cp's OEIS Frontend

Index of smallest 3-almost prime gap equal to n. This is to 3-almost primes A014612 as A123375 is to semiprimes A001358. a(25) = 0 or a(25) > 10000. Conjecture: a(n) > 0 for all n.

3-almost prime analog of A065415.

3-almost prime analog of A065416.

Also referred to as T(n) or C(n+1, 2) or binomial(n+1, 2) (preferred).

Also generalized hexagonal numbers: n*(2*n-1), n=0, +-1, +-2, +-3, ... Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. In this case k = 6. - Omar E. Pol, Sep 13 2011 and Aug 04 2012

Number of edges in complete graph of order n+1, K_{n+1}.

Number of legal ways to insert a pair of parentheses in a string of n letters. E.g., there are 6 ways for three letters: (a)bc, (ab)c, (abc), a(b)c, a(bc), ab(c). Proof: there are C(n+2,2) ways to choose where the parentheses might go, but n + 1 of them are illegal because the parentheses are adjacent. Cf. A002415.

For n >= 1, a(n) is also the genus of a nonsingular curve of degree n+2, such as the Fermat curve x^(n+2) + y^(n+2) = 1. - Ahmed Fares (ahmedfares(AT)my_deja.com), Feb 21 2001

From Harnack's theorem (1876), the number of branches of a nonsingular curve of order n is bounded by a(n-1)+1, and the bound can be achieved. See also A152947. - Benoit Cloitre, Aug 29 2002. Corrected by Robert McLachlan, Aug 19 2024

Number of tiles in the set of double-n dominoes. - Scott A. Brown, Sep 24 2002

Number of ways a chain of n non-identical links can be broken up. This is based on a similar problem in the field of proteomics: the number of ways a peptide of n amino acid residues can be broken up in a mass spectrometer. In general, each amino acid has a different mass, so AB and BC would have different masses. - James A. Raymond, Apr 08 2003

Triangular numbers - odd numbers = shifted triangular numbers; 1, 3, 6, 10, 15, 21, ... - 1, 3, 5, 7, 9, 11, ... = 0, 0, 1, 3, 6, 10, ... - Xavier Acloque, Oct 31 2003 [Corrected by Derek Orr, May 05 2015]

Centered polygonal numbers are the result of [number of sides * A000217 + 1]. E.g., centered pentagonal numbers (1,6,16,31,...) = 5 * (0,1,3,6,...) + 1. Centered heptagonal numbers (1,8,22,43,...) = 7 * (0,1,3,6,...) + 1. - Xavier Acloque, Oct 31 2003

Maximum number of lines formed by the intersection of n+1 planes. - Ron R. King, Mar 29 2004

Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - Mike Zabrocki, Aug 26 2004

Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012

Number of ways two different numbers can be selected from the set {0,1,2,...,n} without repetition, or, number of ways two different numbers can be selected from the set {1,2,...,n} with repetition.

Conjecturally, 1, 6, 120 are the only numbers that are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005

Binomial transform is {0, 1, 5, 18, 56, 160, 432, ...}, A001793 with one leading zero. - Philippe Deléham, Aug 02 2005

Each pair of neighboring terms adds to a perfect square. - Zak Seidov, Mar 21 2006

Number of transpositions in the symmetric group of n+1 letters, i.e., the number of permutations that leave all but two elements fixed. - Geoffrey Critzer, Jun 23 2006

With rho(n):=exp(i*2*Pi/n) (an n-th root of 1) one has, for n >= 1, rho(n)^a(n) = (-1)^(n+1). Just use the triviality a(2*k+1) == 0 (mod (2*k+1)) and a(2*k) == k (mod (2*k)).

a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3)^(n-1). - Sergio Falcon, Feb 12 2007

a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 2 variables. - Richard Barnes, Sep 06 2017

The number of distinct handshakes in a room with n+1 people. - Mohammad K. Azarian, Apr 12 2007 [corrected, Joerg Arndt, Jan 18 2016]

