cp's OEIS Frontend

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2 - 2n + 2 (Wielandt).

a(n) = Phi_4(n), where Phi_k is the k-th cyclotomic polynomial.

As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}. - Benoit Cloitre, Dec 07 2001

a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n-1) + a(n+1))/2 - 1. E.g., 2 = (1+5)/2 - 1, 5 = (2+10)/2 - 1. - Amarnath Murthy, Jul 29 2003

Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,...). - Franz Vrabec, Jan 23 2006

Number of {12,1*2*,21}-avoiding signed permutations in the hyperoctahedral group.

The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^2-2n+2. - Sébastien Dumortier, Jun 16 2005

Also, numbers m such that m^3 - m^2 is a square, (n*(1 + n^2))^2. - Zak Seidov

1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006

For n >= 1, a(n-1) is the minimal number of choices from an n-set such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards). - Rick L. Shepherd, Nov 18 2007

Positive X values of solutions to the equation X^3 + (X - 1)^2 + X - 2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X - 1)^2 + X - 2 = X^3 + X^2 - X - 1 = (X - 1)(X^2 + 2X + 1) = (X - 1)*(X + 1)^2 it means: (X - 1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2). - Mohamed Bouhamida, Nov 29 2007

{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 2009

Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... . - Johannes W. Meijer, Jun 12 2010

For n > 0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26. - Gary W. Adamson, Jul 15 2010

The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[-1](x), x >= 0, m >= 1, with f^(m) the m-th derivative and f^[-1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= -(sqrt(a(n))-n) and A=A(n)=(fallfac(p(n),2*n))^(-p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (there are also two solutions for positive p, see the corresponding comment in A087475). - Wolfdieter Lang, Oct 21 2010

n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.

a(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011

Also numbers k such that 4*k-4 is a square. Hence this sequence is the union of A053755 and A069894. - Arkadiusz Wesolowski, Aug 02 2011

a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)-cage. - Jason Kimberley, Oct 17 2011

Left edge of the triangle in A195437: a(n+1) = A195437(n,0). - Reinhard Zumkeller, Nov 23 2011

If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 is divisible by 24. - Vincenzo Librandi, Apr 14 2014

The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as A005899(n)^2 - a(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014

a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014

a(n-1) is the maximum number of stages in the Gale-Shapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.). - Melvin Peralta, Feb 07 2016

Because of Fermat's little theorem, a(n) is never divisible by 3. - Altug Alkan, Apr 08 2016

For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less. - Melvin Peralta, Jan 21 2017

Also the limit as q->1^- of the unimodal polynomial (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials after making the restriction to partitions of size <= 1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s->G_infinity=G: the q-binomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984. - Bryan T. Ek, Apr 11 2018

a(n) is the smallest number congruent to both 1 (mod n) and 2 (mod n+1). - David James Sycamore, Apr 04 2019

a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition. - Richard Stanley, Dec 22 2022

From Klaus Purath, Apr 03 2025: (Start)

The odd prime factors of these terms are always of the form 4*k + 1.

All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.

It follows from the above that this sequence is a subsequence of A031396. (End)

Number of edges in the join of two star graphs, each of order n, S_n * S_n. - Roberto E. Martinez II, Jan 07 2002

Number of vertices in the hexagonal triangle T(n-2) (see the He et al. reference). - Emeric Deutsch, Nov 14 2014

Positive X values of solutions to the equation X^3 + (X - 3)^2 + X - 6 = Y^2. To prove that X = n^2 + 4n + 1: Y^2 = X^3 + (X - 3)^2 + X - 6 = X^3 + X^2 - 5X + 3 = (X + 3)(X^2 - 2X + 1) = (X + 3)*(X - 1)^2 it means: X = 1 or (X + 3) must be a perfect square, so X = k^2 - 3 with k >= 2. we can put: k = n + 2, which gives: X = n^2 + 4n + 1 and Y = (n + 2)(n^2 + 4n). - Mohamed Bouhamida, Nov 29 2007

Equals binomial transform of [1, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008

Let C = 2 + sqrt(3) = 3.732...; and 1/C = 0.267...; then a(n) = (n - 2 + C) * (n - 2 + 1/C). Example: a(5) = 46 = (5 + 3.732...)*(5 + 0.267...). - Gary W. Adamson, Jul 29 2009

a(n), n >= 0, with a(0) = -3 and a(1) = -2, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 12 for b = 2*n. In general D = b^2 - 4*a*c > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013

