cp's OEIS Frontend

Offset to match A332979.

Let n_2 be the binary expansion of n with a length of b bits. Let W(n) = A000120(n) the binary weight of n, i.e., the number of ones in n_2, while Z(n) = b - W(n) be the number of zeros in n_2. Let Q be the number of runs of ones in n_2, L(k) be the run length of the k-th least significant run of ones, and P(k) the partial sum of the number of zeros to the right of the k-th run of ones.

Define the Doudna function f(n) = Product_{k=1..Q} prime(P(k)+1)^L(k). The Doudna sequence A005940(n) = s(n) = f(n-1) with s(1) = 1.

Theorem 1: the maximum of s(n) for n = 2^(k-1)+1..2^k is a prime power.

Proof. s(n) corresponds to f(m), m = n-1, hence m = 2^(k-1)..2^k. The number m in this domain has k bits. Binary numbers involve zeros and ones, thus, k = W(m) + Z(m). It is clear that W(m) = Omega(f(m)) and Z(m) = pi(gpf(f(m)))-1. Hence we have k = Omega(f(m)) + pi(gpf(f(m)))-1. We maximize f(m) for a k-bit number m when the greatest prime factor of f(m) has maximum multiplicity, therefore the maximum of s(n) for n = 2^(k-1)+1..2^k is a prime power.

Theorem 2: s(2^k-2^j+1) = prime(j+1)^(k-j), j=1..k-1.

Proof: We write (2^k-2^j)_2 as (k-j) ones followed by j zeros, a k-bit binary number. We have Q=1 run of ones in (2^k-2^j)_2. Therefore f(2^k-2^j) = prime(P(1)+1)^L(1) = prime(j+1)^(k-j), that is, row k of A180944.

The maximum of s(n) for n = 2^(k-1)+1..2^k is tantamount to the maximum of row k of A180944.

Maxima of row n > 0 of A005940, A182944, and A182945 are powers of these primes.

Indices k of primes, A000040(k), listed here show an interesting correlation with the function f(k) = A000040(k) - A302334(k). - Peter Munn, Sep 29 2022

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

We alternatively refer to this sequence as a triangle T(.,.), with T(n,k) = A(k,n-k+1) = prime(k)^(n-k+1).

The monotonic ordering of this sequence, prefixed by 1, is A000961.

The joint-rank array of this sequence is A182869.

Main diagonal gives A062457. - Omar E. Pol, Sep 11 2018

Integer m > 0 is listed in row n if the index of the largest prime factor of m (or 0 for empty prime factor set) plus the cardinality of the other prime factors of m (counted with multiplicity) equals n.

Row n+k-1 contains prime(n)^k (for all n, k >= 1).

The concatenation of all rows (with offset 1) gives a permutation of the natural numbers A000027 with fixed points 1, 2, 3, 4, 5, 6, 10, ... and inverse permutation A332990.

This is a variant with sorted rows of A005940 (offset differs) or A163511.