Equal to the rank (minimal cardinality of a generating set) of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, May 03 2007

a(n) gives the total number of triangles found when cevians are drawn from a single vertex on a triangle to the side opposite that vertex, where n = the number of cevians drawn+1. For instance, with 1 cevian drawn, n = 1+1 = 2 and a(n)= 2*(2+1)/2 = 3 so there is a total of 3 triangles in the figure. If 2 cevians are drawn from one point to the opposite side, then n = 1+2 = 3 and a(n) = 3*(3+1)/2 = 6 so there is a total of 6 triangles in the figure. - Noah Priluck (npriluck(AT)gmail.com), Apr 30 2007

For n >= 1, a(n) is the number of ways in which n-1 can be written as a sum of three nonnegative integers if representations differing in the order of the terms are considered to be different. In other words, for n >= 1, a(n) is the number of nonnegative integral solutions of the equation x + y + z = n-1. - Amarnath Murthy, Apr 22 2001 (edited by Robert A. Beeler)

a(n) is the number of levels with energy n + 3/2 (in units of h*f0, with Planck's constant h and the oscillator frequency f0) of the three-dimensional isotropic harmonic quantum oscillator. See the comment by A. Murthy above: n = n1 + n2 + n3 with positive integers and ordered. Proof from the o.g.f. See the A. Messiah reference. - Wolfdieter Lang, Jun 29 2007

From Hieronymus Fischer, Aug 06 2007: (Start)

Numbers m >= 0 such that round(sqrt(2m+1)) - round(sqrt(2m)) = 1.

Numbers m >= 0 such that ceiling(2*sqrt(2m+1)) - 1 = 1 + floor(2*sqrt(2m)).

Numbers m >= 0 such that fract(sqrt(2m+1)) > 1/2 and fract(sqrt(2m)) < 1/2, where fract(x) is the fractional part of x (i.e., x - floor(x), x >= 0). (End)

If Y and Z are 3-blocks of an n-set X, then, for n >= 6, a(n-1) is the number of (n-2)-subsets of X intersecting both Y and Z. - Milan Janjic, Nov 09 2007

Equals row sums of triangle A143320, n > 0. - Gary W. Adamson, Aug 07 2008

a(n) is also an even perfect number in A000396 iff n is a Mersenne prime A000668. - Omar E. Pol, Sep 05 2008. Unnecessary assumption removed and clarified by Rick L. Shepherd, Apr 14 2025

Equals row sums of triangle A152204. - Gary W. Adamson, Nov 29 2008

The number of matches played in a round robin tournament: n*(n-1)/2 gives the number of matches needed for n players. Everyone plays against everyone else exactly once. - Georg Wrede (georg(AT)iki.fi), Dec 18 2008

-a(n+1) = E(2)*binomial(n+2,2) (n >= 0) where E(n) are the Euler numbers in the enumeration A122045. Viewed this way, a(n) is the special case k=2 in the sequence of diagonals in the triangle A153641. - Peter Luschny, Jan 06 2009

Equivalent to the first differences of successive tetrahedral numbers. See A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009

The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-3)(P(2,1)-(-1)^k P(2,2k+1))|. - Peter Luschny, Jul 12 2009

a(n) is the smallest number > a(n-1) such that gcd(n,a(n)) = gcd(n,a(n-1)). If n is odd this gcd is n; if n is even it is n/2. - Franklin T. Adams-Watters, Aug 06 2009

Partial sums of A001477. - Juri-Stepan Gerasimov, Jan 25 2010. [A-number corrected by Omar E. Pol, Jun 05 2012]

The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - Paul Muljadi, Jan 25 2010

From Charlie Marion, Dec 03 2010: (Start)

More generally, a(2k+1) == j*(2j-1) (mod 2k+2j+1) and

a(2k) == [-k + 2j*(j-1)] (mod 2k+2j).

Column sums of:

1 3 5 7 9 ...

1 3 5 ...

1 ...

...............

---------------

1 3 6 10 15 ...

Sum_{n>=1} 1/a(n)^2 = 4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).

(End)

A004201(a(n)) = A000290(n); A004202(a(n)) = A002378(n). - Reinhard Zumkeller, Feb 12 2011

1/a(n+1), n >= 0, has e.g.f. -2*(1+x-exp(x))/x^2, and o.g.f. 2*(x+(1-x)*log(1-x))/x^2 (see the Stephen Crowley formula line). -1/(2*a(n+1)) is the z-sequence for the Sheffer triangle of the coefficients of the Bernoulli polynomials A196838/A196839. - Wolfdieter Lang, Oct 26 2011

From Charlie Marion, Feb 23 2012: (Start)

a(n) + a(A002315(k)*n + A001108(k+1)) = (A001653(k+1)*n + A001109(k+1))^2. For k=0 we obtain a(n) + a(n+1) = (n+1)^2 (identity added by N. J. A. Sloane on Feb 19 2004).

a(n) + a(A002315(k)*n - A055997(k+1)) = (A001653(k+1)*n - A001109(k))^2.

