cp's OEIS Frontend

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

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005

Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006

A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007

Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007

Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008

a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009

sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009

When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010

a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010

The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011

Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012

Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012

Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013

a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013

a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014

A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015

An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015

Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015

Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017

The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017

From Klaus Purath, Mar 18 2019: (Start)

Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with

x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.

But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)

a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019

a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021

Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021

(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022

The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

A253607(a(n)) < 0. - Reinhard Zumkeller, Jan 05 2015

Integers m such that A000196(m) = A079643(m). - Firas Melaih, Dec 10 2020

Also possible values of floor(x*floor(x)) for real x >= 1. - Jianing Song, Feb 16 2021

A253607(a(n)) > 0. - Reinhard Zumkeller, Jan 05 2015

Also possible values of floor(x*floor(x)) for real x < 1. - Jianing Song, Feb 16 2021

If all pairs of consecutive terms (i,i+1) (such as (0,1), (1,2), (2,3), ...) are erased, the original sequence appears; see also A253607.

T(n,n-k) + T(n,n+k) = 4*k - 1 for k = 1..n;

T(n+m,k) = T(n,k) for m > 0, k = 0 .. 2*n.