cp's OEIS Frontend

Old name: For n >= 2, let h=[ (n+1)/2 ], L=n-h, R=n+h; a(L)=n if a(L) not yet defined, else a(R)=n; thus |a(n)-n| = [ (a(n)+1)/2 ].

From Peter Munn, Jan 07 2022: (Start)

A value m occurs at an index n, n < m if and only if m has the form 3^i*4 or 3^i*(6k+2), k >= 1.

Proof:

For k >= 1, values of the form 2k+1 occur at index 2k+1 + [(2k+1+1)/2] = 3k+2 and not at index 2k+1 - [(2k+1+1)/2] = k, because a(k) can take the value 2k and 2k cannot occur earlier.

So, for k >= 1, values of the form 6k+4 occur at index 6k+4 + [(6k+4+1)/2] = 9k+6, and not at index 6k+4 - [(6k+4+1)/2] = 3k+2 because a(3k+2) takes the value 2k+1.

For k >= 1, values of the form 6k+2 occur at index 6k+2 - [(6k+2+1)/2] = 3k+1, because numbers of the form 3k+1 do not have the form m+[(m+1)/2] for any m > 0. (Note that for k = 0, 1 occurs at index 1 due to an explicit exemption from the definition's constraining rule.)

A value of the form 6k occurs at index 6k - [(6k+1)/2] = 3k if and only if 2k occurs at index k rather than occupying index 3k.

From the characterization above of cases 6k+4, 6k+2 and 6k we see the following: an even number 2j > 4 occurs before or after position 2j depending on the base 3 representation of j with its trailing zeros removed. (With respect to the statement being proved j = 3^i*2 or 3^i*(3k+1).)

(End)

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)