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A296062
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Base-2 logarithm of the number of different shapes of balanced binary trees with n nodes (A110316).
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8
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0, 0, 1, 0, 2, 2, 2, 0, 3, 4, 5, 4, 5, 4, 3, 0, 4, 6, 8, 8, 10, 10, 10, 8, 10, 10, 10, 8, 8, 6, 4, 0, 5, 8, 11, 12, 15, 16, 17, 16, 19, 20, 21, 20, 21, 20, 19, 16, 19, 20, 21, 20, 21, 20, 19, 16, 17, 16, 15, 12, 11, 8, 5, 0, 6, 10, 14, 16, 20, 22, 24, 24, 28
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OFFSET
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0,5
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COMMENTS
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Since terms of A110316 are always powers of 2, it seems natural to have a sequence of the exponents too. Also, it conveys the same information as A110316 but is shorter and more readable.
Also, sum of absolute distances from (n+1) to the nearest multiple of 2^k for all 2^k < n+1. - Ivan Neretin, Jul 03 2018
Also, the minimum cost of connecting n+1 nodes when the cost of joining two connected components is the absolute difference of their sizes. In particular, connecting two equal sized components has zero cost. For example, in the case of n=4 there are 5 nodes. Connecting nodes 1 and 2 costs zero, connecting nodes 3 and 4 costs zero, then connecting {5} to {3,4} costs 1 and finally connecting {1,2} to {3,4,5} costs 1 giving a total cost of 2. Because this solution is optimal a(4) = 2. - Qingnian Su, Nov 03 2018
Also, the minimum Colless index of a rooted bifurcating tree with n leaves. - Francesc Rosselló, Apr 08 2019
Also, dilations of the Takagi function restricted to dyadic rationals in [0,1]. The number of points of a(n) in each dilation is 2^k and the scale of each dilation in both the x and y directions is 2^k, where k = floor(log_2(n+1)). See Allaart et. al (2012), Equation 4.7, attributed to Kruppel (2007). Also see Coronado et.al (2020), Corollary 4. - Laura Monroe, Oct 23 2020
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REFERENCES
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Hsien-Kuei Hwang, S, Janson, T.-H. Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968, 2022.
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LINKS
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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.
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FORMULA
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a(0) = a(1) = 0; a(2*n) = a(n) + a(n-1) + 1; a(2*n+1) = 2*a(n).
G.f. g(x) satisfies g(x) = (1+x)^2*g(x^2) + x^2/(1-x^2). - Robert Israel, Dec 04 2017
a(n) = Sum_{j=2..k} (m_1-m_j-2*(j-2))*2^m_j where m_1 > ... > m_k are the exponents in the binary expansion of n. - Francesc Rosselló, Apr 08 2019
a(n+1) = (2^k)*tau(x/(2^k)), where tau is the Takagi function, and n = (2^k) + x with x < 2^k.
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MAPLE
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a:= proc(n) option remember; local r; `if`(n<2, 0,
`if`(irem(n, 2, 'r')=0, 1+a(r)+a(r-1), a(r)*2))
end:
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MATHEMATICA
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Fold[Append[#1, If[EvenQ@ #2, #1[[#2/2 + 1]] + #1[[#2/2]] + 1, 2 #1[[(#2 - 1)/2 + 1]]]] &, {0, 0}, Range[2, 72]] (* Michael De Vlieger, Dec 04 2017 *)
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PROG
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(PARI) seq(n)={my(v=vector(n)); for(m=2, #v, v[m]=vecmin(vector(m\2, i, v[i] + v[m-i] + m-2*i))); v} \\ Andrew Howroyd, Nov 04 2018
(PARI) seq(n)={my(v=vector(n)); for(n=1, n-1, v[n+1]=if(n%2, 2*v[(n+1)/2], v[n/2] + v[n/2+1] + 1)); v} \\ Andrew Howroyd, Nov 04 2018
(Python)
def A296062(n): return (k:=n+1)-(sum(i.bit_count() for i in range(1, k))<<1)+k*(m:=k.bit_length())-(1<<m) # Chai Wah Wu, Mar 02 2023
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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