A026142 -id:A026142 - cp's OEIS frontend

A026136 Lexicographically earliest permutation of the positive integers such that |a(n)-n| = [a(n)/2].

1, 3, 2, 7, 9, 4, 5, 15, 6, 19, 21, 8, 25, 27, 10, 11, 33, 12, 13, 39, 14, 43, 45, 16, 17, 51, 18, 55, 57, 20, 61, 63, 22, 23, 69, 24, 73, 75, 26, 79, 81, 28, 29, 87, 30, 31, 93, 32, 97, 99, 34, 35, 105, 36, 37, 111, 38, 115, 117, 40, 41, 123

Offset: 1

Clark Kimberling

nonn

Old name was: For n >= 2, let L=n-[ n/2 ], R=n+[ n/2 ]; then a(L)=n if a(L) not yet defined, else a(R)=n; thus |a(n)-n|=[ (1/2)*a(n) ].
Also can be defined as follows. For n >= 2, let h=[ n/2 ], L=n-h, R=n+h; then a(R)=n if n even or a(L) already defined, else a(L)=n. For a proof that the two definitions are the same, see my paper "Permutations of N generated by left-right filling algorithms". - Michel Dekking, Jan 30 2020
From Peter Munn, Dec 30 2021: (Start)
A value m occurs at an index n, n < m if and only if m has the form 3^i*(6k+2)+1.
Proof:
Values of the form 2k occur at index 2k + [2k/2] = 3k and not at index 2k - [2k/2] = k, because a(k) can take the value 2k-1 and 2k-1 cannot occur earlier.
So, values of the form 6k+5 occur at index 6k+5 + [(6k+5)/2] = 9k+7, and not at index 6k+5 - [(6k+5)/2] = 3k+3 because a(3k+3) takes the value 2k+2.
Values of the form 6k+3 occur at index 6k+3 - [(6k+3)/2] = 3k+2, because numbers of the form 3k+2 do not have the form m+[m/2] for any m > 0.
A value of the form 6k+1 occurs at index 6k+1 - [(6k+1)/2] = 3k+1 if and only if 2k+1 occurs at index k+1 rather than occupying index 3k+1.
From the characterization above of cases 6k+5, 6k+3 and 6k+1 we see the following: an odd number 2j+1 > 2 occurs before or after position 2j+1 depending on the base 3 representation of j with its trailing zeros removed. (With respect to the statement being proved j = 3^i*(3k+1).)
(End)

Sean A. Irvine, 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.
Sean A. Irvine, Java program (github)
Index entries for sequences that are permutations of the natural numbers

Cf. A026142, A026224.

Block[{a, nn = 123}, a[1] = 1; Do[If[! IntegerQ[a[#1]], Set[a[#1], i], Set[a[#2], i]] & @@ {i - #, i + #} &@ Floor[i/2], {i, nn}]; TakeWhile[Array[a[#] &, nn], IntegerQ]] (* Michael De Vlieger, Apr 16 2020 *)

seq(n)={my(a=vector(n)); a[1]=1; for(i=2, 2*n, my(h=i\2); if(!a[i-h], a[i-h]=i, if(i+h<=n, a[i+h]=i))); a} \\ Andrew Howroyd, Oct 15 2019

import math
A026136 ={1:1}
for n in range(2,3000):
    h=math.floor(n/2)
    L=n-h
    R=n+h
    if not L in A026136 :
        A026136[L]=n
    else :
        A026136[R]=n
for n in range(1,2000):
    if n in A026136:
       print(str(n) + " "+ str(A026136[n]))
    else:
       break # R. J. Mathar, Aug 26 2019

Edited by N. J. A. Sloane, Jan 31 2020

New name from Peter Munn, Dec 30 2021

A026177 For n >= 2, let h=floor(n/2), L=n-h, R=n+h; then a(R)=n if n odd or a(L) already defined, otherwise a(L)=n.

Original entry on oeis.org

1, 4, 2, 3, 10, 12, 5, 16, 6, 7, 22, 8, 9, 28, 30, 11, 34, 36, 13, 40, 14, 15, 46, 48, 17, 52, 18, 19, 58, 20, 21, 64, 66, 23, 70, 24, 25, 76, 26, 27, 82, 84, 29, 88, 90, 31, 94, 32, 33, 100, 102, 35, 106, 108, 37, 112, 38, 39, 118, 120, 41

Offset: 1

Clark Kimberling

nonn

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)

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.

Cf. A026178 (inverse), A026179 (indices of new highs), A026180 (values of new highs), A026136, A026142.

Array[If[Mod[#/3^IntegerExponent[#, 3], 3] == 1, Ceiling[2 #/3], 2 #] &, 61] (* Michael De Vlieger, Feb 08 2020 *)

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

a(n) = if((n/3^valuation(n,3))%3==1, ceil(2*n/3), 2*n); \\ Kevin Ryde, Feb 06 2020

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)

Edited by N. J. A. Sloane, Feb 05 2020

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A026136 Lexicographically earliest permutation of the positive integers such that |a(n)-n| = [a(n)/2].

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A026177 For n >= 2, let h=floor(n/2), L=n-h, R=n+h; then a(R)=n if n odd or a(L) already defined, otherwise a(L)=n.

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