cp's OEIS Frontend

Number of points in simple cubic lattice at most n steps from origin.

If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 6-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007

Equals binomial transform of [1, 6, 12, 8, 0, 0, 0, ...] where (1, 6, 12, 8) = row 3 of the Chebyshev triangle A013609. - Gary W. Adamson, Jul 19 2008

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 4, a(n-2) = -coeff(charpoly(A,x),x^(n-3)). - Milan Janjic, Jan 26 2010

a(n) = A005408(n) * A097080(n-1) / 3. - Reinhard Zumkeller, Dec 15 2013

a(n) = D(3,n) where D are the Delannoy numbers (A008288). As such, a(n) gives the number of grid paths from (0,0) to (3,n) using steps that move one unit north, east, or northeast. - David Eppstein, Sep 07 2014

The first comment above can be re-expressed and generalized as follows: a(n) is the number of points in Z^3 that are L1 (Manhattan) distance <= n from any given point. Equivalently, due to a symmetry that is easier to see in the Delannoy numbers array (A008288), as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= 3 from any given point. - Shel Kaphan, Jan 02 2023

Equals the triangular numbers convolved with [1, 3, 1, 0, 0, 0, ...]. - Gary W. Adamson and Alexander R. Povolotsky, May 29 2009

From Ant King, Jun 15 2012: (Start)

a(n) == 1 (mod 5) for all n.

The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 6, 7, 4, 6, 4, 7, 6, 1.

The units' digits of the a(n) form a purely periodic palindromic 4-cycle 1, 6, 6, 1.

(End)

Binomial transform of (1, 5, 5, 0, 0, 0, ...) and second partial sum of (1, 4, 5, 5, 5, ...). - Gary W. Adamson, Sep 09 2015

a(n) = a(-1-n) for all n in Z. - Michael Somos, Jan 25 2019

On the plane start with a single regular pentagon, and repeat the following procedure, "For each edge of any pentagon not already connected to an existing pentagon create a mirror image such that the mirror image does not overlap with an existing pentagon." a(n) is the number of pentagons occupying the plane after n repetitions. - Torlach Rush, Sep 14 2022

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}. - Henry Bottomley, Dec 12 2000

Surround numbers of an n X n square. - Jason Earls, Apr 16 2001

Numbers n such that 2*n + 2 is a perfect square. - Cino Hilliard, Dec 18 2003, Juri-Stepan Gerasimov, Apr 09 2016

The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2 + ... + 2(n+1)^2-1 = (1/2)(2(n+1)^2 - 1 - 2n^2 + 1)(2(n+1)^2 - 1 + 2n^2) = (2n+1)^3. E.g., 2+3+4+5+6+7 = 27 = 3^3, then 8+9+10+...+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005

X values (other than 0) of solutions to the equation 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n) = 2n*(2*n^2 - 1). - Mohamed Bouhamida, Nov 06 2007

Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2. - Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008

Equals row sums of triangle A143593 and binomial transform of [1, 6, 4, 0, 0, 0, ...] with n > 1. - Gary W. Adamson, Aug 26 2008

Start a spiral of square tiles. Trivially the first tile fits in a 1 X 1 square. 7 tiles fit in a 3 X 3 square, 17 tiles fit in a 5 X 5 square and so on. - Juhani Heino, Dec 13 2009

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jan 26 2010

For each n > 0, the recursive series, formula S(b) = 6*S(b-1) - S(b-2) - 2*a(n) with S(0) = 4n^2-4n+1 and S(1) = 2n^2, has the property that every even term is a perfect square and every odd term is twice a perfect square. - Kenneth J Ramsey, Jul 18 2010

Fourth diagonal of A154685 for n > 2. - Vincenzo Librandi, Aug 07 2010

First integer of (2*n)^2 consecutive integers, where the last integer is 3 times the first + 1. As example, n = 2: term = 7; (2*n)^2 = 16; 7, 8, 9, ..., 20, 21, 22: 7*3 + 1 = 22. - Denis Borris, Nov 18 2012

Chebyshev polynomial of the first kind T(2,n). - Vincenzo Librandi, May 30 2014

For n > 0, number of possible positions of a 1 X 2 rectangle in a (n+1) X (n+2) rectangular integer lattice. - Andres Cicuttin, Apr 07 2016

This sequence also represents the best solution for Ripà's n_1 X n_2 X n_3 dots problem, for any 0 < n_1 = n_2 < n_3 = floor((3/2)*(n_1 - 1)) + 1. - Marco Ripà, Jul 23 2018