(End)

Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - J. M. Bergot, May 04 2012

The sum of four consecutive triangular numbers, beginning with a(n)=n*(n+1)/2, minus 2 is 2*(n+2)^2. a(n)*a(n+2)/2 = a(a(n+1)-1). - J. M. Bergot, May 17 2012

(a(n)*a(n+3) - a(n+1)*a(n+2))*(a(n+1)*a(n+4) - a(n+2)*a(n+3))/8 = a((n^2+5*n+4)/2). - J. M. Bergot, May 18 2012

a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - J. M. Bergot, May 22 2012

In general, a(n)*a(n+1) + a(n+k)*a(n+k+1) + a(k-1)*a(k) = a(n^2 + (k+2)*n + k*(k+1)). - Charlie Marion, Sep 11 2012

a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - J. M. Bergot, May 22 2012

In general, a(n)*a(n+k) + a(n+1)*a(n+k-1) = a(n^2 + (k+1)*n + k-1). - Charlie Marion, Sep 11 2012

a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - J. M. Bergot, May 22 2012

Three points (a(n),a(n+1)), (a(n+1),a(n)) and (a(n+2),a(n+3)) form a triangle with area 4*a(n+1). - J. M. Bergot, May 23 2012

a(n) + a(n+k) = (n+k)^2 - (k^2 + (2n-1)*k -2n)/2. For k=1 we obtain a(n) + a(n+1) = (n+1)^2 (see below). - Charlie Marion, Oct 02 2012

In n-space we can define a(n-1) nontrivial orthogonal projections. For example, in 3-space there are a(2)=3 (namely point onto line, point onto plane, line onto plane). - Douglas Latimer, Dec 17 2012

From James East, Jan 08 2013: (Start)

For n >= 1, a(n) is equal to the rank (minimal cardinality of a generating set) and idempotent rank (minimal cardinality of an idempotent generating set) of the semigroup P_n\S_n, where P_n and S_n denote the partition monoid and symmetric group on [n].

For n >= 3, a(n-1) is equal to the rank and idempotent rank of the semigroup T_n\S_n, where T_n and S_n denote the full transformation semigroup and symmetric group on [n].

(End)

For n >= 3, a(n) is equal to the rank and idempotent rank of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, Jan 15 2013

Conjecture: For n > 0, there is always a prime between A000217(n) and A000217(n+1). Sequence A065383 has the first 1000 of these primes. - Ivan N. Ianakiev, Mar 11 2013

The formula, a(n)*a(n+4k+2)/2 + a(k) = a(a(n+2k+1) - (k^2+(k+1)^2)), is a generalization of the formula a(n)*a(n+2)/2 = a(a(n+1)-1) in Bergot's comment dated May 17 2012. - Charlie Marion, Mar 28 2013

The series Sum_{k>=1} 1/a(k) = 2, given in a formula below by Jon Perry, Jul 13 2003, has partial sums 2*n/(n+1) (telescopic sum) = A022998(n)/A026741(n+1). - Wolfdieter Lang, Apr 09 2013

For odd m = 2k+1, we have the recurrence a(m*n + k) = m^2*a(n) + a(k). Corollary: If number T is in the sequence then so is 9*T+1. - Lekraj Beedassy, May 29 2013

Euler, in Section 87 of the Opera Postuma, shows that whenever T is a triangular number then 9*T + 1, 25*T + 3, 49*T + 6 and 81*T + 10 are also triangular numbers. In general, if T is a triangular number then (2*k + 1)^2*T + k*(k + 1)/2 is also a triangular number. - Peter Bala, Jan 05 2015

