A181132 a(0)=0; thereafter a(n) = total number of 0's in binary expansions of 1, ..., n.
0, 0, 1, 1, 3, 4, 5, 5, 8, 10, 12, 13, 15, 16, 17, 17, 21, 24, 27, 29, 32, 34, 36, 37, 40, 42, 44, 45, 47, 48, 49, 49, 54, 58, 62, 65, 69, 72, 75, 77, 81, 84, 87, 89, 92, 94, 96, 97, 101, 104, 107, 109, 112, 114, 116, 117, 120, 122, 124, 125, 127, 128, 129, 129, 135, 140, 145
Offset: 0
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
- Hsien-Kuei Hwang, S. Janson, and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.
- Jeffrey C. Lagarias, The Takagi function and its properties, arXiv:1112.4205 [math.CA], 2011-2012.
- Jeffrey C. Lagarias, The Takagi function and its properties, In Functions in number theory and their probabilistic aspects, 153--189, RIMS Kôkyûroku Bessatsu, B34, Res. Inst. Math. Sci. (RIMS), Kyoto, 2012. MR3014845.
Programs
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Magma
[ n eq 1 select 0 else Self(n-1)+(#B-&+B) where B is Intseq(n, 2): n in [1..70] ]; // Klaus Brockhaus, Oct 08 2010
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Maple
a:= proc(n) option remember; `if`(n=0, 0, a(n-1)+add(1-i, i=Bits[Split](n))) end: seq(a(n), n=0..66); # Alois P. Heinz, Nov 12 2024
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Mathematica
Accumulate[Count[IntegerDigits[#,2],0]&/@Range[70]] (* Harvey P. Dale, May 16 2012 *) Join[{0},Accumulate[DigitCount[Range[70],2,0]]] (* Harvey P. Dale, Jun 09 2016 *)
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PARI
a(n)=my(m=logint(n,2)); 1 + (m+1)*(n+1) - 2^(m+1) + sum(j=1,m+1,my(t=floor(n/2^j + 1/2));(n>>j)*(2*n + 2 - (1 + (n>>j))<
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Python
def A181132(n): return 1+(n+1)*(m:=(n+1).bit_length())-(1<
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Python
def A181132(n): return 1+(n+1)*((t:=(n+1).bit_length())-n.bit_count())-(1<
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Word
microsoft word macro: Sub concatcount() Dim base As Integer Dim digit As Integer Dim counter As Integer Dim standin As Integer Dim max As Integer Let base = 2 Let digit = 0 Let max = 100 Let counter = 0 For n = 1 To max Let standin = n Do While standin > 0 If standin Mod base = digit Then Let counter = counter + 1 Let standin = standin - (standin Mod base) If standin > 0 Then Let standin = standin / base Loop Selection.TypeText Text:=Str(counter) Next n End Sub
Formula
a(n) = A059015(n)-1. - Klaus Brockhaus, Oct 08 2010
From N. J. A. Sloane, Mar 10 2016: (Start)
a(0)=0; thereafter a(2n) = a(n)+a(n-1)+n, a(2n+1) = 2a(n)+n.
G.f.: (1/(1-x)^2) * Sum_{k >= 0} x^(2^(k+1))/(1+x^(2^k)). (End)
Extensions
a(0)=0 added by N. J. A. Sloane, Mar 10 2016 (simplifies the recurrence, makes entry consistent with A059015 and other closely related sequences).
Comments