cp's OEIS Frontend

All terms are multiples of 3 because for all n >= 2, A006862(n) = 1 + A002110(n) == +1 (mod 3), see A369252.

Although the first thirteen terms appear in the ascending order, this might not be true for all the later terms, if they exist.

Sometimes called "triprimes" or "3-almost primes".

See also A001358 for product of two primes (sometimes called semiprimes).

If you graph a(n)/n for n up to 10000 (and probably quite a bit higher), it appears to be converging to something near 3.9. In fact the limit is infinite. - Franklin T. Adams-Watters, Sep 20 2006

Meng shows that for any sufficiently large odd integer n, the equation n = a + b + c has solutions where each of a, b, c is 3-almost prime. The number of such solutions is (log log n)^6/(16 (log n)^3)*n^2*s(n)*(1 + O(1/log log n)), where s(n) = Sum_{q >= 1} Sum_{a = 1..q, (a, q) = 1} exp(i*2*Pi*n*a/q)*mu(n)/phi(n)^3 > 1/2. - Jonathan Vos Post, Sep 16 2005, corrected & rewritten by M. F. Hasler, Apr 24 2019

Also, a(n) are the numbers such that exactly half of their divisors are composite. For the numbers in which exactly half of the divisors are prime, see A167171. - Ivan Neretin, Jan 12 2016

Note the distinctions between this and "n has exactly three prime factors" (A014612) or "n has exactly three distinct prime factors." (A033992). The word "sphenic" also means "shaped like a wedge" [American Heritage Dictionary] as in dentation with "sphenic molars." - Jonathan Vos Post, Sep 11 2005

Also the volume of a sphenic brick. A sphenic brick is a rectangular parallelepiped whose sides are components of a sphenic number, namely whose sides are three distinct primes. Example: The distinct prime triple (3,5,7) produces a 3x5x7 unit brick which has volume 105 cubic units. 3-D analog of 2-D A037074 Product of twin primes, per Cino Hilliard's comment. Compare with 3-D A107768 Golden 3-almost primes = Volumes of bricks (rectangular parallelepipeds) each of whose faces has golden semiprime area. - Jonathan Vos Post, Jan 08 2007

Sum(n>=1, 1/a(n)^s) = (1/6)*(P(s)^3 - P(3*s) - 3*(P(s)*P(2*s)-P(3*s))), where P is prime zeta function. - Enrique Pérez Herrero, Jun 28 2012

Also numbers n with A001222(n)=3 and A001221(n)=3. - Enrique Pérez Herrero, Jun 28 2012

n = 265550 is the smallest n with a(n) (=1279789) < A006881(n) (=1279793). - Peter Dolland, Apr 11 2020

The old name was "Odd numbers with exactly 3 distinct prime factors", which is slightly ambiguous, since it might be interpreted to include 315 = 3^2*5*7. Cf. A278569. - N. J. A. Sloane, Nov 27 2016

For a guide to related sequences, see A211422.

Also the number of partitions of n+1 into three parts, where each part > 1. - Peter Woodward, May 25 2015

a(n) is also equal to the number of partitions of n+4 into three distinct parts, where each part > 1. - Giovanni Resta, May 26 2015

Number of different distributions of n+1 identical balls in 3 boxes as x,y,z where 0 < x < y < z. - Ece Uslu and Esin Becenen, Dec 31 2015

After the first three terms, partial sums of A008615. - Robert Israel, Dec 31 2015

For n >= 2, also the number of partitions of n - 2 into 3 parts. The Heinz numbers of these partitions are given by A014612. - Gus Wiseman, Oct 11 2020

Number of solutions to n = x', where x' is the arithmetic derivative of x (A003415), and x is a product of three odd primes (not all necessarily distinct, A046316).

See the conjecture in A369055.

Number of solutions to 4n-1 = x', where x' is the arithmetic derivative of x (A003415), and x is a product of three odd primes, A046316.

The number of 0's in range [1..10^n], for n=1..7 are: 8, 46, 288, 2348, 21330, 206355, 2079925, etc.

