cp's OEIS Frontend

Question: Is it possible for a(n)/A331733(n) to be an integer when n is a square > 1? This is equivalent to the question whether there are odd Harmonic numbers (A001599) larger than one.

Squares s for which A331733(s) is two times an odd number, i.e., squares s such that A007814(A331733(s)) == 1.

For each term k present, A006519(k) = 2^(2^e), with A000040(1+e) == 1 (mod 4). See A191218, A228058.

Equal to the sequence A225546(A191218(n)), for n >= 1, when its elements are sorted into ascending order.

Term a(16) has 309 digits (1025 bits).

This is not multiplicative. Even though a(2*3) = a(2)*a(3), a(2*7) = a(2)*a(7), and a(3*5) = a(3)*a(5), we still have a(2*5) = 18 <> a(2)*a(5) = 4*8 = 32.

Multiplicative with a(p^e) = p.

Product of the distinct prime factors of n.

a(k)=k for k=squarefree numbers A005117. - Lekraj Beedassy, Sep 05 2006

A note on square roots of numbers: we can write sqrt(n) = b*sqrt(c) where c is squarefree. Then b = A000188(n) is the "inner square root" of n, c = A007913(n), b*c = A019554(n) = "outer square root" of n, and a(n) = lcm(a(b),c). Unless n is biquadrateful (A046101), a(n) = lcm(b,c). [Edited by Jeppe Stig Nielsen, Oct 10 2021, and Andrey Zabolotskiy, Feb 12 2025]

a(n) = A128651(A129132(n-1) + 2) for n > 1. - Reinhard Zumkeller, Mar 30 2007

Also the least common multiple of the prime factors of n. - Peter Luschny, Mar 22 2011

The Mobius transform of the sequence generates the sequence of absolute values of A097945. - R. J. Mathar, Apr 04 2011

Appears to be the period length of k^n mod n. For example, n^12 mod 12 has period 6, repeating 1,4,9,4,1,0, so a(12)= 6. - Gary Detlefs, Apr 14 2013

a(n) differs from A014963(n) when n is a term of A024619. - Eric Desbiaux, Mar 24 2014

a(n) is also the smallest base (also termed radix) for which the representation of 1/n is of finite length. For example a(12) = 6 and 1/12 in base 6 is 0.03, which is of finite length. - Lee A. Newberg, Jul 27 2016

a(n) is also the divisor k of n such that d(k) = 2^omega(n). a(n) is also the smallest divisor u of n such that n divides u^n. - Juri-Stepan Gerasimov, Apr 06 2017

Meyers (see Guy reference) conjectures that for all r >= 1, the least odd number not in the set {a(i): i < prime(r)} is prime(r+1). - N. J. A. Sloane, Jan 08 2021

Meyers' conjecture would be refuted if and only if for some r there were such a large gap between prime(r) and prime(r+1) that there existed a composite c for which prime(r) < c < a(c) < prime(r+1), in which case (by Bertrand's postulate) c would necessarily be a term of A246281. - Antti Karttunen, Mar 29 2021

a(n) is odd for all n and for each odd m there exists a k with a(k) = m (see A064216). a(n) > n for n > 1: bijection between the odd and all numbers. - Reinhard Zumkeller, Sep 26 2001

a(n) and n have the same number of distinct primes with (A001222) and without multiplicity (A001221). - Michel Marcus, Jun 13 2014

From Antti Karttunen, Nov 01 2019: (Start)

More generally, a(n) has the same prime signature as n, A046523(a(n)) = A046523(n). Also A246277(a(n)) = A246277(n) and A287170(a(n)) = A287170(n).

Many permutations and other sequences that employ prime factorization of n to encode either polynomials, partitions (via Heinz numbers) or multisets in general can be easily defined by using this sequence as one of their constituent functions. See the last line in the Crossrefs section for examples.

(End)

A permutation of the natural numbers. - Robert G. Wilson v, Feb 22 2005

Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006

The even bisection, when halved, gives the sequence back. - Antti Karttunen, Jun 28 2014

From Antti Karttunen, Dec 21 2014: (Start)

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

1

|

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

3 4

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

7 10 15 12 25 18 27 16

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

etc.

Sequence A163511 is obtained by scanning the same tree level by level, from right to left. 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.

A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 2 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is larger than the right child, A252750 the difference between left and right child for each node from node 2 onward.

