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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.

(End)

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 whose digits in base 2 are in nonincreasing order.

Might be called "nialpdromes".

Subset of A077436. Proof: Since a(n) is of the form (2^i-1)*2^j, i,j >= 0, a(n)^2 = (2^(2i) - 2^(i+1))*2^(2j) + 2^(2j) where the first sum term has i-1 one bits and its 2j-th bit is zero, while the second sum term switches the 2j-th bit to one, giving i one bits, as in a(n). - Ralf Stephan, Mar 08 2004

Numbers whose binary representation contains no "01". - Benoit Cloitre, May 23 2004

Every polynomial with coefficients equal to 1 for the leading terms and 0 after that, evaluated at 2. For instance a(13) = x^4 + x^3 + x^2 at 2, a(14) = x^4 + x^3 + x^2 + x at 2. - Ben Paul Thurston, Jan 11 2008

From Gary W. Adamson, Jul 18 2008: (Start)

As a triangle by rows starting:

1;

2, 3;

4, 6, 7;

8, 12, 14, 15;

16, 24, 28, 30, 31;

...,

equals A000012 * A130123 * A000012, where A130123 = (1, 0,2; 0,0,4; 0,0,0,8; ...). Row sums of this triangle = A000337 starting (1, 5, 17, 49, 129, ...). (End)

First differences are A057728 = 1; 1; 1; 1; 2,1; 1; 4,2,1; 1; 8,4,2,1; 1; ... i.e., decreasing powers of 2, separated by another "1". - M. F. Hasler, May 06 2009

Apart from first term, numbers that are powers of 2 or the sum of some consecutive powers of 2. - Omar E. Pol, Feb 14 2013

From Andres Cicuttin, Apr 29 2016: (Start)

Numbers that can be digitally generated with twisted ring (Johnson) counters. This is, the binary digits of a(n) correspond to those stored in a shift register where the input bit of the first bit storage element is the inverted output of the last storage element. After starting with all 0’s, each new state is obtained by rotating the stored bits but inverting at each state transition the last bit that goes to the first position (see link).

Examples: for a(n) represented by three bits

Binary

a(5)= 4 -> 100 last bit = 0

a(6)= 6 -> 110 first bit = 1 (inverted last bit of previous number)

a(7)= 7 -> 111

and for a(n) represented by four bits

Binary

a(8) = 8 -> 1000

a(9) = 12 -> 1100 last bit = 0

a(10)= 14 -> 1110 first bit = 1 (inverted last bit of previous number)

a(11)= 15 -> 1111

(End)

Powers of 2 represented in bases which are terms of this sequence must always contain at least one digit which is also a power of 2. This is because 2^i mod (2^i - 2^j) = 2^j, which means the last digit always cycles through powers of 2 (or if i=j+1 then the first digit is a power of 2 and the rest are trailing zeros). The only known non-member of this sequence with this property is 5. - Ely Golden, Sep 05 2017

Numbers k such that k = 2^(1 + A000523(k)) - 2^A007814(k). - Daniel Starodubtsev, Aug 05 2021

A002260(n) = v(a(n)/2^v(a(n))+1) and A002024(n) = A002260(n) + v(a(n)) where v is the dyadic valuation (i.e., A007814). - Lorenzo Sauras Altuzarra, Feb 01 2023

Let f(n) = A000265(n) be the odd part of n. Let p be the largest prime factor of k, and say k = p * m. Suppose that k is not a power of 2, i.e., p > 2, then f(k) = p * f(m). The iteration is k -> k + k/p = p*m + m = (p+1) * m. So, p * f(m) -> f(p+1) * f(m). Since for p > 2, f(p+1) < p, the odd part in each iteration decreases, until it becomes 1, i.e., until we reach a power of 2. - Amiram Eldar, Feb 19 2020

Any odd prime factor of k can be used at any step of the iteration, and the result will be same. Thus, like A329697, this is also fully additive sequence. - Antti Karttunen, Apr 29 2020

If and only if a(n) is equal to A005087(n), then sigma(2n) - sigma(n) is a power of 2. (See A336923, A046528). - Antti Karttunen, Mar 16 2021

Fixed points of the Doudna sequence: A005940(a(n)) = A005941(a(n)) = a(n). - Reinhard Zumkeller, Aug 23 2006

Subsequence of A103969. - R. J. Mathar, Mar 06 2010

Question: Is there a simple proof that A005940(c) = c would never allow an odd composite c as a solution? See also my comments in A163511 and in A335431 concerning similar problems, also A364551 and A364576. - Antti Karttunen, Jul 28 & Aug 11 2023

2007. For example, 37 = 18 + 19; 48 = 15 + 16 + 17; 56 = 5 + 6 + 7 + 8 + 9 + 10 + 11. (Edited by M. F. Hasler, Aug 29 2020: "positive" was missing here. If nonnegative integers are allowed, none of the triangular numbers 3, 6, 10, ... would be in the corresponding sequence. If negative integers are also allowed, it would only have powers of 2 (A000079) which are the only positive integers not the sum of more than one consecutive positive integers, since any x > 0 is the sum of 1-x through x.)

Numbers that are the difference of two triangular numbers in exactly two ways.

Numbers with largest odd divisor a prime number. - Juri-Stepan Gerasimov, Aug 16 2016

Numbers k for which A001222(A000265(k)) = 1. - Antti Karttunen, Jul 09 2020

Any Mersenne prime (A000668) times any power of 2, i.e., sequence A335431, is fixed by this map (note the indexing), including also all even perfect numbers. It is not currently known whether there are any additional fixed points.

Because a(n) has the same prime signature as A163511(n), it implies that applying A046523 and A052409 to this sequence gives the same results as with A163511, namely, sequences A278531 and A365805. - Antti Karttunen, Oct 09 2023

Any Mersenne prime (A000668) times any power of 2 (i.e., 2^k, for k>=0) is fixed by this sequence, including also all even perfect numbers.

From Antti Karttunen, Jul 10 2020: (Start)

This is a "tuned variant" of A243071, and has many of the same properties.

For example, for n > 1, A007814(a(n)) = A007814(n) - A209229(n), that is, this map preserves the 2-adic valuation of n, except when n is a power of two, in which cases that value is decremented by one, and in particular, a(2^k * n) = 2^k * a(n) for all n > 1. Also, like A243071, this bijection maps primes to the terms of A000225 (binary repunits). However, the "tuning" (A332213) has a specific effect that each Mersenne prime (A000668) is mapped to itself. Therefore the terms of A335431 are fixed by this map. Furthermore, I conjecture that there are no other fixed points. For the starters, see the proof in A335879, which shows that at least none of the terms of A335882 are fixed.

(End)

Numbers k such that A000265(k) is either in A144482 or in A335874.

Each term is either of the form A335874(n)*2^k, for some n >= 2, and k >= 0, or a product of two terms of A335431, whether distinct or not.

Array is read by descending antidiagonals with (n,k) = (0,0), (0,1), (1,0), (0,2), (1,1), (2,0), ... where A(n,k) is the (k+1)-th solution x to A331410(x) = n. The row indexing (n) starts from 0, and column indexing (k) also from 0.

For any odd prime p that appears on row n, p+1 appears on row n-1.

The e-th powers of the terms on row n form a subset of terms on row (e*n). More generally, a product of terms that occur on rows i_1, i_2, ..., i_k can be found at row (i_1 + i_2 + ... + i_k), because A331410 is completely additive.