If A(n) is a 3 X 3 Khovanski matrix having 1 below the main diagonal, n on the main diagonal, and n^3 above the main diagonal, then (Det(A(n)) - 2*n^3) / n^4 = a(n). - Gary Detlefs, Nov 12 2013

Imagine a large square containing four smaller square "holes" of equal size: Let x = large square side and y = smaller square side; considering instances where the area of this shape [x^2 - 4*y^2] equals the length of its perimeter, [4*(x + 4*y)]. When y is an integer n, the above equation is satisfied by x = 2 + 2*sqrt(a(n)). - Peter M. Chema, Apr 10 2016

a(n+1) is the number of distinct linear partitions of 2 X n grid points. A linear partition is a way to partition given points by a line into two nonempty subsets. Details can be found in Pan's link. - Ran Pan, Jun 06 2016

Numbers represented as 141 in number base B: 141(5) = 46, 141(6) = 61 and, if 'digits' larger than (B-1) are allowed, 141(2) = 13, 141(3) = 22, 141(4) = 33. - Ron Knott, Nov 14 2017

Equivalently, numbers k such that both k/2 and k+1 are squares. - Karl-Heinz Hofmann, Sep 20 2022

Equivalently, numbers k such that the k-dimensional volume and total (k-1)-dimensional volume are equal, with side length being a positive integer, for all regular polyhedra constructible in k dimensions. - Matt Moir, Jul 09 2024

a(n) = number of row of Pascal's triangle in which three consecutive entries appear in the ratio n : n+1 : n+2 (valid for n = 0 if you consider a position of -1 to have value 0). E.g., entries in the ratio 1:2:3 appear in row 14 (1001, 2002, 3003); entries in the ratio 2:3:4 appear in row 34 (927983760, 1391975640, 1855967520); and so on. (The position within the row is given by A091823). - Howard A. Landman, Mar 08 2004

a(n)*(a(n)+1) is an oblong number (Cf. A002378) with the property that the product with the oblong numbers n*(n+1) or (n+1)*(n+2) both are again oblong numbers. Example: For n=3 we have (62*63)*(3*4) = 216*217 and (62*63)*(4*5) = 279*280. - Herbert Kociemba, Apr 13 2008

For n > 0, Hermite polynomial H_2(n) = 4*n^2 - 2. - Vincenzo Librandi, Aug 07 2010

The identity (4*n^2-2)^2 - (n^2-1)*(4*n)^2 = 4 can be written as a(n+1)^2 - A132411(n+2)*A008586(n+2)^2 = 4. - Vincenzo Librandi, Jun 16 2014

Equivalently: positive integers k congruent to 2 mod 4 (A016825) such that k$ / (k/2+1)! is a square when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692, A349496 and A349766 for further information). Integers k multiple of 4 such that that k$ / (k/2+1)! is a square are in A035008. - Bernard Schott, Dec 05 2021

The integers are written in blocks of lengths 1, 3, 5, 7, 9, ... . The first number in the block is moved to the center of the block, and then the numbers are written alternately to the left and the right. The block of length 2n-1 ends with n^2, which is not moved.

Let S = Sum_{i >= 1} s(i) be a not necessarily converging series and let T = Sum_{i >= 1} s(a(i)). Then if S converges so does T. On the other hand there are examples where T converges but S does not (for example S = 1 + 1 + 0 - 1 + 1/2 + 1/2 + 0 - 1/2 - 1/2 + 1/3 (3 times) + 0 - 1/3 (3 times) + 1/5 (5 times) + 0 - 1/5 (5 times) + ...). [Conway]

From Reinhard Zumkeller, Mar 08 2010: (Start)

a(n + 2*A003059(n)) = a(n) + 2*A003059(n) - 1;

a(A002522(n-1)) = A132411(n); a(A002061(n)) = A002522(n-1). (End)

The sum of the rows is n^3+(n+1)^3 [A005898] (1,9,35,91,189,...). - Vincenzo Librandi, Feb 22 2010

Every positive integer occurs exactly once; hence, as a sequence, A185725 is a permutation of the positive integers. The square with corners T(0,0)=1 and T(n,n)=n^2 is occupied by the numbers 1,2,...,n^2.

T(n,1)=n^2 (A000290)

T(n,n)=(n-1)^2+1 (A002522)

T(1,k)=k^2-1 (A132411).