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A255743
a(1) = 1; for n > 1, a(n) = 9*8^{A000120(n-1)-1}.
5
1, 9, 9, 72, 9, 72, 72, 576, 9, 72, 72, 576, 72, 576, 576, 4608, 9, 72, 72, 576, 72, 576, 576, 4608, 72, 576, 576, 4608, 576, 4608, 4608, 36864, 9, 72, 72, 576, 72, 576, 576, 4608, 72, 576, 576, 4608, 576, 4608, 4608, 36864, 72, 576, 576, 4608, 576, 4608, 4608
OFFSET
1,2
COMMENTS
Also, this is a row of the square array A255740.
Partial sums give A255764.
LINKS
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 33.
EXAMPLE
Written as an irregular triangle in which the row lengths are the terms of A011782, the sequence begins:
1;
9;
9, 72;
9, 72, 72, 576;
9, 72, 72, 576, 72, 576, 576, 4608;
...
MATHEMATICA
MapAt[Floor, Array[9*8^(DigitCount[# - 1, 2, 1] - 1) &, 55], 1] (* Michael De Vlieger, Nov 03 2022 *)
PROG
(PARI) a(n) = if (n==1, 1, 9*8^(hammingweight(n-1)-1)); \\ Michel Marcus, Mar 15 2015
(Python 3.10+)
def A255743(n): return 1 if n == 1 else 9*(1<<((n-1).bit_count()-1)*3) # Chai Wah Wu, Nov 15 2022
KEYWORD
nonn
AUTHOR
Omar E. Pol, Mar 05 2015
EXTENSIONS
More terms from Michel Marcus, Mar 15 2015
STATUS
approved