Using 1/b and 1/(b+2) will give a Pythagorean triangle with sides 2*b + 2, b^2 + 2*b, and b^2 + 2*b + 2. Set b=n-1 to give a triangle with sides of lengths 2*n,n^2-1, and n^2 + 1. One-fourth the perimeter = a(n) for n > 1. - J. M. Bergot, Jul 24 2013

a(n) = A028896(n)/6, where A028896(n) = s(n) - s(n-1) are the first differences of s(n) = n^3 + 3*n^2 + 2*n - 8. s(n) can be interpreted as the sum of the 12 edge lengths plus the sum of the 6 face areas plus the volume of an n X (n-1) X (n-2) rectangular prism. - J. M. Bergot, Aug 13 2013

Dimension of orthogonal group O(n+1). - Eric M. Schmidt, Sep 08 2013

Number of positive roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013

A formula for the r-th successive summation of k, for k = 1 to n, is binomial(n+r,r+1) [H. W. Gould]. - Gary Detlefs, Jan 02 2014

Also the alternating row sums of A095831. Also the alternating row sums of A055461, for n >= 1. - Omar E. Pol, Jan 26 2014

For n >= 3, a(n-2) is the number of permutations of 1,2,...,n with the distribution of up (1) - down (0) elements 0...011 (n-3 zeros), or, the same, a(n-2) is up-down coefficient {n,3} (see comment in A060351). - Vladimir Shevelev, Feb 14 2014

a(n) is the dimension of the vector space of symmetric n X n matrices. - Derek Orr, Mar 29 2014

Non-vanishing subdiagonal of A132440^2/2, aside from the initial zero. First subdiagonal of unsigned A238363. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of complete graphs. - Tom Copeland, Apr 05 2014

The number of Sidon subsets of {1,...,n+1} of size 2. - Carl Najafi, Apr 27 2014

Number of factors in the definition of the Vandermonde determinant V(x_1,x_2,...,x_n) = Product_{1 <= i < k <= n} x_i - x_k. - Tom Copeland, Apr 27 2014

Number of weak compositions of n into three parts. - Robert A. Beeler, May 20 2014

Suppose a bag contains a(n) red marbles and a(n+1) blue marbles, where a(n), a(n+1) are consecutive triangular numbers. Then, for n > 0, the probability of choosing two marbles at random and getting two red or two blue is 1/2. In general, for k > 2, let b(0) = 0, b(1) = 1 and, for n > 1, b(n) = (k-1)*b(n-1) - b(n-2) + 1. Suppose, for n > 0, a bag contains b(n) red marbles and b(n+1) blue marbles. Then the probability of choosing two marbles at random and getting two red or two blue is (k-1)/(k+1). See also A027941, A061278, A089817, A053142, A092521. - Charlie Marion, Nov 03 2014

Let O(n) be the oblong number n(n+1) = A002378 and S(n) the square number n^2 = A000290(n). Then a(4n) = O(3n) - O(n), a(4n+1) = S(3n+1) - S(n), a(4n+2) = S(3n+2) - S(n+1) and a(4n+3) = O(3n+2) - O(n). - Charlie Marion, Feb 21 2015

Consider the partition of the natural numbers into parts from the set S=(1,2,3,...,n). The length (order) of the signature of the resulting sequence is given by the triangular numbers. E.g., for n=10, the signature length is 55. - David Neil McGrath, May 05 2015

a(n) counts the partitions of (n-1) unlabeled objects into three (3) parts (labeled a,b,c), e.g., a(5)=15 for (n-1)=4. These are (aaaa),(bbbb),(cccc),(aaab),(aaac),(aabb),(aacc),(aabc),(abbc),(abcc),(abbb),(accc),(bbcc),(bccc),(bbbc). - David Neil McGrath, May 21 2015

Conjecture: the sequence is the genus/deficiency of the sinusoidal spirals of index n which are algebraic curves. The value 0 corresponds to the case of the Bernoulli Lemniscate n=2. So the formula conjectured is (n-1)(n-2)/2. - Wolfgang Tintemann, Aug 02 2015

Conjecture: Let m be any positive integer. Then, for each n = 1,2,3,... the set {Sum_{k=s..t} 1/k^m: 1 <= s <= t <= n} has cardinality a(n) = n*(n+1)/2; in other words, all the sums Sum_{k=s..t} 1/k^m with 1 <= s <= t are pairwise distinct. (I have checked this conjecture via a computer and found no counterexample.) - Zhi-Wei Sun, Sep 09 2015

