cp's OEIS Frontend

T(n,k) is the triangle in A174909 with reversed row order. (See that sequence for additional comments).

Row sums = A053144(n) = A002110(n) - T(n+1,n+1).

T(n,k) = number of terms with smallest prime factor prime(k) contained in primorial(n) consecutive numbers, k <= n. For example, T(5,4) = 88, so there are 88 terms with smallest prime factor 7 in any sequence of 2310 consecutive numbers.

Alternatively, numerators of k*A060753(n)/A038110(n) for 1 <= k <= A005867(n).

Alternatively, denominators of k*A060753(n)/A038110(n) for 1 <= k <= A005867(n).

Let m = A038110(n). For row n, the primitive denominators d | m.

The mean of row n is related to the mean of row n of A309497: A060753(n+1)/(A038110(n+1)*2) = Mean(A309497(n))/A038110(n+1).

A005867 can be generated recursively using A006093 and A000165 can be generated recursively using the even numbers. Since all primes after 2 are odd, (A005867 - A000165)/96 is an integer.

For any interval I of length P(n) that starts beyond prime(n), divisibility by one or more of the first n primes limits the maximum number of primes in I to Product_{j=1..n} (prime(j) - 1) = A005867(n). Beyond this, for n >= 3, divisibility by primes larger than prime(n) (but smaller than P(n)) ensures that T(n,k)=0 for one or more terms at the end of the row. E.g., for every value of m mod 7, at least one of the A005867(3) = 8 numbers in [30*m+1, 30*(m+1)] that is not divisible by 2, 3, or 5 -- i.e., at least one of the 8 numbers {M+1, M+7, M+11, M+13, M+17, M+19, M+23, M+29} where M=30*m -- will be divisible by 7, so T(3,8)=0. A similar argument shows that T(4,46) = T(4,47) = T(4,48) = 0.

Partial sums of A010051 (characteristic function of primes). - Jeremy Gardiner, Aug 13 2002

pi(n) and prime(n) are inverse functions: a(A000040(n)) = n and A000040(n) is the least number m such that A000040(a(m)) = A000040(n). A000040(a(n)) = n if (and only if) n is prime. - Jonathan Sondow, Dec 27 2004

See the additional references and links mentioned in A143227. - Jonathan Sondow, Aug 03 2008

A lower bound that gets better with larger N is that there are at least T prime numbers less than N, where the recursive function T is: T = N - N*Sum_{i=0..T(sqrt(N))} A005867(i)/A002110(i). - Ben Paul Thurston, Aug 23 2010

Number of partitions of 2n into exactly two parts with the smallest part prime. - Wesley Ivan Hurt, Jul 20 2013

Equivalent to the Riemann hypothesis: abs(a(n) - li(n)) < sqrt(n)*log(n)/(8*Pi), for n >= 2657, where li(n) is the logarithmic integral (Lowell Schoenfeld). - Ilya Gutkovskiy, Jul 05 2016

The second Hardy-Littlewood conjecture, that pi(x) + pi(y) >= pi(x + y) for integers x and y with min{x, y} >= 2, is known to hold for (x, y) sufficiently large (Udrescu 1975). - Peter Luschny, Jan 12 2021

See A034386 for the second definition of primorial numbers: product of primes in the range 2 to n.

a(n) is the least number N with n distinct prime factors (i.e., omega(N) = n, cf. A001221). - Lekraj Beedassy, Feb 15 2002

Phi(n)/n is a new minimum for each primorial. - Robert G. Wilson v, Jan 10 2004

Smallest number stroked off n times after the n-th sifting process in an Eratosthenes sieve. - Lekraj Beedassy, Mar 31 2005

Apparently each term is a new minimum for phi(x)*sigma(x)/x^2. 6/Pi^2 < sigma(x)*phi(x)/x^2 < 1 for n > 1. - Jud McCranie, Jun 11 2005

Let f be a multiplicative function with f(p) > f(p^k) > 1 (p prime, k > 1), f(p) > f(q) > 1 (p, q prime, p < q). Then the record maxima of f occur at n# for n >= 1. Similarly, if 0 < f(p) < f(p^k) < 1 (p prime, k > 1), 0 < f(p) < f(q) < 1 (p, q prime, p < q), then the record minima of f occur at n# for n >= 1. - David W. Wilson, Oct 23 2006

Wolfe and Hirshberg give ?, ?, ?, ?, ?, 30030, ?, ... as a puzzle.

