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URL: https://oeis.org/A151920

⇱ A151920 - OEIS

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A151920
a(n) = (Sum_{i=1..n+1} 3^wt(i))/3, where wt() = A000120().
15
1, 2, 5, 6, 9, 12, 21, 22, 25, 28, 37, 40, 49, 58, 85, 86, 89, 92, 101, 104, 113, 122, 149, 152, 161, 170, 197, 206, 233, 260, 341, 342, 345, 348, 357, 360, 369, 378, 405, 408, 417, 426, 453, 462, 489, 516, 597, 600, 609, 618, 645, 654, 681, 708, 789, 798, 825, 852, 933, 960
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
FORMULA
a(n) = (A147562(n+2) - 1)/4 = (A151917(n+2) - 1)/2. - Omar E. Pol, Mar 13 2011
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
KEYWORD
nonn,easy,look
AUTHOR
N. J. A. Sloane, Aug 05 2009, Aug 06 2009
STATUS
approved