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A159912 Partial sums of A159913(k) = 2^bitcount(2k+1)-1 = A038573(2k+1), bitcount=A000120. 5
0, 1, 4, 7, 14, 17, 24, 31, 46, 49, 56, 63, 78, 85, 100, 115, 146, 149, 156, 163, 178, 185, 200, 215, 246, 253, 268, 283, 314, 329, 360, 391, 454, 457, 464, 471, 486, 493, 508, 523, 554, 561, 576, 591, 622, 637, 668, 699, 762, 769, 784, 799, 830, 845, 876, 907 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,3
COMMENTS
More precisely, a(n)=sum(i<n, A159913(i)), since we want the sequence to start with a(0)=0 and not with A159913(0)=1.
a(n) is also the total number of ON cells after n generations in the outward corner version of the Ulam-Warburton cellular automaton of A147562, and a(n) is also the total number of Y-toothpicks after n generations in the outward corner version of the Y-toothpick structure of A160120. - David Applegate and Omar E. Pol, Jan 24 2016
LINKS
David Applegate, The movie version
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, pp. 6, 30.
FORMULA
a(n) = sum( i=0...n-1, A159913(i)) = sum(i=0..n-1, 2^A000120(i))*2-n
a(n) = n + (A160720(n) - 1)/2 = n + 2*(A266532(n) - 1)/3 = n + 2*A267700(n-1), n >= 1. - Omar E. Pol, Jan 25 2016
MATHEMATICA
Accumulate@ Table[2^(DigitCount[n, 2][[1]] + 1) - 1, {n, 0, 54}] (* Michael De Vlieger, Jan 25 2016 *)
PROG
(PARI) A159912(n)=sum(i=0, n-1, 1<<norml2(binary(i)))*2-n
CROSSREFS
Sequence in context: A310906 A310907 A310908 * A183060 A310909 A310910
KEYWORD
nonn
AUTHOR
M. F. Hasler, May 03 2009
STATUS
approved

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Last modified March 28 05:02 EDT 2024. Contains 371235 sequences. (Running on oeis4.)