cp's OEIS Frontend

Poly-Bernoulli numbers B_n^(k) with k=-2.

Binomial transform of A007051, if both sequences start at 0. Binomial transform of A000225(n+1). - Paul Barry, Mar 24 2003

Euler expands (1-z)/(1-5z+6z^2) and finds the general term. Section 226 of the Introductio indicates that he could have written down the recursion relation: a(n) = 5 a(n-1)-6 a(n-2). - V. Frederick Rickey (fred-rickey(AT)usma.edu), Feb 10 2006

Let R be a binary relation on the power set P(A) of a set A having n = |A| elements such that for every element x, y of P(A), xRy if x is a subset of y or y is a subset of x. Then a(n) = |R|. - Ross La Haye, Dec 22 2006

With regard to the comment by Ross La Haye: For proper subsets see A056182. - For nonempty subsets see A091344. - For nonempty proper subsets see a(n+1) in A260217. - Manfred Boergens, Aug 02 2023

If x, y are two n-bit binary strings then a(n) gives the number of pairs (x,y) such that XOR(x, y) = ABS(x - y). - Ramasamy Chandramouli, Feb 15 2009

Equals row sums of the triangular version of A038573. - Gary W. Adamson, Jun 04 2009

Inverse binomial transform of A085350. - Paul Curtz, Nov 14 2009

Related to the number of even a's in a nontrivial cycle (should one exist) in the 3x+1 Problem, where a <= floor(log_2(2*(3^n) - 2^n)). The value n correlates to the number of odds in such a nontrivial cycle. See page 1288 of Crandall's paper. Also, this relation gives another proof that the number of odds divided by the number of evens in a nontrivial cycle is bounded by log 2 / log 3 (this observation does not resolve the finite cycles conjecture as the value could be arbitrarily close to this bound). However, the same argument gives that log 2 / log 3 is less than or equal to the number of odds divided by the number of evens in a divergent sequence (should one exist), as log 2 / log 3 is the limit value for a cycle of an arbitrarily large length, where the length is given by the value n. - Jeffrey R. Goodwin, Aug 04 2011

Row sums of Riordan triangle A106516. - Wolfdieter Lang, Jan 09 2015

Number of restricted barred preferential arrangements having 3 bars in which the sections are all restricted sections such that (for fixed sections i and j) section i or section j is empty. - Sithembele Nkonkobe, Oct 12 2015

This is also row 2 of A281891: for n >= 1, when consecutive positive integers are written as a product of primes in nondecreasing order, a factor of 2 or 3 occurs in n-th position a(n) times out of every 6^n. - Peter Munn, May 18 2017

Also row sums of A124929. - Omar E. Pol, Jun 15 2017

This is the sum of A318921(n) for n in the range 2^(k+1) to 2^(k+2)-1. See A318921 for proof. - N. J. A. Sloane, Sep 25 2018

a(n) is also the number of acyclic orientations of the complete bipartite graph K_{2,n}. - Vincent Pilaud, Sep 15 2020

a(n-1) is also the number of n-digit numbers whose largest decimal digit is 2. - Stefano Spezia, Nov 15 2023

The n chosen integers need not be distinct.

By "more than half of all integers" we mean more precisely "more than half of the integers in -m..m, for all sufficiently large m (depending on n)", and similarly with 1..m for "more than half of all positive integers".

Equivalently, a(n) is the least prime p such that more than half of all positive integers can be written as a product of primes of which n or more are not greater than p. (In this sense, a(n) might be called the median n-th least prime factor of the integers.)

The number of integers that satisfy the "product of primes" criterion for p = prime(m) is the same in every interval of primorial(m)^n integers and is A281891(m,n). Primorial(m) = A002110(m), product of the first m primes.

a(n) is the least k = prime(m) such that 2 * A281891(m,n) > A002110(m)^n.

a(n) is the least k such that more than half of all positive integers equate to the volume of an orthotope with integral sides at least n of which are orthogonal with length between 2 and k inclusive.

The next term is estimated to be a(5) ~ 3*10^18.

Square array read by descending antidiagonals: A(n,k) with rows n >= 1, columns k >= 1. Primorial(n) = A002110(n): product of first n primes.

Visualize the prime factorizations of the positive integers as a table with row headings giving each successive integer, and the primes of which the row heading is the product listed across the columns in nondecreasing order, repeated when necessary. Except for 1, which lacks prime factors, column 1 has the row heading's least prime factor, column 2 has a value for composite numbers but is blank for primes, and so on. This sequence measures precisely how frequently the various primes occur in each column. This is possible because any given prime occurs cyclically in any given column, for the reason following.

The occurrence pattern of up to k factors of prime(n) in such prime factorizations has a fundamental period over the positive integers of prime(n)^k. The least common period for up to k factors of each of the first n primes is Primorial(n)^k, and this covers everything that can affect the occurrence of prime(n) in the least k factors. Thus prime(n) is k-th least prime factor of integer m if and only if it is k-th least prime factor of m+Primorial(n)^k.

Intermediate values in the calculation of this sequence appear in A281891.

A(n,1) = A005867(n-1) in accordance with the comment on A005867 dated Jul 16 2006.

A(2,k) = A001047(k) = 3^k - 2^k.

a(n) > A005367(n), a(n) > A002110(n)/2.

Limit_{n->oo} a(n)/A002110(n) = 1 because (in the limit) the quotient is the probability that a randomly selected integer contains at least one of the first n primes in its factorization. - Geoffrey Critzer, Apr 08 2010

In every interval of 44100 integers, exactly 22164 are divisible by a 7-smooth composite number. 44100 = (2*3*5*7)^2 = A002110(4)^2 and 22164 = A281891(4,2). See A281891 for more details.

The sequence is finite with largest term a(136) = 1406.