Also, the number of regions the plane can be cut into by two overlapping concave (2n)-gons. - Joshua Zucker, Nov 05 2002

If X is an n-set and Y_i (i=1,2,3) are mutually disjoint 2-subsets of X then a(n-5) is equal to the number of 5-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007

Binomial transform of a(n) is A055580(n). - Wesley Ivan Hurt, Apr 15 2014

The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as a(n)^2 - A002522(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014

Also the least number of unit cubes required, at the n-th iteration, to surround a 3D solid built from unit cubes, in order to hide all its visible faces, starting with a unit cube. - R. J. Cano, Sep 29 2015

Also, coordination sequence for "tfs" 3D uniform tiling. - N. J. A. Sloane, Feb 10 2018

Also, the number of n-th order specular reflections arriving at a receiver point from an emitter point inside a cuboid with reflective faces. - Michael Schutte, Sep 18 2018

Subsequence of A004613: all numbers in this sequence have all prime factors of the form 4k+1. E.g., 40001 = 13*17*181, 13 = 4*3 + 1, 17 = 4*4 + 1, 181 = 4*45 + 1. - Cino Hilliard, Aug 26 2006, corrected by Franklin T. Adams-Watters, Mar 22 2011

A000466(n), A008586(n) and a(n) are Pythagorean triples. - Zak Seidov, Jan 16 2007

Solutions x of the Mordell equation y^2 = x^3 - 3a^2 - 1 for a = 0, 1, 2, ... - Michel Lagneau, Feb 12 2010

Ulam's spiral (NW spoke). - Robert G. Wilson v, Oct 31 2011

For n >= 1, a(n) is numerator of radius r(n) of circle with sagitta = n and cord length = 1. The denominator is A008590(n). - Kival Ngaokrajang, Jun 13 2014

a(n)+6 is prime for n = 0..6 and for n = 15..20. - Altug Alkan, Sep 28 2015

If X is an n-set and Y a fixed (n-3)-subset of X then a(n-3) is equal to the number of 2-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007

Bisection of A165157. - Jaroslav Krizek, Sep 05 2009

a(n) is the number of (w,x,y) having all terms in {0,...,n} and w=x+y-1. - Clark Kimberling, Jun 02 2012

Numbers m >= 0 such that 8m+25 is a square. - Bruce J. Nicholson, Jul 26 2017

a(n-1) = 3*(n-1) + (n-1)*(n-2)/2 is the number of connected, loopless, non-oriented, multi-edge vertex-labeled graphs with n edges and 3 vertices. Labeled multigraph analog of A253186. There are 3*(n-1) graphs with the 3 vertices on a chain (3 ways to label the middle graph, n-1 ways to pack edges on one of connections) and binomial(n-1,2) triangular graphs (one way to label the graphs, pack 1 or 2 or ...n-2 on the 1-2 edge, ...). - R. J. Mathar, Aug 10 2017

a(n) is also the number of vertices of the quiver for PGL_{n+1} (see Shen). - Stefano Spezia, Mar 24 2020

Starting from a(2) = 7, this is the 4th column of the array: natural numbers written by antidiagonals downwards. See the illustration by Kival Ngaokrajang and the cross-references. - Andrey Zabolotskiy, Dec 21 2021

The A_n lattice consists of all vectors v = (x_1,...,x_(n+1)) in Z^(n+1) such that x_1 + ... + x_(n+1) = 0. The lattice is equipped with the norm ||v|| = 1/2*(|x_1| + ... + |x_(n+1)|). Pairs of lattice points (v,w) in the product lattice A_n x A_n have norm ||(v,w)|| = ||v|| + ||w||. Then the k-th term in the crystal ball sequence for the A_n x A_n lattice gives the number of such pairs (v,w) for which ||(v,w)|| is less than or equal to k.

This array has a remarkable relationship with Apery's constant zeta(3). The row (or column) and main diagonal entries of the array occur in series acceleration formulas for zeta(3). For row n entries there holds zeta(3) = (1+1/2^3+...+1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)). Also, as consequence of Apery's proof of the irrationality of zeta(3), we have a series acceleration formula along the main diagonal of the table: zeta(3) = 6 * sum {n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Apery's result appears to generalize to the other diagonals of the table. Calculation suggests the following result may hold: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2 +3*n*k +k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).

For the corresponding results for the constant zeta(2), related to the crystal ball sequences of the lattices A_n, see A108625. For corresponding results for log(2), coming from either the crystal ball sequences of the hypercubic lattices A_1 x ... x A_1 or the lattices of type C_n, see A008288 and A142992 respectively.