The Pisano period lengths of reading the sequence modulo m seem to be A022998(m). - R. J. Mathar, Nov 29 2015

For n >= 1, a(n) is the number of compositions of n+4 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016

In this sequence only 3 is prime. - Fabian Kopp, Jan 09 2016

Suppose you are playing Bulgarian Solitaire (see A242424 and Chamberland's and Gardner's books) and, for n > 0, you are starting with a single pile of a(n) cards. Then the number of operations needed to reach the fixed state {n, n-1,...,1} is a(n-1). For example, {6}->{5,1}->{4,2}->{3,2,1}. - Charlie Marion, Jan 14 2016

Numbers k such that 8k + 1 is a square. - Juri-Stepan Gerasimov, Apr 09 2016

Every perfect cube is the difference of the squares of two consecutive triangular numbers. 1^2-0^2 = 1^3, 3^2-1^2 = 2^3, 6^2-3^2 = 3^3. - Miquel Cerda, Jun 26 2016

For n > 1, a(n) = tau_n(k*) where tau_n(k) is the number of ordered n-factorizations of k and k* is the square of a prime. For example, tau_3(4) = tau_3(9) = tau_3(25) = tau_3(49) = 6 (see A007425) since the number of divisors of 4, 9, 25, and 49's divisors is 6, and a(3) = 6. - Melvin Peralta, Aug 29 2016

In an (n+1)-dimensional hypercube, number of two-dimensional faces congruent with a vertex (see also A001788). - Stanislav Sykora, Oct 23 2016

Generalizations of the familiar formulas, a(n) + a(n+1) = (n+1)^2 (Feb 19 2004) and a(n)^2 + a(n+1)^2 = a((n+1)^2) (Nov 22 2006), follow: a(n) + a(n+2k-1) + 4a(k-1) = (n+k)^2 + 6a(k-1) and a(n)^2 + a(n+2k-1)^2 + (4a(k-1))^2 + 3a(k-1) = a((n+k)^2 + 6a(k-1)). - Charlie Marion, Nov 27 2016

a(n) is also the greatest possible number of diagonals in a polyhedron with n+4 vertices. - Vladimir Letsko, Dec 19 2016

For n > 0, 2^5 * (binomial(n+1,2))^2 represents the first integer in a sum of 2*(2*n + 1)^2 consecutive integers that equals (2*n + 1)^6. - Patrick J. McNab, Dec 25 2016

Does not satisfy Benford's law (cf. Ross, 2012). - N. J. A. Sloane, Feb 12 2017

Number of ordered triples (a,b,c) of positive integers not larger than n such that a+b+c = 2n+1. - Aviel Livay, Feb 13 2017

Number of inequivalent tetrahedral face colorings using at most n colors so that no color appears only once. - David Nacin, Feb 22 2017

Also the Wiener index of the complete graph K_{n+1}. - Eric W. Weisstein, Sep 07 2017

Number of intersections between the Bernstein polynomials of degree n. - Eric Desbiaux, Apr 01 2018

a(n) is the area of a triangle with vertices at (1,1), (n+1,n+2), and ((n+1)^2, (n+2)^2). - Art Baker, Dec 06 2018

For n > 0, a(n) is the smallest k > 0 such that n divides numerator of (1/a(1) + 1/a(2) + ... + 1/a(n-1) + 1/k). It should be noted that 1/1 + 1/3 + 1/6 + ... + 2/(n(n+1)) = 2n/(n+1). - Thomas Ordowski, Aug 04 2019

Upper bound of the number of lines in an n-homogeneous supersolvable line arrangement (see Theorem 1.1 in Dimca). - Stefano Spezia, Oct 04 2019

For n > 0, a(n+1) is the number of lattice points on a triangular grid with side length n. - Wesley Ivan Hurt, Aug 12 2020

From Michael Chu, May 04 2022: (Start)

Maximum number of distinct nonempty substrings of a string of length n.

