A296062 Base-2 logarithm of the number of different shapes of balanced binary trees with n nodes (A110316).

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

Offset: 0

Author

Katarzyna Matylla, Dec 04 2017

Keywords

• nonn
• look
• easy

Comments

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

References

• Hsien-Kuei Hwang, S, Janson, T.-H. Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968, 2022.

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..16383 (first 10001 terms from Robert Israel)
• Pieter C. Allaart and Kiko Kawamura, The Takagi Function: a Survey, Real Analysis Exchange, pp. 1-54, Vol. 37, No. 1, 2011-12.
• T. M. Coronado and F. Rosselló, The minimum value of the Colless index, arXiv:1903.11670 [q-bio.PE], 2019. - _Francesc Rosselló_, Apr 08 2019
• T. M. Coronado et. al, On the minimum value of the Colless index and the bifurcating trees that achieve it, Journal of Mathematical Biology, pp. 1993-2054, Vol. 80, 2020.
• M. Kruppel, On the extrema and the improper derivatives of Takagi’s continuous nowhere differentiable function, pp. 41-59, Rostock. Math. Kolloq.,Vol. 62, 2007.
• 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.
• Laura Monroe, A Few Identities of the Takagi Function on Dyadic Rationals, arXiv:2111.05996 [math.CO], 2021.
• Laura Monroe, Takagi Function Identities on Dyadic Rationals, J. Int. Seq (2024) Vol. 27, Art. 24.2.7.
• Wikipedia, Blancmange curve

Crossrefs

Cf. A007814, A110316, A268289.

Programs

Maple

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:
seq(a(n), n=0..100);  # Alois P. Heinz, Dec 04 2017

Mathematica

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 *)

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<Chai Wah Wu, Mar 02 2023

Formula

a(0) = a(1) = 0; a(2*n) = a(n) + a(n-1) + 1; a(2*n+1) = 2*a(n).

a(n) = A007814(A110316(n)).

2^a(n) = A110316(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) = min_{i=1..floor((n+1)/2)} n + 1 - 2*i + a(i-1) + a(n-i). - Qingnian Su and Andrew Howroyd, Nov 04 2018

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

From Laura Monroe, Oct 23 2020: (Start)

a(n+1) = (2^k)*tau(x/(2^k)), where tau is the Takagi function, and n = (2^k) + x with x < 2^k.

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