Goldbach's conjecture can be expressed by claiming that each even number > 4 is an arithmetic derivative of an odd semiprime, as (p*q)' = p+q, where p and q are odd primes. One way to extend Goldbach's conjecture to three primes involves applying the arithmetic derivative to all possible products of three odd primes (A046316) as: (p*q*r)' = (p*q) + (p*r) + (q*r), and asking, "Onto which subset of natural numbers does this map surjectively?" Clearly, the above formula can only produce numbers of the form 4m+3, and furthermore, an analysis at A369252 shows that the trisections of this sequence have quite different expected values, being on average the highest in the trisection A369462, which gives the number of representations for the numbers of the form 12m+11. This motivates a new kind of Goldbach-3 conjecture: "All numbers of the form 12*m-1, with m large enough, have at least one representation as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r." Furthermore, empirical data for sequence A369463 suggests that "large enough" in this case might well be 4224080, as 1+(12*4224079) = 50688949 = A369463(285), with the next term of A369463 so far unknown. Similar conjectures can be envisaged for the arithmetic derivatives of products of four or more primes. - Antti Karttunen, Jan 25 2024

The table showing the possible modulo 3 combinations for p, q, r and the sum ((p*q) + (p*r) + (q*r)):

| p | q | r | sum ((p*q) + (p*r) + (q*r)) (mod 3)

--+------+------+------+----------------------------------------

| 0 | 0 | 0 | 0, p=q=r=3, sum is 27.

--+------+------+------+----------------------------------------

| 0 | 0 | +/-1 | 0, p=q=3, r > 3.

--+------+------+------+----------------------------------------

| 0 | +1 | +1 | +1

--+------+------+------+----------------------------------------

| 0 | -1 | -1 | +1

--+------+------+------+----------------------------------------

| 0 | -1 | +1 | -1

--+------+------+------+----------------------------------------

| 0 | +1 | -1 | -1

--+------+------+------+----------------------------------------

| +1 | +1 | +1 | 0

--+------+------+------+----------------------------------------

| -1 | -1 | -1 | 0

--+------+------+------+----------------------------------------

| -1 | +1 | +1 | -1, regardless of the order, thus x3.

--+------+------+------+----------------------------------------

| +1 | -1 | -1 | -1, regardless of the order, thus x3.

--+------+------+------+----------------------------------------

Notably a(n) is a multiple of 3 only when A046316(n) is either a multiple of 9, or all primes p, q and r are either == +1 (mod 3) or all are == -1 (mod 3), and the case a(n) == +1 (mod 3) is only possible when A046316(n) is a multiple of 3, but not of 9, and furthermore, it is required that r == q (mod 3). See how these combinations affects sequences like A369241, A369245, A369450, A369451, A369452.

For n=1..9 the number of terms of the form 3k, 3k+1 and 3k+2 in range [1..10^n-1] are:

6, 2, 1,

39, 22, 38,

291, 209, 499,

2527, 1884, 5588,

23527, 17020, 59452,

227297, 156240, 616462,

2232681, 1453030, 6314288,

22119496, 13661893, 64218610,

220098425, 129624002, 650277572.

It seems that 3k+2 terms are slowly gaining at the expense of 3k+1 terms when n grows, while the density of the multiples of 3 might converge towards a limit.

A multiplicative semigroup: if m and n are in the sequence, then so is m*n. - Antti Karttunen, Jul 02 2024

Note how there are generally less solutions for even n than for odd n. This is explained by the fact that A143293(2n) == 1 (mod 4) and A143293(2n+1) == 3 (mod 4) and the arithmetic derivative A003415 of a product of any three odd primes (A046316) is always of the form 4k+3, therefore the solution set counted by a(2n) does not have any solutions from A046316 that contribute the majority of the solutions counted by a(2n+1). See also A369055.

a(13) >= 1 as there are solutions like 5744093403180469, 12538540924097819, etc., probably thousands or even more in total.

a(14) >= 1 [see examples].