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Here the base-3 reverse has been adjusted so that the maximal suffix of trailing zeros (in base-3 representation A007089) stays where it is at the right side, and only the section from the most significant digit to the least significant nonzero digit is reversed, thus making this sequence a self-inverse permutation of nonnegative integers.

Because successive powers of 3 and 9 modulo 2, 4 and 8 are always either constant 1, 1, 1, ... or alternating 1, -1, 1, -1, ... it implies similar simple divisibility rules for 2, 4 and 8 in base 3 as e.g. 3, 9 and 11 have in decimal base (see the Wikipedia-link). As these rules do not depend on which direction they are applied from, it means that this bijection preserves the fact whether a number is divisible by 2, 4 or 8, or whether it is not. Thus natural numbers are divided to several subsets, each of which is closed with respect to this bijection. See the Crossrefs section for permutations obtained from these sections.

When polynomials over GF(3) are encoded as natural numbers (coefficients presented with the digits of the base-3 expansion of n), this bijection works as a multiplicative automorphism of the ring GF(3)[X]. This follows from the fact that as there are no carries involved, the multiplication (and thus also the division) of such polynomials could be as well performed by temporarily reversing all factors (like they were seen through mirror). This implies also that the sequences A207669 and A207670 are closed with respect to this bijection.

Note that a(A005820(4)) = A005820(4) and a(A005820(6)) = A005820(6), i.e., the fourth and sixth 3-perfect numbers, 459818240 and 51001180160 are among the fixed points of this sequence, precisely because they are also terms of A323653. As the former factorizes as 459818240 = 256 * 5 * 7 * 19 * 37 * 73, it must follow that a(256)/256 * a(5)/5 * a(7)/7 * a(19)/19 * a(37)/37 * a(73)/73 = 1, because ratio a(n)/n is multiplicative. See also comments in A348738.

The symbol <+> removes powers of three of the sum of the two operands.

The process of starting with 1, adding some constant number x = A001651(n-1) and reducing it iteratively with this operation defines a track 1, x<+>1, x<+>(x<+>1), ... which enters a cycle.

The period of this cycle specifies a(n).

Similar iterated reductions can be defined for power bases m other than 3.

a(n) is the greatest divisor of n coprime to 6 if n is not coprime to 6, otherwise a(n) is the greatest divisor of 5*n + 1 coprime to 6.

This is the '5x+1' map with the successive dividing steps removed. The 'Px+1' map with those steps removed: If x is divisible by any prime < P, then divide out those primes; otherwise multiply x by P, add 1, and then divide out the primes < P.

There is a conjecture which states that for any value of n > 0 there is a k such that a^{k}(n) = 1 or a^{k}(n) enters one of a finite number of periodic cycles, where a^{0}(n) = n and a^{k + 1}(n) = a(a^{k}(n)).

b(i,n,k) = (b(i-1,n,k) + b(i-2,n,k))/p^valuation(b(i-1,n,k) + b(i-2,n,k), p), i.e., b(i,n,k) is b(i-1,n,k) + b(i-2,n,k) with all factors of p removed, where p = 3 in this sequence. Therefore, b(i,n,k) is not divisible by 3 for i >= 3.

At least one of b(i,n,k) < b(i-1,n,k) + b(i-2,n,k) and b(i+1,n,k) < b(i-1,n,k) + b(i,n,k) is true for i >= 5.

It appears that all repetends have the form of (1, 1, 2) (the position of 2 possibly changed), multiplied by G = A038502(gcd(n,k)).

Conjecture: T(n,k) >= 0.

This conjecture can be supported by a heuristic argument: Using dynamic programming, we can compute that for p's in A000057, the probability that b(i-1,n,k) + b(i-2,n,k) is not divisible by p is (p-2)/p, and the probability that valuation(b(i-1,n,k) + b(i-2,n,k), p) = x (x >= 1) is 2*(p-1)/p^(x+1). Therefore, the mathematical expectation of a(i) is (a(i-1,n,k) + a(i-2,n,k))*(p-1)/(p+1), which is exactly the average of the earlier two terms when p = 3, and larger when p >= 5.

When n > 0 is written as k*2^j with k odd then k = A000265(n) and j = A007814(n), so: when n is written as k*2^j - 1 with k odd then k = A000265(n+1) and j = A007814(n+1), when n > 1 is written as k*2^j + 1 with k odd then k = A000265(n-1) and j = A007814(n-1).

Also denominator of 2^n/n (numerator is A075101(n)). - Reinhard Zumkeller, Sep 01 2002

Slope of line connecting (o, a(o)) where o = (2^k)(n-1) + 1 is 2^k and (by design) starts at (1, 1). - Josh Locker (joshlocker(AT)macfora.com), Apr 17 2004

Numerator of n/2^(n-1). - Alexander Adamchuk, Feb 11 2005

From Marco Matosic, Jun 29 2005: (Start)

"The sequence can be arranged in a table:

1

1 3 1

1 5 3 7 1

1 9 5 11 3 13 7 15 1

1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31 1

Every new row is the previous row interspaced with the continuation of the odd numbers.

