A037861 (Number of 0's) - (number of 1's) in the base-2 representation of n.
1, -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
Offset: 0
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
- Jonathan Sondow, Double integrals for Euler's constant and ln(4/Pi) and an analog of Hadjicostas's formula, arXiv:math/0211148 [math.CA], 2002-2004; Amer. Math. Monthly 112 (2005), 61-65.
- Jonathan Sondow, New Vacca-Type Rational Series for Euler's Constant and Its "Alternating" Analog ln(4/Pi), arXiv:math/0508042 [math.NT], 2005; Additive Number Theory, Festschrift In Honor of the Sixtieth Birthday of Melvyn B. Nathanson (D. Chudnovsky and G. Chudnovsky, eds.), Springer, 2010, pp. 331-340.
- Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
- Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
- Ralf Stephan, Table of generating functions (ps file).
- Ralf Stephan, Table of generating functions (pdf file).
- Index entries for sequences related to binary expansion of n
Crossrefs
Programs
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Haskell
a037861 n = a023416 n - a000120 n -- Reinhard Zumkeller, Aug 01 2013
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Maple
A037861:= proc(n) local L; L:= convert(n,base,2); numboccur(0,L) - numboccur(1,L) end proc: map(A037861, [$0..100]); # Robert Israel, Mar 08 2016
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Mathematica
Table[Count[ IntegerDigits[n, 2], 0] - Count[IntegerDigits[n, 2], 1], {n, 0, 75}]
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PARI
a(n) = if (n==0, 1, 1 + logint(n, 2) - 2*hammingweight(n)); \\ Michel Marcus, May 15 2020 and Jun 16 2020
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Python
def A037861(n): return 2*format(n,'b').count('0')-len(format(n,'b')) # Chai Wah Wu, Mar 07 2016
Formula
From Henry Bottomley, Oct 27 2000: (Start)
a(2*n) = a(n) + 1; a(2*n + 1) = a(2*n) - 2 = a(n) - 1. (End)
G.f. satisfies A(x) = (1 + x)*A(x^2) - x*(2 + x)/(1 + x). - Franklin T. Adams-Watters, Dec 26 2006
a(n) = b(n) for n > 0 with b(0) = 0 and b(n) = b(floor(n/2)) + (-1)^(n mod 2). - Reinhard Zumkeller, Dec 31 2007
G.f.: 1 + (1/(1 - x))*Sum_{k>=0} x^(2^k)*(x^(2^k) - 1)/(1 + x^(2^k)). - Ilya Gutkovskiy, Apr 07 2018
Comments