cp's OEIS Frontend

For k > 1, if k is a term, then also 2*k is present in this sequence, and vice versa.

Sequence A163511(n^2), n >= 0, sorted into ascending order.

Contains no squares after the initial 1. See A365808 for a proof.

Question: Are there any more odd cubes or odd powers after 1, 27, 35937 ?

From Antti Karttunen, May 12 2014: (Start)

a(A003961(n)) = n for all n. [This is a left inverse function for the injection A003961.]

Bisections are A064216 (the terms at odd indices) and A064989 itself (the terms at even indices), i.e., a(2n) = a(n) for all n.

(End)

From Antti Karttunen, Dec 18-21 2014: (Start)

When n represents an unordered integer partition via the indices of primes present in its prime factorization (for n >= 2, n corresponds to the partition given as the n-th row of A112798) this operation subtracts one from each part. If n is of the form 2^k (a partition having just k 1's as its parts) the result is an empty partition (which is encoded by 1, having an "empty" factorization).

For all odd numbers n >= 3, a(n) tells which number is located immediately above n in square array A246278. Cf. also A246277.

(End)

Alternatively, if numbers are represented as the multiset of indices of prime factors with multiplicity, this operation subtracts 1 from each element and discards the 0's. - M. F. Hasler, Dec 29 2014

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

Same as Pisot sequences E(3, 6), L(3, 6), P(3, 6), T(3, 6). See A008776 for definitions of Pisot sequences.

Numbers k such that A006530(A000010(k)) = A000010(A006530(k)) = 2. - Labos Elemer, May 07 2002

Also least number m such that 2^n is the smallest proper divisor of m which is also a suffix of m in binary representation, see A080940. - Reinhard Zumkeller, Feb 25 2003

Length of the period of the sequence Fibonacci(k) (mod 2^(n+1)). - Benoit Cloitre, Mar 12 2003

The sequence of first differences is this sequence itself. - Alexandre Wajnberg and Eric Angelini, Sep 07 2005

Subsequence of A122132. - Reinhard Zumkeller, Aug 21 2006

Apart from the first term, a subsequence of A124509. - Reinhard Zumkeller, Nov 04 2006

Total number of Latin n-dimensional hypercubes (Latin polyhedra) of order 3. - Kenji Ohkuma (k-ookuma(AT)ipa.go.jp), Jan 10 2007

Number of different ternary hypercubes of dimension n. - Edwin Soedarmadji (edwin(AT)systems.caltech.edu), Dec 10 2005

For n >= 1, a(n) is equal to the number of functions f:{1, 2, ..., n + 1} -> {1, 2, 3} such that for fixed, different x_1, x_2,...,x_n in {1, 2, ..., n + 1} and fixed y_1, y_2,...,y_n in {1, 2, 3} we have f(x_i) <> y_i, (i = 1,2,...,n). - Milan Janjic, May 10 2007

a(n) written in base 2: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, n times 0 (see A003953). - Jaroslav Krizek, Aug 17 2009

Subsequence of A051916. - Reinhard Zumkeller, Mar 20 2010

Numbers containing the number 3 in their Collatz trajectories. - Reinhard Zumkeller, Feb 20 2012

a(n-1) gives the number of ternary numbers with n digits with no two adjacent digits in common; e.g., for n=3 we have 010, 012, 020, 021, 101, 102, 120, 121, 201, 202, 210 and 212. - Jon Perry, Oct 10 2012

If n > 1, then a(n) is a solution for the equation sigma(x) + phi(x) = 3x-4. This equation also has solutions 84, 3348, 1450092, ... which are not of the form 3*2^n. - Farideh Firoozbakht, Nov 30 2013

a(n) is the upper bound for the "X-ray number" of any convex body in E^(n + 2), conjectured by Bezdek and Zamfirescu, and proved in the plane E^2 (see the paper by Bezdek and Zamfirescu). - L. Edson Jeffery, Jan 11 2014

If T is a topology on a set V of size n and T is not the discrete topology, then T has at most 3 * 2^(n-2) many open sets. See Brown and Stephen references. - Ross La Haye, Jan 19 2014

Comment from Charles Fefferman, courtesy of Doron Zeilberger, Dec 02 2014: (Start)

Fix a dimension n. For a real-valued function f defined on a finite set E in R^n, let Norm(f, E) denote the inf of the C^2 norms of all functions F on R^n that agree with f on E. Then there exist constants k and C depending only on the dimension n such that Norm(f, E) <= C*max{ Norm(f, S) }, where the max is taken over all k-point subsets S in E. Moreover, the best possible k is 3 * 2^(n-1).

The analogous result, with the same k, holds when the C^2 norm is replaced, e.g., by the C^1, alpha norm (0 < alpha <= 1). However, the optimal analogous k, e.g., for the C^3 norm is unknown.

