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Term a(n) essentially records the run lengths of numbers of form 4k+3 encountered when starting from that node in binary tree A005940 which contains n, and by then traversing towards the root by iterating the map n -> A252463(n). The actual run lengths can be read from the exponents of primes in the prime factorization of A278222(A292383(m)), where m = min_{k=1..n} for which a(k) = a(n). In compound filter A292584 this is combined with similar information about the run lengths of the numbers of the form 4k+1 (A292585).

From Antti Karttunen, Sep 25 2017: (Start)

For all i, j: a(i) = a(j) => A053866(i) = A053866(j).

This follows from the interpretation of A053866 (A093709) as the characteristic function of squares and twice-squares. In binary tree A005940 each number is "born" by repeated applications of two functions: when we descend leftward we apply A003961, which shifts all prime factors of n one step towards larger primes. On the other hand, when we descend rightward the terms grow by doubling: n -> 2n (A005843). No square is ever of the form 4k+3, and for any square x, A003961(x) is also a square. Multiplying a square by 2 gives twice a square, and then multiplying by 2 again gives 4*square, which is also a square. In general, applying an even number of doubling steps in succession keeps a square as a square, while an odd number of doubling steps gives twice a square. Applying A003961 to any 2*square gives 3*(some square) which is always of the form 4k+3. Moreover, after any such "wrong turn" in A005940-tree no square nor twice a square can ever be encountered under any of the further descendants, because with this process it is impossible to find a pair for the lone prime factor now present. On the other hand, when turning left from any (2^2k)*s (where s is a square), one again gets a square of the form (3^2k)*A003961(s). All this implies that there are no numbers of the form 4k+3 in any trajectory leading to a square or twice a square in A005940-tree, while all trajectories to any other kind of number contain at least one number of the form 4k+3. Because each a(n) in this sequence contains enough information to count the 4k+3 numbers encountered on a A005940-trajectory to n (being 1 iff there are none), this filter matches A053866.

(End)

Because A292383(n) = a(A243071(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are of the form 4k+3 in binary tree A005940 on that trajectory which leads from the root of the tree to the node containing A163511(n). This works because A243071(n) = A054429(A156552(n)), a bit-flipped variant of Leonid Broukhis's unary-binary encoded compressed factorization of natural numbers, A156552(n) being an inverse of Doudna map f(n) = A005940(1+n).

This is a permutation of the positive integers.

From Antti Karttunen, Jun 20 2014: (Start)

Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.

Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.

Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).

(End)

From Antti Karttunen, Dec 30 2017: (Start)

This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:

1

|

...................2...................

4 3

8......../ \........9 6......../ \........5

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

16 27 18 25 12 15 10 7

32 81 54 125 36 75 50 49 24 45 30 35 20 21 14 11

etc.

Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.

A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.

(End)

Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020

From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)

This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).

Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?

Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)

From Antti Karttunen, Jul 25 2023: (Start)

(In the above question, it is assumed that the starting offset would be 1 instead of 0).

Questions:

Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?

It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.

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If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023

Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.

See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

Numbers n such that sum of divisors of n (A000203) is odd.

Also the numbers with an odd number of run sums (trapezoidal arrangements, number of ways of being written as the difference of two triangular numbers). - Ron Knott, Jan 27 2003

Pell(n)*Sum_{k|n} 1/Pell(k) is odd, where Pell(n) is A000129(n). - Paul Barry, Oct 12 2005

Number of odd divisors of n (A001227) is odd. - Vladeta Jovovic, Aug 28 2007

A071324(a(n)) is odd. - Reinhard Zumkeller, Jul 03 2008

Sigma(a(n)) = A000203(a(n)) = A152677(n). - Jaroslav Krizek, Oct 06 2009

Numbers n such that sum of odd divisors of n (A000593) is odd. - Omar E. Pol, Jul 05 2016

A187793(a(n)) is odd. - Timothy L. Tiffin, Jul 18 2016

If k is odd (k = 2m+1 for m >= 0), then 2^k = 2^(2m+1) = 2*(2^m)^2. If k is even (k = 2m for m >= 0), then 2^k = 2^(2m) = (2^m)^2. So, the powers of 2 sequence (A000079) is a subsequence of this one. - Timothy L. Tiffin, Jul 18 2016

Numbers n such that A175317(n) = Sum_{d|n} pod(d) is odd, where pod(m) = the product of divisors of m (A007955). - Jaroslav Krizek, Dec 28 2016

Positions of zeros in A292377 and A292383, positions of ones in A286357 and A292583. (See A292583 for why.) - Antti Karttunen, Sep 25 2017

Numbers of the form A000079(i)*A016754(j), i,j>=0. - R. J. Mathar, May 30 2020

Equivalently, numbers whose odd part is square. Cf. A042968. - Peter Munn, Jul 14 2020

These are the Heinz numbers of the partitions counted by A119620. - Gus Wiseman, Oct 29 2021

Numbers m whose abundance, A033880(m), is odd. - Peter Munn, May 23 2022

Numbers with an odd number of middle divisors (cf. A067742). - Omar E. Pol, Aug 02 2022

Note the indexing: the domain starts from 1, while the range includes also zero.

See also the comments at A163511, which is the inverse permutation to this one.

Variant of A292381. Here the most significant 1-bit is at the one step smaller position.

For numbers > 1, iterate the map x -> A252463(x) which divides even numbers by 2 and shifts every prime in the prime factorization of odd n one index step towards smaller primes. a(n) counts the numbers of the form 4k+3 encountered until 1 has been reached. The count includes also n itself if it is of the form 4k+3 (A004767).

In other words, locate the node which contains n in binary tree A005940 and traverse from that node towards the root, counting all numbers of the form 4k+3 that occur on the path.