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)

Erdős conjectured that n^2 - 1 = k! has a solution if and only if n is 5, 11 or 71 (when k is 4, 5 or 7).

Second-order linear recurrences y(m) = 2y(m-1) + a(n)*y(m-2), y(0) = y(1) = 1, have closed form solutions involving only powers of integers. - Len Smiley, Dec 08 2001

Number of edges in the join of two cycle graphs, both of order n, C_n * C_n. - Roberto E. Martinez II, Jan 07 2002

Let k be a positive integer, M_n be the n X n matrix m_(i,j) = k^abs(i-j) then det(M_n) = (-1)^(n-1)*a(k-1)^(n-1). - Benoit Cloitre, May 28 2002

Also numbers k such that 4*k + 4 is a square. - Cino Hilliard, Dec 18 2003

For each term k, the function sqrt(x^2 + 1), starting with 1, produces an integer after k iterations. - Gerald McGarvey, Aug 19 2004

a(n) mod 3 = 0 if and only if n mod 3 > 0: a(A008585(n)) = 2; a(A001651(n)) = 0; a(n) mod 3 = 2*(1-A079978(n)). - Reinhard Zumkeller, Oct 16 2006

a(n) is the number of divisors of a(n+1) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007

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

Sequence allows us to find X values of the equation: X + (X + 1)^2 + (X + 2)^3 = Y^2. To prove that X = n^2 + 2n: Y^2 = X + (X + 1)^2 + (X + 2)^3 = X^3 + 7*X^2 + 15X + 9 = (X + 1)(X^2 + 6X + 9) = (X + 1)*(X + 3)^2 it means: (X + 1) must be a perfect square, so X = k^2 - 1 with k>=1. we can put: k = n + 1, which gives: X = n^2 + 2n and Y = (n + 1)(n^2 + 2n + 3). - Mohamed Bouhamida, Nov 12 2007

From R. K. Guy, Feb 01 2008: (Start)

Toads and Frogs puzzle:

This is also the number of moves that it takes n frogs to swap places with n toads on a strip of 2n + 1 squares (or positions, or lily pads) where a move is a single slide or jump, illustrated for n = 2, a(n) = 8 by

T T - F F

T - T F F

T F T - F

T F T F -

T F - F T

- F T F T

F - T F T

F F T - T

F F - T T

I was alerted to this by the Holton article, but on consulting Singmaster's sources, I find that the puzzle goes back at least to 1867.

Probably the first to publish the number of moves for n of each animal was Edouard Lucas in 1883. (End)

a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15). - Paul Curtz, Oct 28 2008

Row 3 of array A163280, n >= 1. - Omar E. Pol, Aug 08 2009

Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. - Mohamed Bouhamida, Sep 04 2009 [Comment edited by N. J. A. Sloane, Sep 24 2009]

Let f(x) be a polynomial in x. Then f(x + n*f(x)) is congruent to 0 (mod f(x)); here n belongs to N. There is nothing interesting in the quotients f(x + n*f(x))/f(x) when x belongs to Z. However, when x is irrational these quotients consist of two parts, a) rational integers and b) integer multiples of x. The present sequence represents the non-integer part when the polynomial is x^2 + x + 1 and x = sqrt(2), f(x+n*f(x))/f(x) = A056108(n) + a(n)*sqrt(2). - A.K. Devaraj, Sep 18 2009

For n >= 1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). - Rick L. Shepherd, Sep 27 2009

For n > 0, continued fraction [n, 1, n] = (n+1)/a(n); e.g., [6, 1, 6] = 7/48. - Gary W. Adamson, Jul 15 2010

Starting (3, 8, 15, ...) = binomial transform of [3, 5, 2, 0, 0, 0, ...]; e.g., a(3) = 15 = (1*3 + 2*5 +1*2) = (3 + 10 + 2). - Gary W. Adamson, Jul 30 2010

a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n} ((k-2)*i-(k-3)). Thus P_0(n) = 2*n-n^2 and a(n) = -P_0(n+2). See also A067998 and for the case k=1 A080956. - Peter Luschny, Jul 08 2011

a(n) is the maximal determinant of a 2 X 2 matrix with integer elements from {1, ..., n+1}, so the maximum determinant of a 2x2 matrix with integer elements from {1, ..., 5} = 5^2 - 1 = a(4) = 24. - Aldo González Lorenzo, Oct 12 2011