Maximum cardinality of the sumset A+A, where A is a set of n numbers. (End)

a(n) is the number of parking functions of size n avoiding the patterns 123, 132, and 312. - Lara Pudwell, Apr 10 2023

Suppose two rows, each consisting of n evenly spaced dots, are drawn in parallel. Suppose we bijectively draw lines between the dots of the two rows. For n >= 1, a(n - 1) is the maximal possible number of intersections between the lines. Equivalently, the maximal number of inversions in a permutation of [n]. - Sela Fried, Apr 18 2023

The following equation complements the generalization in Bala's Comment (Jan 05 2015). (2k + 1)^2*a(n) + a(k) = a((2k + 1)*n + k). - Charlie Marion, Aug 28 2023

a(n) + a(n+k) + a(k-1) + (k-1)*n = (n+k)^2. For k = 1, we have a(n) + a(n+1) = (n+1)^2. - Charlie Marion, Nov 17 2023

a(n+1)/3 is the expected number of steps to escape from a linear row of n positions starting at a random location and randomly performing steps -1 or +1 with equal probability. - Hugo Pfoertner, Jul 22 2025

a(n+1) is the number of nonnegative integer solutions to p + q + r = n. By Sylvester's law of inertia, it is also the number of congruence classes of real symmetric n-by-n matrices or equivalently, the number of symmetric bilinear forms on a real n-dimensional vector space. - Paawan Jethva, Jul 24 2025

Maximal number of terms in any factorization of n.

Number of prime powers (not including 1) that divide n.

Sum of exponents in prime-power factorization of n. - Daniel Forgues, Mar 29 2009

Sum_{d|n} 2^(-A001221(d) - a(n/d)) = Sum_{d|n} 2^(-a(d) - A001221(n/d)) = 1 (see Dressler and van de Lune link). - Michel Marcus, Dec 18 2012

Row sums in A067255. - Reinhard Zumkeller, Jun 11 2013

Conjecture: Let f(n) = (x+y)^a(n), and g(n) = x^a(n), and h(n) = (x+y)^A046660(n) * y^A001221(n) with x, y complex numbers and 0^0 = 1. Then f(n) = Sum_{d|n} g(d)*h(n/d). This is proved for x = 1-y (see Dressler and van de Lune link). - Werner Schulte, Feb 10 2018

Let r, s be some fixed integers. Then we have:

(1) The sequence b(n) = Dirichlet convolution of r^bigomega(n) and s^bigomega(n) is multiplicative with b(p^e) = (r^(e+1)-s^(e+1))/(r-s) for prime p and e >= 0. The case r = s leads to b(p^e) = (e+1)*r^e.

(2) The sequence c(n) = Dirichlet convolution of r^bigomega(n) and mu(n)*s^bigomega(n) is multiplicative with c(p^e) = (r-s)*r^(e-1) and c(1) = 1 for prime p and e > 0 where mu(n) = A008683(n). - Werner Schulte, Feb 20 2019

a(n) is also the length of the composition series for every solvable group of order n. - Miles Englezou, Apr 25 2024

Numbers of the form p*q where p and q are primes, not necessarily distinct.

These numbers are sometimes called semiprimes or 2-almost primes.

Numbers n such that Omega(n) = 2 where Omega(n) = A001222(n) is the sum of the exponents in the prime decomposition of n.

Complement of A100959; A064911(a(n)) = 1. - Reinhard Zumkeller, Nov 22 2004

The graph of this sequence appears to be a straight line with slope 4. However, the asymptotic formula shows that the linearity is an illusion and in fact a(n)/n ~ log(n)/log(log(n)) goes to infinity. See also the graph of A066265 = number of semiprimes < 10^n.

For numbers between 33 and 15495, semiprimes are more plentiful than any other k-almost prime. See A125149.

Numbers that are divisible by exactly 2 prime powers (not including 1). - Jason Kimberley, Oct 02 2011

The (disjoint) union of A006881 and A001248. - Jason Kimberley, Nov 11 2015

An equivalent definition of this sequence is a'(n) = smallest composite number which is not divided by any smaller composite number a'(1),...,a'(n-1). - Meir-Simchah Panzer, Jun 22 2016

The above characterization can be simplified to "Composite numbers not divisible by a smaller term." This shows that this is the equivalent of primes computed via Eratosthenes's sieve, but starting with the set of composite numbers (i.e., complement of 1 union primes) instead of all positive integers > 1. It's easy to see that iterating the method (using Eratosthenes's sieve each time on the remaining numbers, complement of the previously computed set) yields numbers with bigomega = k for k = 0, 1, 2, 3, ..., i.e., {1}, A000040, this, A014612, etc. - M. F. Hasler, Apr 24 2019