For the above results, see Y. Brudnyi and P. Shvartsman (1994). (End)

Also, coordination sequence for (infinity, infinity, infinity) tiling of hyperbolic plane. - N. J. A. Sloane, Dec 29 2015

The average of consecutive powers of 2 beginning with 2^1. - Melvin Peralta and Miriam Ong Ante, May 14 2016

For n > 1, a(n) is the smallest Zumkeller number with n divisors that are also Zumkeller numbers (A083207). - Ivan N. Ianakiev, Dec 09 2016

Also, for n >= 2, the number of length-n strings over the alphabet {0,1,2,3} having only the single letters as nonempty palindromic subwords. (Corollary 21 in Fleischer and Shallit) - Jeffrey Shallit, Dec 02 2019

Also, a(n) is the minimum link-length of any covering trail, circuit, path, and cycle for the set of the 2^(n+2) vertices of an (n+2)-dimensional hypercube. - Marco Ripà, Aug 22 2022

The known fixed points of maps n -> A163511(n) and n -> A243071(n). [See comments in A163511]. - Antti Karttunen, Sep 06 2023

The finite subsequence a(3), a(4), a(5), a(6) = 24, 48, 96, 192 is one of only two geometric sequences that can be formed with all interior angles (all integer, in degrees) of a simple polygon. The other sequence is a subsequence of A000244 (see comment there). - Felix Huber, Feb 15 2024

A level 1 Sierpiński triangle is a triangle. Level n+1 is formed from three copies of level n by identifying pairs of corner vertices of each pair of triangles. For n>2, a(n-3) is the radius of the level n Sierpiński triangle graph. - Allan Bickle, Aug 03 2024

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

Inverse of sequence A064216 considered as a permutation of the positive integers. - Howard A. Landman, Sep 25 2001

From Antti Karttunen, Dec 20 2014: (Start)

Permutation of natural numbers obtained by replacing each prime divisor of n with the next prime and mapping the generated odd numbers back to all natural numbers by adding one and then halving.

Note: there is a 7-cycle almost right in the beginning: (6 8 14 17 10 11 7). (See also comments at A249821. This 7-cycle is endlessly copied in permutations like A250249/A250250.)

The only 3-cycle in range 1 .. 402653184 is (2821 3460 5639).

For 1- and 2-cycles, see A245449.

(End)

The first 5-cycle is (1410, 2783, 2451, 2703, 2803). - Robert Israel, Jan 15 2015

From Michel Marcus, Aug 09 2020: (Start)

(5194, 5356, 6149, 8186, 10709), (46048, 51339, 87915, 102673, 137205) and (175811, 200924, 226175, 246397, 267838) are other 5-cycles.

(10242, 20479, 21413, 29245, 30275, 40354, 48241) is another 7-cycle. (End)

From Antti Karttunen, Feb 10 2021: (Start)

Somewhat artificially, also this permutation can be represented as a binary tree. Each child to the left is obtained by multiplying the parent by 3 and subtracting one, while each child to the right is obtained by applying A253888 to the parent:

1

|

................../ \..................

2 3

5......../ \........4 8......../ \........6

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

14 13 11 7 23 9 17 18

41 10 38 12 32 28 20 15 68 25 26 63 50 16 53 19

etc.

Each node's (> 1) parent can be obtained with A253889. Sequences A292243, A292244, A292245 and A292246 are constructed from the residues (mod 3) of the vertices encountered on the path from n to the root (1).

(End)

a((A003961(n) + 1) / 2) = n and A003961(a(n)) = 2*n - 1 for all n. If the sequence is indexed by odd numbers only, it becomes multiplicative. In this variant sequence, denoted b, even indices don't exist, and we get b(1) = a(1) = 1, b(3) = a(2) = 2, b(5) = 3, b(7) = 5, b(9) = 4 = b(3) * b(3), ... , b(15) = 6 = b(3) * b(5), and so on. This property can also be stated as: a(x) * a(y) = a(((2x - 1) * (2y - 1) + 1) / 2) for x, y > 0. - Reinhard Zumkeller [re-expressed by Peter Munn, May 23 2020]

Not multiplicative in usual sense - but letting m=2n-1=product_j (p_j)^(e_j) then a(n)=a((m+1)/2)=product_j (p_(j-1))^(e_j). - Henry Bottomley, Apr 15 2005

From Antti Karttunen, Jul 25 2016: (Start)

Several permutations that use prime shift operation A064989 in their definition yield a permutation obtained from their odd bisection when composed with this permutation from the right. For example, we have:

A243505(n) = A122111(a(n)).

A243065(n) = A241909(a(n)).

A244153(n) = A156552(a(n)).

A245611(n) = A243071(a(n)).

(End)