OFFSET
1,2
COMMENTS
Also a(n) = (1/5)*s(n), where s(n) is the n-th multiple of 5 in A026142.
Inverse is A026178.
From Kevin Ryde, Feb 06 2020: (Start)
Dekking's cases III and IV can be combined as ceiling(2n/3). Theorem 8 determines the case for n by discarding low ternary 0 digits until reaching the lowest ternary non-0 digit of n, LNZ(n) = A060236(n), and hence the formula below for when the bigger a(n) = 2n or smaller a(n) = ceiling(2n/3).
For c odd and LNZ(c)=1, so c = (6j+1)*3^k, this sequence has a self-similarity in that taking the values which are multiples of c, and dividing them by c, gives the full sequence again. (Multiples of 3 divided by 3 this way had been the definition of A026216.) Using the inverse A026178, a(n)=c*m is located at n = A026178(c*m) = c*A026178(m) + (floor(c/2) if m odd) since c*m goes to the same bigger or smaller case in A026178 as m does. Then floor(c/2) < c so values c*m are in the same order as all values m.
For c even and LNZ(c)=1, so c = (6j+4)*3^k = A026180 except initial 1, this sequence has an inverse self-similarity in that taking the values which are multiples of c, and dividing them by c, gives the inverse sequence A026178. c*m is located here at A026178(c*m) and conversely m in A026178 is located at a(m). These locations are related by an identity 4*A026178(c*m) = 3*c*a(m) - (c if m==1 (mod 3)) since c*m is even so goes to the big or small cases in A026178 according to LNZ, the same as here. The cases here and there differ by factor 3/4. So values c*m here are in the same order as all values m in A026178.
For c even and LNZ(c)=2, so c = (6j+2)*3^k = 2*A026225, taking the values which are multiples of c, and dividing them by c, gives A026214. A026214 is defined as the multiples of 2 divided by 2, i.e., c=2, and other c of this form are the same. The locations of c*m and 2*m here are A026178(c*m) = (c/2)*A026178(2*m) since c*m has the same effect as 2*m on the big or small cases in A026178, and so values c*m here are in the same order as values 2*m.
For c odd and LNZ(c)=2, so c = (6j+5)*3^k, taking the values which are multiples of c, and dividing them by c, gives A026215. (Multiples of 5 divided by 5 this way had been the definition of A026220.) Using the formulas in their respective inverses, the location of c*m here and m in A026215 are related by A026178(c*m) = c*A026214(m) - (ceiling(c/2) if m odd). This is since LNZ(c)=2 in c*m flips the sense of the LNZ test in A026178 so it corresponds to A026214. Then ceiling(c/2) < c so values c*m here are in the same order as all values of A026215.
(End)
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..10000
F. M. Dekking, Permutations of N generated by left-right filling algorithms, arXiv:2001.08915 [math.CO], 2020.
FORMULA
From Kevin Ryde, Feb 06 2020: (Start)
a(n) = ceiling(2n/3) if A060236(n)=1, otherwise a(n) = 2n, where A060236(n) is the lowest non-0 ternary digit of n.
a(n) = ceiling(2n / 3^A137893(n)).
a(3n) = 3*a(n) - (1 if n==1 (mod 3)).
(End)
MATHEMATICA
Array[If[Mod[#/3^IntegerExponent[#, 3], 3] == 1, Ceiling[2 #/3], 2 #] &, 61] (* Michael De Vlieger, Feb 08 2020 *)
PROG
(PARI) seq(n)={my(a=vector(n)); a[1]=1; for(i=2, 2*n, my(h=i\2); if(i%2==0&&!a[i-h], a[i-h]=i, if(i+h<=n, a[i+h]=i))); a} \\ Andrew Howroyd, Oct 15 2019
(PARI) a(n) = if((n/3^valuation(n, 3))%3==1, ceil(2*n/3), 2*n); \\ Kevin Ryde, Feb 06 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
EXTENSIONS
Edited by N. J. A. Sloane, Feb 05 2020
STATUS
approved