cp's OEIS Frontend

Number of intersection points of diagonals of convex n-gon where no more than two diagonals intersect at any point in the interior.

Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation. - Ignacio Larrosa Cañestro, Apr 09 2002. [See Les Reid link for proof. - N. J. A. Sloane, Apr 02 2016] [See Peter Kagey link for alternate proof. - Sameer Gauria, Jul 29 2025]

Start from cubane and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink on chemistry. - Robert G. Wilson v, Aug 02 2002

For n>0, a(n) = (-1/8)*(coefficient of x in Zagier's polynomial P_(2n,n)). (Zagier's polynomials are used by PARI/GP for acceleration of alternating or positive series.)

Figurate numbers based on the 4-dimensional regular convex polytope called the regular 4-simplex, pentachoron, 5-cell, pentatope or 4-hypertetrahedron with Schlaefli symbol {3,3,3}. a(n)=((n*(n-1)*(n-2)*(n-3))/4!). - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 07 2009

Maximal number of crossings that can be created by connecting n vertices with straight lines. - Cameron Redsell-Montgomerie (credsell(AT)uoguelph.ca), Jan 30 2007

If X is an n-set and Y a fixed (n-1)-subset of X then a(n) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007

Product of four consecutive numbers divided by 24. - Artur Jasinski, Dec 02 2007

The only prime in this sequence is 5. - Artur Jasinski, Dec 02 2007

For strings consisting entirely of 0's and 1's, the number of distinct arrangements of four 1's such that 1's are not adjacent. The shortest possible string is 7 characters, of which there is only one solution: 1010101, corresponding to a(5). An eight-character string has 5 solutions, nine has 15, ten has 35 and so on, congruent to A000332. - Gil Broussard, Mar 19 2008

For a(n)>0, a(n) is pentagonal if and only if 3 does not divide n. All terms belong to the generalized pentagonal sequence (A001318). Cf. A000326, A145919, A145920. - Matthew Vandermast, Oct 28 2008

Nonzero terms = row sums of triangle A158824. - Gary W. Adamson, Mar 28 2009

Except for the 4 initial 0's, is equivalent to the partial sums of the tetrahedral numbers A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009

If the first 3 zeros are disregarded, that is, if one looks at binomial(n+3, 4) with n>=0, then it becomes a 'Matryoshka doll' sequence with alpha=0: seq(add(add(add(i,i=alpha..k),k=alpha..n),n=alpha..m),m=alpha..50). - Peter Luschny, Jul 14 2009

For n>=1, a(n) is the number of n-digit numbers the binary expansion of which contains two runs of 0's. - Vladimir Shevelev, Jul 30 2010

For n>0, a(n) is the number of crossing set partitions of {1,2,..,n} into n-2 blocks. - Peter Luschny, Apr 29 2011

The Kn3, Ca3 and Gi3 triangle sums of A139600 are related to the sequence given above, e.g., Gi3(n) = 2*A000332(n+3) - A000332(n+2) + 7*A000332(n+1). For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 29 2011

For n > 3, a(n) is the hyper-Wiener index of the path graph on n-2 vertices. - Emeric Deutsch, Feb 15 2012

Except for the four initial zeros, number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Aug 31 2012

a(n+3) is the number of different ways to color the faces (or the vertices) of a regular tetrahedron with n colors if we count mirror images as the same.

a(n) = fallfac(n,4)/4! is also the number of independent components of an antisymmetric tensor of rank 4 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015

Does not satisfy Benford's law [Ross, 2012] - N. J. A. Sloane, Feb 12 2017

Number of chiral pairs of colorings of the vertices (or faces) of a regular tetrahedron with n available colors. Chiral colorings come in pairs, each the reflection of the other. - Robert A. Russell, Jan 22 2020

From Mircea Dan Rus, Aug 26 2020: (Start)

a(n+3) is the number of lattice rectangles (squares included) in a staircase of order n; this is obtained by stacking n rows of consecutive unit lattice squares, aligned either to the left or to the right, which consist of 1, 2, 3, ..., n squares and which are stacked either in the increasing or in the decreasing order of their lengths. Below, there is a staircase or order 4 which contains a(7) = 35 rectangles. [See the Teofil Bogdan and Mircea Dan Rus link, problem 3, under A004320]

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a(n+4) is the number of strings of length n on an ordered alphabet of 5 letters where the characters in the word are in nondecreasing order. E.g., number of length-2 words is 15: aa,ab,ac,ad,ae,bb,bc,bd,be,cc,cd,ce,dd,de,ee. - Jim Nastos, Jan 18 2021

From Tom Copeland, Jun 07 2021: (Start)

Aside from the zeros, this is the fifth diagonal of the Pascal matrix A007318, the only nonvanishing diagonal (fifth) of the matrix representation IM = (A132440)^4/4! of the differential operator D^4/4!, when acting on the row vector of coefficients of an o.g.f., or power series.