Using four consecutive triangular numbers t1, t2, t3 and t4, plot the points (0, 0), (t1, t2), and (t3, t4) to create a triangle. Twice the area of this triangle are the numbers in this sequence beginning with n = 1 to give 8. - J. M. Bergot, May 03 2012

Given a particle with spin S = n/2 (always a half-integer value), the quantum-mechanical expectation value of the square of the magnitude of its spin vector evaluates to = S(S+1) = n(n+2)/4, i.e., one quarter of a(n) with n = 2S. This plays an important role in the theory of magnetism and magnetic resonance. - Stanislav Sykora, May 26 2012

Twice the harmonic mean [H(x, y) = (2*x*y)/(x + y)] of consecutive triangular numbers A000217(n) and A000217(n+1). - Raphie Frank, Sep 28 2012

Number m such that floor(sqrt(m)) = floor(m/floor(sqrt(m))) - 2 for m > 0. - Takumi Sato, Oct 10 2012

The solutions of equation 1/(i - sqrt(j)) = i + sqrt(j), when i = (n+1), j = a(n). For n = 1, 2 + sqrt(3) = 3.732050.. = A019973. For n = 2, 3 + sqrt(8) = 5.828427... = A156035. - Kival Ngaokrajang, Sep 07 2013

The integers in the closed form solution of a(n) = 2*a(n-1) + a(m-2)*a(n-2), n >= 2, a(0) = 0, a(1) = 1 mentioned by Len Smiley, Dec 08 2001, are m and -m + 2 where m >= 3 is a positive integer. - Felix P. Muga II, Mar 18 2014

Let m >= 3 be a positive integer. If a(n) = 2*a(n-1) + a(m-2) * a(n-2), n >= 2, a(0) = 0, a(1) = 1, then lim_{n->oo} a(n+1)/a(n) = m. - Felix P. Muga II, Mar 18 2014

For n >= 4 the Szeged index of the wheel graph W_n (with n + 1 vertices). In the Sarma et al. reference, Theorem 2.7 is incorrect. - Emeric Deutsch, Aug 07 2014

If P_{k}(n) is the n-th k-gonal number, then a(n) = t*P_{s}(n+2) - s*P_{t}(n+2) for s=t+1. - Bruno Berselli, Sep 04 2014

For n >= 1, a(n) is the dimension of the simple Lie algebra A_n. - Wolfdieter Lang, Oct 21 2015

Finding all positive integers (n, k) such that n^2 - 1 = k! is known as Brocard's problem, (see A085692). - David Covert, Jan 15 2016

For n > 0, a(n) mod (n+1) = a(n) / (n+1) = n. - Torlach Rush, Apr 04 2016

Conjecture: When using the Sieve of Eratosthenes and sieving (n+1..a(n)), with divisors (1..n) and n>0, there will be no more than a(n-1) composite numbers. - Fred Daniel Kline, Apr 08 2016

a(n) mod 8 is periodic with period 4 repeating (0,3,0,7), that is a(n) mod 8 = 5/2 - (5/2) cos(n*Pi) - sin(n*Pi/2) + sin(3*n*Pi/2). - Andres Cicuttin, Jun 02 2016

Also for n > 0, a(n) is the number of times that n-1 occurs among the first (n+1)! terms of A055881. - R. J. Cano, Dec 21 2016

The second diagonal of composites (the only prime is number 3) from the right on the Klauber triangle (see Kival Ngaokrajang link), which is formed by taking the positive integers and taking the first 1, the next 3, the following 5, and so on, each centered below the last. - Charles Kusniec, Jul 03 2017

Also the number of independent vertex sets in the n-barbell graph. - Eric W. Weisstein, Aug 16 2017

Interleaving of A000466 and A033996. - Bruce J. Nicholson, Nov 08 2019

a(n) is the number of degrees of freedom in a triangular cell for a Raviart-Thomas or Nédélec first kind finite element space of order n. - Matthew Scroggs, Apr 22 2020

From Muge Olucoglu, Jan 19 2021: (Start)

For n > 1, a(n-2) is the maximum number of elements in the second stage of the Quine-McCluskey algorithm whose minterms are not covered by the functions of n bits. At n=3, we have a(3-2) = a(1) = 1*(1+2) = 3 and f(A,B,C) = sigma(0,1,2,5,6,7).