Except for the ones; the terms (t) in each column are t+t+/-s = t_+1. Starting from the center column of threes and working to the left the values of s are given by A000265 and working to the right by A000265." (End)

This is a fractal sequence. The odd-numbered elements give the odd natural numbers. If these elements are removed, the original sequence is recovered. - Kerry Mitchell, Dec 07 2005

2k + 1 is the k-th and largest of the subsequence of k terms separating two successive equal entries in a(n). - Lekraj Beedassy, Dec 30 2005

It's not difficult to show that the sum of the first 2^n terms is (4^n + 2)/3. - Nick Hobson, Jan 14 2005

In the table, for each row, (sum of terms between 3 and 1) - (sum of terms between 1 and 3) = A020988. - Eric Desbiaux, May 27 2009

This sequence appears in the analysis of A160469 and A156769, which resemble the numerator and denominator of the Taylor series for tan(x). - Johannes W. Meijer, May 24 2009

Indices n such that a(n) divides 2^n - 1 are listed in A068563. - Max Alekseyev, Aug 25 2013

From Alexander R. Povolotsky, Dec 17 2014: (Start)

With regard to the tabular presentation described in the comment by Marco Matosic: in his drawing, starting with the 3rd row, the first term in the row, which is equal to 1 (or, alternatively the last term in the row, which is also equal to 1), is not in the actual sequence and is added to the drawing as a fictitious term (for the sake of symmetry); an actual A000265(n) could be considered to be a(j,k) (where j >= 1 is the row number and k>=1 is the column subscript), such that a(j,1) = 1:

1

1 3

1 5 3 7

1 9 5 11 3 13 7 15

1 17 9 19 5 21 11 23 3 25 13 27 7 29 15 31

and so on ... .

The relationship between k and j for each row is 1 <= k <= 2^(j-1). In this corrected tabular representation, Marco's notion that "every new row is the previous row interspaced with the continuation of the odd numbers" remains true. (End)

Partitions natural numbers to the same equivalence classes as A064989. That is, for all i, j: a(i) = a(j) <=> A064989(i) = A064989(j). There are dozens of other such sequences (like A003602) for which this also holds: In general, all sequences for which a(2n) = a(n) and the odd bisection is injective. - Antti Karttunen, Apr 15 2017

From Paul Curtz, Feb 19 2019: (Start)

This sequence is the truncated triangle:

1, 1;

3, 1, 5;

3, 7, 1, 9;

5, 11, 3, 13, 7;

15, 1, 17, 9, 19, 5;

21, 11, 23, 3, 25, 13, 27;

7, 29, 15, 31, 1, 33, 17, 35;

...

The first column is A069834. The second column is A213671. The main diagonal is A236999. The first upper diagonal is A125650 without 0.

c(n) = ((n*(n+1)/2))/A069834 = 1, 1, 2, 2, 1, 1, 4, 4, 1, 1, 2, 2, 1, 1, 8, 8, 1, 1, ... for n > 0. n*(n+1)/2 is the rank of A069834. (End)

As well as being multiplicative, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019

a(n) is also the map n -> A026741(n) applied at least A007814(n) times. - Federico Provvedi, Dec 14 2021

To construct the sequence: start with 1 and concatenate twice: 1,1,1 then tripling the last term gives: 1,1,3. Concatenating those 3 terms twice gives: 1,1,3,1,1,3,1,1,3, triple the last term -> 1,1,3,1,1,3,1,1,9. Concatenating those 9 terms twice gives: 1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9, triple the last term -> 1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,9,1,1,3,1,1,3,1,1,27 etc. - Benoit Cloitre, Dec 17 2002

Also 3-adic value of 1/n, n >= 1. See the Mahler reference, definition on p. 7. This is a non-archimedean valuation. See Mahler, p. 10. Sometimes also called 3-adic absolute value. - Wolfdieter Lang, Jun 28 2014

Number of elliptic points of order 3 for Gamma_0(n).

Equivalently, number of fixed points of Gamma_0(n) of type rho.

Values are 0 or a power of 2.

Shadow transform of central polygonal numbers A002061. - Michel Marcus, Jun 06 2013

Empirical: a(n) == A001615(n) (mod 3) for all natural numbers n. - John M. Campbell, Apr 01 2018

From Jianing Song, Jul 03 2018: (Start)

The comment above is true. Since both a(n) and A001615(n) are multiplicative we just have to verify that for prime powers. Note that A001615(p^e) = (p+1)*p^(e-1). For p == 1 (mod 3), p+1 == 2 (mod 3) so (p+1)*p^(e-1) == 2 (mod 3); for p == 2 (mod 3), p+1 is a multiple of 3 so (p+1)*p^(e-1) == 0 (mod 3). For p = 3, if e = 1 then p+1 == 1 (mod 3); if e > 1 then (p+1)*p^(e-1) == 0 (mod 3).

Equivalently, number of solutions to x^2 + x + 1 == 0 (mod n). (End)

Bennett, Filaseta, & Trifonov show that if n > 8 then a(n^2 + n) > n^0.285. - Charles R Greathouse IV, May 21 2014