cp's OEIS Frontend

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

Does the ratio of adjacent terms converge?

It appears that lim_{n->infinity} a(n+1)/a(n) = 1.7532... - Jon E. Schoenfield, Feb 21 2019

For n > 1, a(n) is smallest prime p = prime(k) such that no fewer than (n-1)/n of any p# consecutive integers are divisible by a prime not greater than p. Cf. A053144(k)/A002110(k). - Peter Munn, Apr 29 2017

Also, the smallest prime p such that the sum of the reciprocals of the p-smooth numbers converges to at least n. - Keith F. Lynch, Apr 29 2023

Also, if m is a random integer much larger than the square of a(n), and m is not divisible by any prime less than or equal to a(n), the probability that m is prime is n/log(m). - Keith F. Lynch, Dec 17 2023

By rewriting the sequence of sums as 1 - Product_{n>=1} (1 - 1/prime(n)), one can show that the product goes to zero and the sequence of sums converges to 1. This is interesting because the terms approach 1/(2*prime(n)) for large n, and a sum of such terms might be expected to diverge, since Sum_{n>=1} 1/(2*prime(n)) diverges.

Denominators appear to be given by A060753(n+1). - Peter Kagey, Jun 08 2019

A254196 appears to be a duplicate of this sequence. - Michel Marcus, Aug 05 2019

A001611 is similar but different.

Equal to A005579 except for n = 2 and n = 3. The following argument shows that they are equal for n > 3. First note that b(k+1) > b(k). Next, Product_{i=1..k} p_i is 2 times an odd number, i.e., it is not divisible by 4. Similarly since p_i - 1 is even for i > 1, Product_{i=1..k} (p_i - 1) is divisible by 2^(k-1), i.e., it is divisible by 4 for k >= 3. Thus b(k) is not an integer for k >= 3. Since b(3) = 15/4 > 3, this means that a(n) = A005579(n) for n > 3 - Chai Wah Wu, Apr 17 2015

This sequence does not include the sextet (7,11,13,17,19,23). It is a proper subset of A014561 in a certain sense.

Row n has length A000010(n).

Row n > 1 has sum = n*A076512(n)/2.

First value on row(n) = A076511(n).

Last value on row(n) = A076512(n) for n > 1.

For n > 1, A109395(n) = Max(row) + Min(row).

For values x and y on row n > 1 at positions a and b on the row:

x + y = A109395(n), where a = A000010(n) - (b-1).

For n > 2 the penultimate value on row A002110(n) is given by

A038110(n)*A000040(n)-A060753(n).

From Charlie Neder, Jun 05 2019: (Start)

If p is a prime dividing n, then row p*n consists of p copies of row n.

Conjecture: If n is odd, then row 2n can be obtained from row n by interchanging the first and second halves. (End)

Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117.

There are 2^(n-1) numbers k with gpf(k) = prime(n), since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which phi(k)/k - 1/2 is positive and minimal.

We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but more than 1/2.

Apart from a(1) = 2, all terms are odd. For n > 1 and k even, phi(k)/k - 1/2 is negative.

Let gpf(k) = A006530(k) and let phi(n) = A000010(n) for k in A005117. There are 2^(n-1) numbers k with gpf(k) = n, since we can only either have p_i^0 or p_i^1 where p_i | k and i <= n. For example, for n = 2, there are only 2 squarefree numbers k with prime(2) = 3 as greatest prime factor. These are 3 = 2^0 * 3^1, and 6 = 2^1 * 3^1. We observe that we can write multiplicities of the primes as A067255(k), and thus for the example derive 3 = "0,1" and 6 = "1,1". Thus for n = 3, we have 5 = "0,0,1", 15 = "0,1,1", 10 = "1,0,1", and 30 = "1,1,1". This establishes the possible values of k with respect to n. We choose the value of k in n for which 1/2 - phi(k)/k is positive and minimal.

We know that prime k (in A000040) have phi(k)/k = A006093(n)/A000040(n) and represent maxima in n. We likewise know primorials k (in A002110) have phi(k)/k = A038110(n)/A060753(n) and represent minima in n. This sequence shows squarefree numbers k with gpf(k) = n such that their value phi(k)/k is closest to but less than 1/2.

Conjecture: for n > 3, k is always odd. This assertion is reliant upon phi(2 prime(n))/2 prime(n) = phi(2)/2 * phi(prime(n))/prime(n) = 1/2 * (prime(n) - 1)/prime(n), and it is clear that 1/2 is an asymptote for even k.

Looking for prime factors > 5=prime(3) in 8=A005867(3) candidates mod 30=A002110(3) two candidates in the form 30k +- 7 with k > 0 never belong to a twin prime pair. Twin primes can be (30k-13, 30k-11) A331840, (30k-1, 30k +1) A176114, or (30k+11, 30k+13) A089160.