Final digits of a(n), i.e., a(n) mod 10, are repeated periodically with period of length 5 {1,5,5,5,9}. There is a symmetry in this list since the sum of two numbers equally distant from the ends is equal to 10 = 1 + 9 = 5 + 5 = 2*5. Last two digits of a(n), i.e., a(n) mod 100, are repeated periodically with period of length 50. - Alexander Adamchuk, Aug 11 2006

a(n) = VarScheme(n,2) in the scheme displayed in A128195. - Peter Luschny, Feb 26 2007

If Y is a 3-subset of a 2n-set X then, for n >= 2, a(n-2) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007

The numbers are the constant number found in magic squares of order n, where n is an odd number, see the comment in A006003. A Magic Square of side 1 is 1; 3 is 15; 5 is 65 and so on. - David Quentin Dauthier, Nov 07 2008

Two times the area of the triangle with vertices at (0,0), ((n - 1)^2, n^2), and (n^2, (n - 1)^2). - J. M. Bergot, Jun 25 2013

Bisection of A006003. - Omar E. Pol, Sep 01 2018

Construct an array M with M(0,n) = 2*n^2 + 4*n + 1 = A056220(n+1), M(n,0) = 2*n^2 + 1 = A058331(n) and M(n,n) = 2*n*(n+1) + 1 = A001844(n). Row(n) begins with all the increasing odd numbers from A058331(n) to A001844(n) and column(n) begins with all the decreasing odd numbers from A056220(n+1) to A001844(n). The sum of the terms in row(n) plus those in column(n) minus M(n,n) equals a(n+1). The first five rows of array M are [1, 7, 17, 31, 49, ...]; [3, 5, 15, 29, 47, ...]; [9, 11, 13, 27, 45, ...]; [19, 21, 23, 25, 43, ...]; [33, 35, 37, 39, 41, ...]. - J. M. Bergot, Jul 16 2013 [This contribution was moved here from A047926 by Petros Hadjicostas, Mar 08 2021.]

For n>=2, these are the primitive sides s of squares of type 2 described in A344332. - Bernard Schott, Jun 04 2021

(a(n) + 1) / 2 = A212133(n) is the number of cells in the n-th rhombic-dodecahedral polycube. - George Sicherman, Jan 21 2024

Also, primes of the form 4*k+1 which are the hypotenuse of one and only one right triangle with integral legs. - Cino Hilliard, Mar 16 2003

Centered square primes (i.e., prime terms of centered squares A001844). - Lekraj Beedassy, Jan 21 2005

Primes of the form 2*k*(k-1)+1. - Juri-Stepan Gerasimov, Apr 27 2010

Equivalently, primes of the form (m^2+1)/2 (take m=2*j+1). These primes a(n) have nontrivial solutions of x^2 == 1 (Modd a(n)) given by x=x(n)=A002731(n). For Modd n see a comment on A203571. See also A206549 for such solutions for primes of the form 4*k+1, given in A002144.

E.g., a(3)=41, A002731(3)=9, 9^2=81, floor(81/41)=1 (odd),

-81 = -2*41 + 1 == 1 (mod 2*41), hence 9^2 == 1 (Modd 41). - Wolfdieter Lang, Feb 24 2012

Also primes of the form 4*k+1 that are the smallest side length of one and only one integer Soddyian triangle (see A230812). - Frank M Jackson, Mar 13 2014

Also, primes of the form (m^2+1)/2. - Zak Seidov, May 01 2014

Note that ((2n+1)^2 + 1)/2 = n^2 + (n+1)^2. - Thomas Ordowski, May 25 2015

Primes p such that 2p-1 is a square. - Thomas Ordowski, Aug 27 2016

Primes in the main diagonal of A000027 when represented as an array read by antidiagonals. - Clark Kimberling, Mar 12 2023

The diophantine equation x^2 + ... + (x + r)^2 = p may be rewritten to A*x^2 + B*x + C = p, where A = (r + 1), B = r*(r + 1), C = r*(r + 1)*(2*r + 1)/6. If gcd(A, B, C) > 1 no solution for a prime p exists. The gcd(A, B, C) = 1 holds only for r = 1, 2, 5 (gcd is the greatest common divisor). For r = 1 we have x^2 + (x + 1)^2 = p, thus for x from A027861 we calculate primes p from A027862. For r = 2 we have x^2 + (x + 1)^2 + (x + 2)^2 = p, thus for x from A027863 we calculate primes p from A027864. For r = 5 we have x^2 + ... + (x + 5)^2 = p, thus for x from A027866 we calculate primes p from A027867. - Ctibor O. Zizka, Oct 04 2023