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return sum(2*bin(i).count('1')-len(bin(i))+2 for i in range(n+1)) # Chai Wah Wu, Sep 03 2020

(Python)

def A301336(n):

return sum(2*bin(i).count('1')-len(bin(i))+2 for i in range(n+1)) # Chai Wah Wu, Sep 03 2020

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Cf. A000079, A000120, A000295, A000788, A001855, A023416, A037861, A059015, A083652, A145037, A268289, A301896.

allocated for Ilya Gutkovskiya(n) = total number of 1's minus total number of 0's in binary expansions of 0, ..., n.

-1, 0, 0, 2, 1, 2, 3, 6, 4, 4, 4, 6, 6, 8, 10, 14, 11, 10, 9, 10, 9, 10, 11, 14, 13, 14, 15, 18, 19, 22, 25, 30, 26, 24, 22, 22, 20, 20, 20, 22, 20, 20, 20, 22, 22, 24, 26, 30, 28, 28, 28, 30, 30, 32, 34, 38, 38, 40, 42, 46, 48, 52, 56, 62, 57, 54, 51, 50, 47, 46, 45, 46, 43, 42, 41, 42

0,4

<a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

G.f.: -1/(1 - x) + (1/(1 - x)^2)*Sum_{k>=0} x^(2^k)*(1 - x^(2^k))/(1 + x^(2^k)).

a(n) = A000788(n) - A059015(n).

a(n) = A268289(n) - 1.

a(A000079(n)) = A000295(n).

+---+-----+---+---+---+---+------------+

| n | bin.|1's|sum|0's|sum| a(n) |

+---+-----+---+---+---+---+------------+

| 0 | 0 | 0 | 0 | 1 | 1 | 0 - 1 =-1 |

| 1 | 1 | 1 | 1 | 0 | 1 | 1 - 1 = 0 |

| 2 | 10 | 1 | 2 | 1 | 2 | 2 - 2 = 0 |

| 3 | 11 | 2 | 4 | 0 | 2 | 4 - 2 = 2 |

| 4 | 100 | 1 | 5 | 2 | 4 | 5 - 4 = 1 |

| 5 | 101 | 2 | 7 | 1 | 5 | 7 - 5 = 2 |

| 6 | 110 | 2 | 9 | 1 | 6 | 9 - 6 = 3 |

+---+-----+---+---+---+---+------------+

bin. - n written in base 2;

1's - number of 1's in binary expansion of n;

0's - number of 0's in binary expansion of n;

sum - total number of 1's (or 0's) in binary expansions of 0, ..., n.

Accumulate[DigitCount[Range[0, 75], 2, 1] - DigitCount[Range[0, 75], 2, 0]]

Cf. A000079, A000120, A000295, A000788, A001855, A023416, A037861, A059015, A083652, A145037, A268289.

allocated

sign,base

Ilya Gutkovskiy, Mar 28 2018

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