cp's OEIS Frontend

The corresponding primes n!-1 are often called factorial primes.

Let S = {x(1),...,x(k)} be a multiset whose distinct elements are y(1),...,y(h). Let f(i) be the frequency of y(i) in S. Define F(S) = {f(1),..,f(h)}, F(1,S) = F(S), and F(m,S) = F(F(m-1),S) for m>1. Then lim(F(m,S)) = {1} for every S, so that there is a least positive integer i for which F(i,S) = {1}, which we call the frequency depth of S.

Equivalently, the frequency depth of an integer partition is the number of times one must take the multiset of multiplicities to reach (1). For example, the partition (32211) has frequency depth 5 because we have: (32211) -> (221) -> (21) -> (11) -> (2) -> (1). - Gus Wiseman, Apr 19 2019

From Clark Kimberling, Sep 26 2023: (Start)

Below, m^n abbreviates the sum m+...+m of n terms. In the following list, the numbers p_1,...,p_k are distinct, m >= 1, and k >= 1. The forms of the partitions being counted are as follows:

column 1: [n],

column 2: [m^k],

column 3: [p_1^m,...,p_k^m],

column 4: [(p_1^m_1)^m,..., (p_k^m_k)^m], distinct numbers m_i.

Column 3 is of special interest. Assume first that m = 1, so that the form of partition being counted is p = [p_1,...,p_k], with conjugate given by [q_1,...,q_m] where q_i is the number of parts of p that are >= i. Since the p_i are distinct, the distinct parts of q are the integers 1,2,...,k. For the general case that m >= 1, the distinct parts of q are the integers m,...,km. Let S(n) denote the set of partitions of n counted by column 3. Then if a and b are in the set S*(n) of conjugates of partitions in S(n), and if a > b, then a - b is also in S*(n). Call this the subtraction property. Conversely, if a partition q has the subtraction property, then q must consist of a set of numbers m,..,km for some m. Thus, column 3 counts the partitions of n that have the subtraction property. (End)

Erdős proved that there exist two constants c1, c2 > 0 such that c1 (n / log(n))^(1/2) < a(n) < c2 (n / log(n))^(1/2). - Carlo Sanna, May 28 2019

R. Heyman and R. Miraj proved that the cardinality of the set { floor(n/p) : p <= n, p prime } is same as the number of distinct exponents in the prime factorization of n!. - Md Rahil Miraj, Apr 05 2024

We define the omega-sequence of n (row n of A323023) to have length A323014(n) = adjusted frequency depth of n, and the k-th term is Omega(red^{k-1}(n)), where Omega = A001222 and red^{k} is the k-th functional iteration of red = A181819, defined by red(n = p^i*...*q^j) = prime(i)*...*prime(j) = product of primes indexed by the prime exponents of n. For example, we have 180 -> 18 -> 6 -> 4 -> 3, so the omega-sequence of 180 is (5,3,2,2,1).

The prime omicron of n (A304465) is 0 if n is 1, 1 if n is prime, and otherwise the second-to-last part of the omega-sequence of n. For example, the prime omicron of 180 is 2.

Conjecture: all terms after a(10) = 4 are less than 4.

From James Rayman, Apr 17 2021: (Start)

The conjecture is false. a(3804) = 4. In fact, there are 91 values of n < 10000 such that a(n) = 4.

The first value of n such that a(n) = 5 is 37934. For any other n < 5*10^5, a(n) < 5. (End)

The factorization of n! is n! = 2^T(n,1)*3^T(n,2)*...*p_(pi(n))^T(n,pi(n)) where p_k = k-th prime, pi(n) = A000720(n).

Nonzero terms of A085604; T(n,k) = A085604(n,k), k = 1..A000720(n). - Reinhard Zumkeller, Nov 01 2013

For n=2, 3, 4 and 5, all terms of the n-th row are odd. Are there other such rows? - Michel Marcus, Nov 11 2018

From Gus Wiseman, May 15 2019: (Start)

Differences between successive rows are A067255, so row n is the sum of the first n row-vectors of A067255 (padded with zeros on the right so that all n row-vectors have length A000720(n)). For example, the first 10 rows of A067255 are

{}

1

0 1

2 0

0 0 1

1 1 0

0 0 0 1

3 0 0 0

0 2 0 0

1 0 1 0

with column sums (8,4,2,1), which is row 10.

(End)

For all prime p > 7, 3*p > 2*nextprime(p), so for any n > 21 there will always be a prime p dividing n! with exponent 2 and there are no further rows with all entries odd. - Charlie Neder, Jun 03 2019

A number is a perfect power iff it is 1 or its prime exponents (signature) are not relatively prime.

Number of permutations of the multiset of prime factors of n!.

A number has distinct prime multiplicities iff its prime signature is strict.

The prime indices of a(n) are the signature of n!, which is row n of A115627.

Equivalently, n divided by the largest factorial divisor of n.

Also, the smallest r such that n/r is a factorial number.

Positions of 1's are the factorial numbers A000142. Is every positive integer in this sequence? - Gus Wiseman, May 15 2019

Let m = A055874(n), the largest integer such that 1,2,...,m divides n. Then a(n*m!) = n since m+1 does not divide n, showing that every integer is part of the sequence. - Etienne Dupuis, Sep 19 2020