M = e^{IM} is the matrix of coefficients of the Appell sequence p_n(x) = e^{D^4/4!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^4/4!}, which is that for A025036 aerated with triple zeros, the first column of M.

See A099174 and A000292 for analogous relationships for the third and fourth diagonals of the Pascal matrix. (End)

For integer m and positive integer r >= 3, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (3 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

Row sums of A157703. - Johannes W. Meijer, Mar 07 2009

Number of bottom-row-increasing column-strict arrays of size 3 X n. - Ran Pan, Apr 10 2015

a(n) is the number of rooted semi-labeled or phylogenetic trees with n interior vertices and each interior vertex having out-degree 3. - Albert Alejandro Artiles Calix, Aug 12 2016

The Copeland link gives the associations of this entry with the operator calculus of Appell Sheffer polynomials, the combinatorics of simple set partitions encoded in the Faa di Bruno formula for composition of analytic functions (formal Taylor series), the Pascal matrix, and the geometry of the n-dimensional simplices (hypertriangles, or hypertetrahedra). These, in turn, are related to simple instances of the application of the exponential formula / principle / schema giving the number of not-necessarily-connected objects composed from an ensemble of connected objects. - Tom Copeland, Jun 09 2021

A set partition of m-shape is a partition of a set with cardinality m*n for some n >= 0 such that the sizes of the blocks are m times the parts of the integer partitions of n.

If m = 0, all possible sizes are zero. Thus the number of set partitions of 0-shape is the number of integer partitions of n (partition numbers A000041).

If m = 1, the set is {1, 2, ..., n} and the set of all possible sizes are the integer partitions of n. Thus the number of set partitions of 1-shape is the number of set partitions (Bell numbers A000110).

If m = 2, the set is {1, 2, ..., 2n} and the number of set partitions of 2-shape is the number of set partitions into even blocks A005046.

From Petros Hadjicostas, Aug 06 2019: (Start)

Irwin (1916) proved the following combinatorial result: Assume r_1, r_2, ..., r_n are positive integers and we have r_1*r_2*...*r_n objects. We divide them into r_1 classes of r_2*r_3*...*r_n objects each, then each class into r_2 subclasses of r_3*...*r_n objects each, and so on. We call each such classification, without reference to order, a "classification" par excellence. He proved that the total number of classifications is (r_1*r_2*...*r_n)!/( r1! * (r_2!)^(r_1) * (r_3!)^(r_1*r_2) * ... (r_n!)^(r_1*r_2*...*r_{n-1}) ).

Apparently, this problem appeared in Carmichael's "Theory of Numbers".

This result can definitely be used to prove some special cases of my conjecture below. (End)

In this generalized game of memory n indistinguishable quadruples of matched cards are placed on the vertices of the path of length 4n. A polyomino is a quadruple on four adjacent vertices.

T(n,k) is the number of set partitions of {1..4n} into n sets of 4 with k of the sets being a contiguous set of elements. - Andrew Howroyd, Apr 16 2020

If k is not a factor of n, T(n,k) = 0. If k is a factor of n, T(n,k) = (n!/k!)/(n/k)!^k. If n is a multiple of k, we may obtain T(n,k) by arranging all n people in an ordered line, which can be done in n! ways. Peel off the first n/k people for "group 1", the next n/k people for "group 2", ..., and the last n/k people for "group k". Since the k groups are actually unlabeled, we must divide n! by k! Also, since the ordering of the n/k people within each of the k groups is not of importance, we must now divide by (n/k)!^k. Therefore, T(n,k) = (n!/k!)/(n/k)!^k.

Also, T(2n,n) provide the sequence consisting of the products of consecutive odd integers.

T(n,k) depends on n only through its prime signature. (See A118914.) For example, for fixed k, T(p*q^2,k) will be the same for any pair of distinct primes p and q. Hence we may define T(n,k) = R(s(n),k), where s(n) is the prime signature of n.

Also the number of Krasner factorizations of (x^(n^k)-1) / (x-1) into k polynomials each having n nonzero terms all with coefficient +1. (Krasner and Ranulac, 1937)