(End)

-A008836(a(1+n)) gives the corresponding numerator for A323505(n). - Antti Karttunen, Jan 19 2019

(a(2n+1)-1)/2 [= A244154(n)-1, for n >= 0] is a permutation of the natural numbers. - George Beck and Antti Karttunen, Dec 08 2019

From Peter Munn, Oct 04 2020: (Start)

Each term has the same even part (equivalently, the same 2-adic valuation) as its index.

Using the tree depicted in Antti Karttunen's 2014 comment:

Numbers are on the right branch (4 and descendants) if and only if divisible by the square of their largest prime factor (cf. A070003).

Numbers on the left branch, together with 2, are listed in A102750.

(End)

According to Kutz (1981), he learned of this sequence from American mathematician Byron Leon McAllister (1929-2017) who attributed the invention of the sequence to a graduate student by the name of Doudna (first name Paul?) in the mid-1950's at the University of Wisconsin. - Amiram Eldar, Jun 17 2021

From David James Sycamore, Sep 23 2022: (Start)

Alternative (recursive) definition: If n is a power of 2 then a(n)=n. Otherwise, if 2^j is the greatest power of 2 not exceeding n, and if k = n - 2^j, then a(n) is the least m*a(k) that has not occurred previously, where m is an odd prime.

Example: Use recursion with n = 77 = 2^6 + 13. a(13) = 25 and since 11 is the smallest odd prime m such that m*a(13) has not already occurred (see a(27), a(29),a(45)), then a(77) = 11*25 = 275. (End)

The odd bisection, when transformed by replacing all prime(k)^e in a(2*n - 1) with prime(k-1)^e, returns a(n), and thus gives the sequence back. - David James Sycamore, Sep 28 2022

A permutation of the squarefree numbers A005117. The missing positive numbers are in A013929. - Alois P. Heinz, Sep 06 2014

From Antti Karttunen, Apr 18 & 19 2017: (Start)

Because a(n) toggles the parity of n there are neither fixed points nor any cycles of odd length.

Conjecture: there are no finite cycles of any length. My grounds for this conjecture: any finite cycle in this sequence, if such cycles exist at all, must have at least one member that occurs somewhere in A285319, the terms that seem already to be quite rare. Moreover, any such a number n should satisfy in addition to A019565(n) < n also that A048675^{k}(n) is squarefree, not just for k=0, 1 but for all k >= 0. As there is on average a probability of only 6/(Pi^2) = 0.6079... that any further term encountered on the trajectory of A048675 is squarefree, the total chance that all of them would be squarefree (which is required from the elements of A019565-cycles) is soon minuscule, especially as A048675 is not very tightly bounded (many trajectories seem to skyrocket, at least initially). I am also assuming that usually there is no significant correlation between the binary expansions of n and A048675(n) (apart from their least significant bits), or, for that matter, between their prime factorizations.

See also the slightly stronger conjecture in A285320, which implies that there would neither be any two-way infinite cycles.

If either of the conjectures is false (there are cycles), then certainly neither sequence A285332 nor its inverse A285331 can be a permutation of natural numbers. (End)

The conjecture made in A087207 (see also A288569) implies the two conjectures mentioned above. A further constraint for cycles is that in any A019565-trajectory which starts from a squarefree number (A005117), every other term is of the form 4k+2, while every other term is of the form 6k+3. - Antti Karttunen, Jun 18 2017

The sequence satisfies the exponential function identity, a(x + y) = a(x) * a(y), whenever x and y do not have a 1-bit in the same position, i.e., when A004198(x,y) = 0. See also A283475. - Antti Karttunen, Oct 31 2019

The above identity becomes unconditional if binary exclusive OR, A003987(.,.), is substituted for addition, and A059897(.,.), a multiplicative equivalent of A003987, is substituted for multiplication. This gives us a(A003987(x,y)) = A059897(a(x), a(y)). - Peter Munn, Nov 18 2019

Also the Heinz number of the binary indices of n, where the Heinz number of a sequence (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and a number's binary indices (A048793) are the positions of 1's in its reversed binary expansion. - Gus Wiseman, Dec 28 2022

Least positive k such that m^k + 1 divides m^n + 1 (with fixed base m). - Vladimir Baltic, Mar 25 2002

To construct the sequence: start with 1, concatenate 1, 1 and double last term gives 1, 2. Concatenate those 2 terms, 1, 2, 1, 2 and double last term 1, 2, 1, 2 -> 1, 2, 1, 4. Concatenate those 4 terms: 1, 2, 1, 4, 1, 2, 1, 4 and double last term -> 1, 2, 1, 4, 1, 2, 1, 8, etc. - Benoit Cloitre, Dec 17 2002

a(n) = gcd(seq(binomial(2*n, 2*m+1)/2, m = 0 .. n - 1)) (odd numbered entries of even numbered rows of Pascal's triangle A007318 divided by 2), where gcd() denotes the greatest common divisor of a set of numbers. Due to the symmetry of the rows it suffices to consider m = 0 .. floor((n-1)/2). - Wolfdieter Lang, Jan 23 2004

