A061392 a(n) = a(floor(n/3)) + a(ceiling(n/3)) with a(0) = 0 and a(1) = 1.
0, 1, 1, 2, 2, 2, 2, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 10, 10, 10, 10, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 14, 14, 14, 14, 15, 15, 16, 16, 16, 16, 16, 16, 16
Offset: 0
Links
- Rémy Sigrist, Table of n, a(n) for n = 0..6561
- Sam Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.
- Wikipedia, Cantor function.
Crossrefs
Programs
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Maple
A061392 := proc(n) option remember; local a : if n <=1 then n else A061392(floor(n/3)) + A061392(ceil(n/3)) fi: end: seq(A061392(n), n=0..87); # Johannes W. Meijer, Jun 05 2011
Formula
a(n+1) + A081609(n) = n+1. - Reinhard Zumkeller, Mar 23 2003; corrected by Henry Bottomley, Mar 24 2003
From Johannes W. Meijer, Jun 05 2011: (Start)
a(3*n+1) = a(n+1) + a(n), a(3*n+2) = a(n+1) + a(n) and a(3*n+3) = 2*a(n+1), for n>=1, with a(0)=0, a(1)=1, a(2)=1 and a(3)=2. [Northshield]
G.f.: x*Product_{n>=0} (1 + x^(3^n) + 2*x^(2*3^n) + x^(3*3^n) + x^(4*3^n)). [Northshield] (End)
Apparently, for any n >= 0 and k such that n < 3^k, a(n) = 2^k * c(n / 3^k) where c is the Cantor function. - Rémy Sigrist, Jul 12 2019
Comments