OFFSET
0,2
COMMENTS
Partial sums of A147610 (but with offset changed to 0).
It appears that the first bisection gives the positive terms of A147562. - Omar E. Pol, Mar 07 2015
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10000
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. 31.
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
FORMULA
a(n) = (A130665(n+1) - 1)/3. - Omar E. Pol, Mar 07 2015
a(n) = a(n-1) + 3^A000120(n+1))/3. - David A. Corneth, Mar 21 2015
EXAMPLE
n=3: (3^1+3^1+3^2+3^1)/3 = 18/3 = 6.
n=18: the binary expansion of 18+1 is 10011, i.e., 19 = 2^4 + 2^1 + 2^0.
The exponents of these powers of 2 (4, 1 and 0) reoccur as exponents in the powers of 4: a(19) = 3^0 * [(4^4 - 1) / 3 + 1] + 3^1 * [(4^1 - 1) / 3 + 1] + 3^2 * [(4^0 - 1)/3 + 1] = 1 * 86 + 3 * 2 + 9 * 1 = 101. - David A. Corneth, Mar 21 2015
MATHEMATICA
t = Nest[Join[#, # + 1] &, {0}, 14]; Table[Sum[3^t[[i + 1]], {i, 1, n}]/3, {n, 60}] (* Michael De Vlieger, Mar 21 2015 *)
PROG
(PARI) a(n) = sum(i=1, n+1, 3^hammingweight(i))/3; \\ Michel Marcus, Mar 07 2015
(PARI) a(n)={b=binary(n+1); t=#b; e=-1; sum(i=1, #b, e+=(b[i]==1); (b[i]==1)*3^e*((4^(#b-i)-1)/3+1))} \\ David A. Corneth, Mar 21 2015
CROSSREFS
KEYWORD
AUTHOR
N. J. A. Sloane, Aug 05 2009, Aug 06 2009
STATUS
approved