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%I #34 Nov 17 2022 05:25:24
%S 1,9,9,72,9,72,72,576,9,72,72,576,72,576,576,4608,9,72,72,576,72,576,
%T 576,4608,72,576,576,4608,576,4608,4608,36864,9,72,72,576,72,576,576,
%U 4608,72,576,576,4608,576,4608,4608,36864,72,576,576,4608,576,4608,4608
%N a(1) = 1; for n > 1, a(n) = 9*8^{A000120(n-1)-1}.
%C Also, this is a row of the square array A255740.
%C Partial sums give A255764.
%H Felix Fröhlich, <a href="/A255743/b255743.txt">Table of n, a(n) for n = 1..10000</a>
%H Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="https://arxiv.org/abs/2210.10968">Identities and periodic oscillations of divide-and-conquer recurrences splitting at half</a>, arXiv:2210.10968 [cs.DS], 2022, p. 33.
%e Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
%e 1;
%e 9;
%e 9, 72;
%e 9, 72, 72, 576;
%e 9, 72, 72, 576, 72, 576, 576, 4608;
%e ...
%t MapAt[Floor, Array[9*8^(DigitCount[# - 1, 2, 1] - 1) &, 55], 1] (* _Michael De Vlieger_, Nov 03 2022 *)
%o (PARI) a(n) = if (n==1, 1, 9*8^(hammingweight(n-1)-1)); \\ _Michel Marcus_, Mar 15 2015
%o (Python 3.10+)
%o def A255743(n): return 1 if n == 1 else 9*(1<<((n-1).bit_count()-1)*3) # _Chai Wah Wu_, Nov 15 2022
%Y Cf. A000120, A151787, A147582, A151789, A151779, A151791, A151782, A255740, A255741, A255744, A255745, A255764.
%K nonn
%O 1,2
%A _Omar E. Pol_, Mar 05 2015
%E More terms from _Michel Marcus_, Mar 15 2015
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