   1;
   0,  2;
  -1,  1,  1, 3;
  -2,  0,  0, 2,  0, 2, 2, 4;
  -3, -1, -1, 1, -1, 1, 1, 3, -1, 1, 1, 3, 1, 3, 3, 5;
  ... (End)
		

Links

• Table of n, a(n) for n = 0..2^14

Crossrefs

Cf. A037861 (negated), A031443 (indices of 0's), A144912, A000120.

Cf. A269735 (first differences), A268289 (partial sums).

Column k=1 of A360099.

Programs

Haskell

a145037 0 = 0
a145037 n = a145037 n' + 2*m - 1 where (n', m) = divMod n 2
-- Reinhard Zumkeller, Jun 16 2011

Maple

a:= n-> add(2*i-1, i=Bits[Split](n)):
seq(a(n), n=0..90);  # Alois P. Heinz, Jan 18 2022

Mathematica

Join[{0}, Table[Count[#, 1] - Count[#, 0] &[IntegerDigits[n, 2]], {n, 1, 90}]] (* Robert P. P. McKone, Feb 12 2022 *)

PARI

A145037(n)=hammingweight(n)*2-logint(n<<1+!n,2) \\ M. F. Hasler, Mar 08 2018

Python

result = [0]
for n in range (1, 2**14 + 1):
    result.append(bin(n)[2:].count("1") - bin(n)[2:].count("0"))
print(result[0:129]) # Karl-Heinz Hofmann, Jan 18 2022

Python

def a(n): return (n.bit_count()<<1) - n.bit_length()
print([a(n) for n in range(1, 2**14+1)]) # Michael S. Branicky, May 14 2024
(C#)
int A145037(int n)  {
    int result = 0;
    while(n > 0)  {
        result += 2 * (n % 2) - 1;
        n /= 2;
    }
    return result;
} \\ Frank Hollstein, Dec 08 2022

Formula

a(n) = -A037861(n) for n >= 1.

a(n) = Sum_{i=1..k} (2*b[i] - 1) where b is the binary expansion of n and k is the number of bits in this binary expansion. - Michel Marcus, Jun 28 2021

From Aayush Soni Feb 12 2022: (Start)

Upper bound: a(n) <= floor(log_2(n+1)).

Lower bound: For n > 0, a(n) >= 1 - floor(log_2(n)).

If n is even, a(2^n) to a(2^(n+1)-1) inclusive are all odd and vice versa. (End)

Extensions

Renamed (using a Mar 08 2018 comment from M. F. Hasler) and edited by Jon E. Schoenfield, Jun 29 2021