Equals the continued fraction expansion of a constant x (cf. A100338) such that the continued fraction expansion of 2*x interleaves this sequence with 2's: contfrac(2*x) = [2; 1, 2, 2, 2, 1, 2, 4, 2, 1, 2, 2, 2, 1, 2, 8, 2, ...].

Simon Plouffe observes that this sequence and A003484 (Radon function) are very similar, the difference being all zeros except for every 16th term (see A101119 for nonzero differences). Dec 02 2004

This sequence arises when calculating the next odd number in a Collatz sequence: Next(x) = (3*x + 1) / A006519, or simply (3*x + 1) / BitAnd(3*x + 1, -3*x - 1). - Jim Caprioli, Feb 04 2005

a(n) = n if and only if n = 2^k. This sequence can be obtained by taking a(2^n) = 2^n in place of a(2^n) = n and using the same sequence building approach as in A001511. - Amarnath Murthy, Jul 08 2005

Also smallest m such that m + n - 1 = m XOR (n - 1); A086799(n) = a(n) + n - 1. - Reinhard Zumkeller, Feb 02 2007

Number of 1's between successive 0's in A159689. - Philippe Deléham, Apr 22 2009

Least number k such that all coefficients of k*E(n, x), the n-th Euler polynomial, are integers (cf. A144845). - Peter Luschny, Nov 13 2009

In the binary expansion of n, delete everything left of the rightmost 1 bit. - Ralf Stephan, Aug 22 2013

The equivalent sequence for partitions is A194446. - Omar E. Pol, Aug 22 2013

Also the 2-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 called 2-adic absolute value of 1/n. - Wolfdieter Lang, Jun 28 2014

First 2^(k-1) - 1 terms are also the heights of the successive rectangles and squares of width 2 that are adjacent to any of the four sides of the toothpick structure of A139250 after 2^k stages, with k >= 2. For example: if k = 5 the heights after 32 stages are [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1] respectively, the same as the first 15 terms of this sequence. - Omar E. Pol, Dec 29 2020

Factor n; replace each prime(i) with i, take the conjugate partition, replace parts i with prime(i) and multiply out.

From Antti Karttunen, May 12-19 2014: (Start)

For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).

Because the partition conjugation doesn't change the partition's total sum, this permutation preserves A056239, i.e., A056239(a(n)) = A056239(n) for all n.

(Similarly, for all n, A001221(a(n)) = A001221(n), because the number of steps in the Ferrers/Young-diagram stays invariant under the conjugation. - Note added Apr 29 2022).

Because this permutation commutes with A241909, in other words, as a(A241909(n)) = A241909(a(n)) for all n, from which follows, because both permutations are self-inverse, that a(n) = A241909(a(A241909(n))), it means that this is also induced when partitions are conjugated in the partition enumeration system A241918. (Not only in A112798.)

(End)

From Antti Karttunen, Jul 31 2014: (Start)

Rows in arrays A243060 and A243070 converge towards this sequence, and also, assuming no surprises at the rate of that convergence, this sequence occurs also as the central diagonal of both.

Each even number is mapped to a unique term of A102750 and vice versa.

Conversely, each odd number (larger than 1) is mapped to a unique term of A070003, and vice versa. The permutation pair A243287-A243288 has the same property. This is also used to induce the permutations A244981-A244984.

Taking the odd bisection and dividing out the largest prime factor results in the permutation A243505.

Shares with A245613 the property that each term of A028260 is mapped to a unique term of A244990 and each term of A026424 is mapped to a unique term of A244991.

Conversely, with A245614 (the inverse of above), shares the property that each term of A244990 is mapped to a unique term of A028260 and each term of A244991 is mapped to a unique term of A026424.

(End)

The Maple program follows the steps described in the first comment. The subprogram C yields the conjugate partition of a given partition. - Emeric Deutsch, May 09 2015

The Heinz number of the partition that is conjugate to the partition with Heinz number n. The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r). Example: a(3) = 4. Indeed, the partition with Heinz number 3 is [2]; its conjugate is [1,1] having Heinz number 4. - Emeric Deutsch, May 19 2015