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(m1) + A005563(n)y(m2), 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(ij) then det(M_n) = (1)^(n1)*a(k1)^(n1).  Benoit Cloitre, May 28 2002
Also numbers n such that 4n + 4 is a square.  Cino Hilliard, Dec 18 2003
The function sqrt(x^2 + 1), starting with 1, produces an integer after n(n+2) 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*(1A079978(n)).  Reinhard Zumkeller, Oct 16 2006
A129296(n) = number of divisors of a(n+1) that are not greater than n.  Reinhard Zumkeller, Apr 09 2007
Sequence allows us to find X values of the equation: X^3 + X^2 = Y^2. To find Y values: b(n) = n(n+1)(n+2).  Mohamed Bouhamida (bhmd95(AT)yahoo.fr), 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 (bhmd95(AT)yahoo.fr), 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 lilypads) 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)
For n > 0: A143053(a(n)) = A000290(n+1).  Reinhard Zumkeller, Jul 20 2008
a(n+1) = terms of rank 0, 1, 3, 6, 10 = A000217 of A120072 (3, 8, 5, 15).  Paul Curtz, Oct 28 2008
A053186(a(n)) = 2*n.  Reinhard Zumkeller, May 20 2009
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 (bhmd95(AT)yahoo.fr), 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 noninteger 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} ((k2)*i(k3)). Thus P_0(n) = 2*nn^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) = A002378(n) + floor(sqrt(A002378(n))); Pronic number + its root.  Fred Daniel Kline, Sep 16 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
a(n1) = A008833(n) * A068310(n) for n > 1.  Reinhard Zumkeller, Nov 26 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 halfinteger value), the quantummechanical expectation value of the square of the magnitude of its spin vector evaluates to <S^2> = 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(n1) + A005563(m2)*a(n2), 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(n1) + A005563(m2) * a(n2), n >= 2, a(0) = 0, a(1) = 1, then lim a(n+1)/a(n) = m as n approaches infinity.  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 nth kgonal number, then a(n) = t*P_{s}(n+2)  s*P_{t}(n+2) for s=t+1.  Bruno Berselli, Sep 04 2014
A253607(a(n)) = 1.  Reinhard Zumkeller, Jan 05 2015
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
n > 0, a(n) % (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(n1) 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 n1 occurs among the first (n+1)! terms of A055881.  R. J. Cano, Dec 21 2016