For all n except n = 2, a(n) is a deficient number. - Amrit Awasthi, Sep 10 2024

It is reasonable to assume that the "comforting numbers" which John T. Williams found in Chapter 3 of Milne's book "The House at Pooh Corner" are these semiprimes. Winnie-the-Pooh wonders whether he has 14 or 15 honey pots and concludes: "It's sort of comforting." To arrange a semiprime number of honey pots in a rectangular way, let's say on a shelf, with the larger divisor parallel to the wall, there is only one solution and this is for a simple mind like Winnie-the-Pooh comforting. - Ruediger Jehn, Dec 12 2024

The natural numbers 1,2,... are divided into three sets: 1 (the unit), the primes (A000040) and the composite numbers (A002808).

The number of composite numbers <= n (A065855) = n - pi(n) (A000720) - 1.

n is composite iff sigma(n) + phi(n) > 2n. This is a nice result of the well known theorem: For all positive integers n, n = Sum_{d|n} phi(d). For the proof see my contribution to puzzle 76 of Carlos Rivera's Primepuzzles. - Farideh Firoozbakht, Jan 27 2005, Jan 18 2015

The composite numbers have the semiprimes A001358 as primitive elements.

A211110(a(n)) > 1. - Reinhard Zumkeller, Apr 02 2012

A060448(a(n)) > 1. - Reinhard Zumkeller, Apr 05 2012

A086971(a(n)) > 0. - Reinhard Zumkeller, Dec 14 2012

Composite numbers n which are the product of r=A001222(n) prime numbers are sometimes called r-almost primes. Sequences listing r-almost primes are: A000040 (r = 1), A001358 (r = 2), A014612 (r = 3), A014613 (r = 4), A014614 (r = 5), A046306 (r = 6), A046308 (r = 7), A046310 (r = 8), A046312 (r = 9), A046314 (r = 10), A069272 (r = 11), A069273 (r = 12), A069274 (r = 13), A069275 (r = 14), A069276 (r = 15), A069277 (r = 16), A069278 (r = 17), A069279 (r = 18), A069280 (r = 19), A069281 (r = 20). - Jason Kimberley, Oct 02 2011

a(n) = A056608(n) * A160180(n). - Reinhard Zumkeller, Mar 29 2014

Degrees for which there are irreducible polynomials which are reducible mod p for all primes p, see Brandl. - Charles R Greathouse IV, Sep 04 2014

An integer is composite if and only if it is the sum of strictly positive integers in arithmetic progression with common difference 2: 4 = 1 + 3, 6 = 2 + 4, 8 = 3 + 5, 9 = 1 + 3 + 5, etc. - Jean-Christophe Hervé, Oct 02 2014

This statement holds since k+(k+2)+...+k+2(n-1) = n*(n+k-1) = a*b with arbitrary a,b (taking n=a and k=b-a+1 if b>=a). - M. F. Hasler, Oct 04 2014

For n > 4, these are numbers n such that n!/n^2 = (n-1)!/n is an integer (see A056653). - Derek Orr, Apr 16 2015

Let f(x) = Sum_{i=1..x} Sum_{j=2..i-1} cos((2*Pi*x*j)/i). It is known that the zeros of f(x) are the prime numbers. So these are the numbers n such that f(n) > 0. - Michel Lagneau, Oct 13 2015

Numbers n that can be written as solutions of the Diophantine equation n = (x+2)(y+2) where {x,y} in N^2, pairs of natural numbers including zero (cf. Mathematica code and Davis). - Ron R Spencer and Bradley Klee, Aug 15 2016

Numbers n with a partition (containing at least two summands) so that its summands also multiply to n. If n is prime, there is no way to find those two (or more) summands. If n is composite, simply take a factor or several, write those divisors and fill with enough 1's so that they add up to n. For example: 4 = 2*2 = 2+2, 6 = 1*2*3 = 1+2+3, 8 = 1*1*2*4 = 1+1+2+4, 9 = 1*1*1*3*3 = 1+1+1+3+3. - Juhani Heino, Aug 02 2017