| Erdos conjectured that n^2 - 1 = k! has a solution iff n is 5, 11 or 71 (when k is 4, 5 or 7).
Second order linear recurrences y(m)=2y(m-1)+A005563(n)y(m-2),y(0)=y(1)=1, have closed form solutions involving only powers of integers. - Len Smiley (smiley(AT)math.uaa.alaska.edu), 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 (remartin(AT)fas.harvard.edu), 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 n such that 4n + 4 is a square. - Cino Hilliard (hillcino368(AT)gmail.com), 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 iff 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)=A067725/3 - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 06 2007
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
Comment from R. K. Guy, Feb 01 2008 (Start):
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,). [From Paul Curtz, Oct 28 2008]
A053186(a(n)) = 2*n. [From Reinhard Zumkeller, May 20 2009]
Final digit belongs to a periodic sequence: 0, 3, 8, 5, 4, 5, 8, 3, 0, 9. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009] [Comment edited by N. J. A. Sloane, Sep 24 2009]
Comment from A. K. Devaraj, Sep 18 2009: (Start)
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).
(End)
For n>=1, a(n) is the number for which 1/a(n) = 0.0101... (A000035) in base (n+1). [From 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 [From Gary W. Adamson, Jul 15 2010]
Contribution from Gary W. Adamson, Jul 30 2010: (Start)
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). (End)
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)=A002378(n)+Floor(Sqrt(A002378(n))); Pronic number + its root [From Fred Daniel Kline, Sep 16, 2011]
a(n) = maximal determinant of a 2x2 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(n-1) = A008833(n) * A068310(n) for n > 1. [Reinhard Zumkeller, Nov 26 2011]
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