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a(n) = A000788(n)^2.
1

%I #22 Mar 03 2023 01:15:56

%S 0,1,4,16,25,49,81,144,169,225,289,400,484,625,784,1024,1089,1225,

%T 1369,1600,1764,2025,2304,2704,2916,3249,3600,4096,4489,5041,5625,

%U 6400,6561,6889,7225,7744,8100,8649,9216,10000,10404,11025,11664,12544,13225

%N a(n) = A000788(n)^2.

%H Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, <a href="http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf">Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications</a>, Preprint 2016.

%H Hsien-Kuei Hwang, S. Janson, T.-H. Tsai, <a href="https://doi.org/10.1145/3127585">Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications</a>, ACM Transactions on Algorithms, 13:4 (2017), #47; DOI: 10.1145/3127585.

%p read("transforms") : A000788 := proc(n) add( wt(j),j=0..n) ; end: A163095 := proc(n) A000788(n)^2 ; end: seq(A163095(n),n=0..100) ; # _R. J. Mathar_, Feb 22 2010

%t Accumulate@ DigitCount[Range[0, 44], 2, 1]^2 (* _Michael De Vlieger_, Jan 23 2019 *)

%o (Python)

%o def A163095(n): return sum(i.bit_count() for i in range(1,n+1))**2 # _Chai Wah Wu_, Mar 02 2023

%Y Cf. A000788.

%K easy,nonn,base

%O 0,3

%A _Omar E. Pol_, Aug 06 2009

%E Extended by _R. J. Mathar_, Feb 22 2010