Records in number of distinct prime divisors. - Artur Jasinski, Apr 06 2008

For n >= 2, the digital roots of a(n) are multiples of 3. - Parthasarathy Nambi, Aug 19 2009 [with corrections by Zak Seidov, Aug 30 2015]

Denominators of the sum of the ratios of consecutive primes (see A094661). - Vladimir Joseph Stephan Orlovsky, Oct 24 2009

Where record values occur in A001221. - Melinda Trang (mewithlinda(AT)yahoo.com), Apr 15 2010

It can be proved that there are at least T prime numbers less than N, where the recursive function T is: T = N - N*Sum_{i = 0..T(sqrt(N))} A005867(i)/A002110(i). This can show for example that at least 0.16*N numbers are primes less than N for 29^2 > N > 23^2. - Ben Paul Thurston, Aug 23 2010

The above comment from Parthasarathy Nambi follows from the observation that digit summing produces a congruent number mod 9, so the digital root of any multiple of 3 is a multiple of 3. prime(n)# is divisible by 3 for n >= 2. - Christian Schulz, Oct 30 2013

The peaks (i.e., local maximums) in a graph of the number of repetitions (i.e., the tally of values) vs. value, as generated by taking the differences of all distinct pairs of odd prime numbers within a contiguous range occur at regular periodic intervals given by the primorial numbers 6 and greater. Larger primorials yield larger (relative) peaks, however the range must be >50% larger than the primorial to be easily observed. Secondary peaks occur at intervals of those "near-primorials" divisible by 6 (e.g., 42). See A259629. Also, periodicity at intervals of 6 and 30 can be observed in the local peaks of all possible sums of two, three or more distinct odd primes within modest contiguous ranges starting from p(2) = 3. - Richard R. Forberg, Jul 01 2015

If a number k and a(n) are coprime and k < (prime(n+1))^b < a(n), where b is an integer, then k has fewer than b prime factors, counting multiplicity (i.e., bigomega(k) < b, cf. A001222). - Isaac Saffold, Dec 03 2017

If n > 0, then a(n) has 2^n unitary divisors (A034444), and a(n) is a record; i.e., if k < a(n) then k has fewer unitary divisors than a(n) has. - Clark Kimberling, Jun 26 2018

Unitary superabundant numbers: numbers k with a record value of the unitary abundancy index, A034448(k)/k > A034448(m)/m for all m < k. - Amiram Eldar, Apr 20 2019

Psi(n)/n is a new maximum for each primorial (psi = A001615) [proof in link: Patrick Sole and Michel Planat, proposition 1 page 2]; compare with comment 2004: Phi(n)/n is a new minimum for each primorial. - Bernard Schott, May 21 2020

The term "primorial" was coined by Harvey Dubner (1987). - Amiram Eldar, Apr 16 2021

a(n)^(1/n) is approximately (n log n)/e. - Charles R Greathouse IV, Jan 03 2023

Subsequence of A267124. - Frank M Jackson, Apr 14 2023

Also, the largest number of distinct integers such that all their pairwise differences are coprime to n. - Max Alekseyev, Mar 17 2006

The unit 1 is not a prime number (although it has been considered so in the past). 1 is the empty product of prime numbers, thus 1 has no least prime factor. - Daniel Forgues, Jul 05 2011

a(n) = least m > 0 for which n! + m and n - m are not relatively prime. - Clark Kimberling, Jul 21 2012

For n > 1, a(n) = the smallest k > 1 that divides n. - Antti Karttunen, Feb 01 2014

For n > 1, records are at prime indices. - Zak Seidov, Apr 29 2015

The initials "lpf" might be mistaken for "largest prime factor" (A009190), using "spf" for "smallest prime factor" would avoid this. - M. F. Hasler, Jul 29 2015

n = 89 is the first index > 1 for which a(n) differs from the smallest k > 1 such that (2^k + n - 2)/k is an integer. - M. F. Hasler, Aug 11 2015

From Stanislav Sykora, Jul 29 2017: (Start)

For n > 1, a(n) is also the smallest k, 1 < k <= n, for which the binomial(n,k) is not divisible by n.

Proof: (A) When k and n are relatively prime then binomial(n,k) is divisible by n because k*binomial(n,k) = n*binomial(n-1,k-1). (B) When gcd(n,k) > 1, one of its prime factors is the smallest; let us denote it p, p <= k, and consider the binomial(n,p) = (1/p!)*Product_{i=0..p-1} (n-i). Since p is a divisor of n, it cannot be a divisor of any of the remaining numerator factors. It follows that, denoting as e the largest e > 0 such that p^e|n, the numerator is divisible by p^e but not by p^(e+1). Hence, the binomial is divisible by p^(e-1) but not by p^e and therefore not divisible by n. Applying (A), (B) to all considered values of k completes the proof. (End)

From Bob Selcoe, Oct 11 2017, edited by M. F. Hasler, Nov 06 2017: (Start)

a(n) = prime(j) when n == J (mod A002110(j)), n, j >= 1, where J is the set of numbers <= A002110(j) with smallest prime factor = prime(j). The number of terms in J is A005867(j-1). So:

a(n) = 2 when n == 0 (mod 2);

a(n) = 3 when n == 3 (mod 6);

a(n) = 5 when n == 5 or 25 (mod 30);

a(n) = 7 when n == 7, 49, 77, 91, 119, 133, 161 or 203 (mod 210);

etc. (End)

For n > 1, a(n) is the leftmost term, other than 0 or 1, in the n-th row of A127093. - Davis Smith, Mar 05 2019