.

0 1 2 5 6 7

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

*(0,1)| X X

(0,2)| X X

(1,5)| X X

*(2,6)| X X

*(5,7)| X X

(6,7)| X X

.

*: represents the elements that are covered. (End)

1/a(n) is the ratio of the sum of the first k odd numbers and the sum of the next n*k odd numbers. - Melvin Peralta, Jul 15 2021

For n >= 1, the continued fraction expansion of sqrt(a(n)) is [n; {1, 2n}]. - Magus K. Chu, Sep 09 2022

Number of diagonals parallel to an edge in a regular (2*n+4)-gon (cf. A367204). - Paolo Xausa, Nov 21 2023

For n >= 1, also the number of minimum cyclic edge cuts in the (n+2)-trapezohedron graph. - Eric W. Weisstein, Nov 21 2024

For n >= 1, a(n) is the sum of the interior angles of a polygon with n+2 sides, in radians, multiplied by (n+2)/Pi. - Stuart E Anderson, Aug 06 2025

For n >= 2, least m >= 1 such that f(m, n) = 0 where f(m,n) = Sum_{i=0..m} Sum_{k= 0..i} (-1)^k*(floor(i/n^k) - n*floor(i/n^(k+1))). - Benoit Cloitre, May 02 2004

For n >= 3, the a(n)-th row of Pascal's triangle always contains a triple forming an arithmetic progression. - Lekraj Beedassy, Jun 03 2004

Let C = 1 + sqrt(2) = 2.414213...; and 1/C = 0.414213... Then a(n) = (n + 1 + 1/C) * (n + 1 - C). Example: a(6) = 34 = (7 + 0.414...) * (7 - 2.414...). - Gary W. Adamson, Jul 29 2009

The sequence (n-4)^2-2, n = 7, 8, ... enumerates the number of non-isomorphic sequences of length n, with entries from {1, 2, 3} and no two adjacent entries the same, that minimally contain each of the thirteen rankings of three players (111, 121, 112, 211, 122, 212, 221, 123, 132, 213, 231, 312, 321) as embedded order isomorphic subsequences. By "minimally", we mean that the n-th symbol is necessary for complete inclusion of all thirteen words. See the arXiv paper below for proof. If n = 7, these sequences are 1213121, 1213212, 1231213, 1231231, 1231321, 1232123, and 1232132, and for each case, there are 3! = 6 isomorphs. - Anant Godbole, Feb 20 2013

a(n), n >= 0, with a(0) = -2, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 8 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

With a different offset, this is 2*n^2 - (n + 1)^2, which arises in one explanation of why Bertrand's postulate does not automatically prove Legendre's conjecture: as n gets larger, so does the range of numbers that can have primes that satisfy Bertrand's postulate yet do nothing for Legendre's conjecture. - Alonso del Arte, Nov 06 2013

x*(x + r*y)^2 + y*(y + r*x)^2 can be written as (x + y)*(x^2 + s*x*y + y^2). For r >= 0, the sequence gives the values of s: in fact, s = (r + 1)^2 - 2. - Bruno Berselli, Feb 20 2019

For n >= 2, the continued fraction expansion of sqrt(a(n)) is [n-1; {1, n-2, 1, 2n-2}]. For n=2, this collapses to [1; {2}]. - Magus K. Chu, Sep 06 2022

Permutation of the natural numbers, inverse: A143052;

A143053(n) = a(a(n));

a(A000290(n)) = A005563(n-1);

a(A002522(n)) = A000290(n+1);

for n>1: GCD(a(n^2),a(n^2+1)) = A050873(A000290(n),A002522(n)) =

A022998(n+1).

Permutation of the natural numbers, inverse: A143053;

For n>1: a(A008865(n))=A000290(n),

a(A000290(n))=A000290(n)-A005408(n-2).