A077071 Row sums of A077070.

0, 2, 8, 16, 30, 46, 66, 88, 118, 150, 186, 224, 268, 314, 364, 416, 478, 542, 610, 680, 756, 834, 916, 1000, 1092, 1186, 1284, 1384, 1490, 1598, 1710, 1824, 1950, 2078, 2210, 2344, 2484, 2626, 2772, 2920, 3076, 3234, 3396, 3560, 3730, 3902, 4078, 4256

Offset: 0

Author

Michael Somos, Oct 25 2002

Keywords

• nonn

Comments

Conjecture: given n balls, all of which are initially in the first of n numbered boxes, a(n-1) is the number of steps of the following process required to move them all to the last box. A step consists of first identifying j, the lowest numbered box which has at least one ball. If it has only one ball then move it to box j+1; otherwise move half its balls rounded down to box j+1 and (unless it's the first box) half its balls rounded down to box j-1. See also A356254. - Mikhail Kurkov, Nov 24 2022

Links

• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi 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, Svante Janson, and Tsung-Hsi 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.
• Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple generating functions, 2004.
• Ralf Stephan, Table of generating functions (ps file).
• Ralf Stephan, Table of generating functions (pdf file).

Crossrefs

Cf. A077070, A356254.

Programs

PARI

{a(n) = sum( k=0, n, -valuation( polcoeff( pollegendre(2*n), 2*k), 2))}

PARI

a(n)=my(P=pollegendre(2*n)); -sum(k=0,n,valuation(polcoeff(P,2*k), 2)) \\ Charles R Greathouse IV, Apr 12 2012

Python

def A077071(n): return ((n+1)*(n-n.bit_count())<<1)-sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1)) # Chai Wah Wu, Nov 12 2024

Formula

a(n) is asymptotic to 2*n^2 and it seems that a(n) = 2*n^2 + O(n^(3/2)) (where O(n^(3/2))/n^(3/2) is bounded and O(n^(3/2)) < 0). - Benoit Cloitre, Oct 30 2002

G.f.: (1/(1-x)^2) * Sum_{k>=0} t/(1-t) where t = x^2^k. Twice the value of the partial sum of A005187. a(0) = 0, a(2n) = a(n) + a(n-1) + 4*n^2 + 2*n, a(2n+1) = 2*a(n) + 4*n^2 + 6*n + 2. - Ralf Stephan, Sep 12 2003

a(n) = 2*n*(n+1) - 2*A000788(n) and therefore asymptotically a(n) = 2*n^2 - n*log_2(n) + O(n). - Peter J. Taylor, Dec 21 2022