
COMMENTS

These are Hogben's central polygonal numbers denoted by the symbol
...2....
....P...
...2.n..
(P with three attachments).
Also the maximal number of 1's that an n X n invertible {0,1} matrix can have. (See Halmos for proof.)  Felix Goldberg (felixg(AT)tx.technion.ac.il), Jul 07 2001
a(n) = Phi_6(n) = Phi_3(n1), where Phi_k is the kth cyclotomic polynomial.
Maximal number of interior regions formed by n intersecting circles, for n >= 1.  Amarnath Murthy, Jul 07 2001
The terms are the smallest of n consecutive odd numbers whose sum is n^3: 1, 3 + 5 = 8 = 2^3, 7 + 9 + 11 = 27 = 3^3, etc.  Amarnath Murthy, May 19 2001
(n*a(n+1)+1)/(n^2+1) is the smallest integer of the form (n*k+1)/(n^2+1).  Benoit Cloitre, May 02 2002
For n >= 3, a(n) is also the number of cycles in the wheel graph W(n) of order n.  Sharon Sela (sharonsela(AT)hotmail.com), May 17 2002
Let b(k) be defined as follows: b(1) = 1 and b(k+1) > b(k) is the smallest integer such that sum(i=b(k), b(k+1), 1/sqrt(i)) > 2; then b(n) = a(n) for n > 0.  Benoit Cloitre, Aug 23 2002
Drop the first three terms. Then n*a(n) + 1 = (n+1)^3. E.g., 7*1 + 1 = 8 = 2^3, 13*2 + 1 = 27 = 3^3, 21*3 + 1 = 64 = 4^3, etc.  Amarnath Murthy, Oct 20 2002
Arithmetic mean of next 2n  1 numbers.  Amarnath Murthy, Feb 16 2004
The nth term of an arithmetic progression with first term 1 and common difference n: a(1) = 1 > 1, 2, 3, 4, 5, ...; a(2) = 3 > 1, 3, ...; a(3) = 7 > 1, 4, 7, ...; a(4) = 13 > 1, 5, 9, 13, ...  Amarnath Murthy, Mar 25 2004
Number of walks of length 3 between any two distinct vertices of the complete graph K_{n+1} (n >= 1). Example: a(2) = 3 because in the complete graph ABC we have the following walks of length 3 between A and B: ABAB, ACAB and ABCB.  Emeric Deutsch, Apr 01 2004
Narayana transform of [1, 2, 0, 0, 0...] = [1, 3, 7, 13, 21...]. Let M = the infinite lower triangular matrix of A001263 and let V = the Vector [1, 2, 0, 0, 0...]. Then A002061 starting (1, 3, 7, ...) = M * V.  Gary W. Adamson, Apr 25 2006
The sequence 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, ... is the trajectory of 3 under repeated application of the map n > n + 2 * square excess of n, cf. A094765.
Also n^3 mod (n^2+1).  Zak Seidov, Aug 31 2006
Also, omitting the first 1, the main diagonal of A081344.  Zak Seidov, Oct 05 2006
Ignoring the first ones, these are rectangular parallelepipeds with integer dimensions that have integer interior diagonals. Using Pythagoras: sqrt(a^2 + b^2 + c^2) = d, an integer; then this sequence: sqrt(n^2 + (n+1)^2 + (n(n+1))^2) = 2T_n + 1 is the first and most simple example. Problem: Are there any integer diagonals which do not satisfy the following general formula? sqrt((k*n)^2 + (k*(n+(2*m+1)))^2 + (k*(n*(n+(2*m+1)) + 4*T_m))^2) = k*d where m >= 0, k >= 1, and T is a triangular number.  Marco Matosic, Nov 10 2006
Numbers n such that a(n) is prime are listed in A055494. Prime a(n) are listed in A002383. All terms are odd. Prime factors of a(n) are listed in A007645. 3 divides a(3*k1), 7 divides a(7*k4) and a(7*k2), 7^2 divides a(7^2*k18) and a(7^2*k+19), 7^3 divides a(7^3*k18) and a(7^3*k+19), 7^4 divides a(7^4*k+1048) and a(7^4*k1047), 7^5 divides a(7^5*k+1354) and a(7^5*k1353), 13 divides a(13*k9) and a(13*k3), 13^2 divides a(13^2*k+23) and a(13^2*k22), 13^3 divides a(13^3*k+1037) and a(13^3*k1036).  Alexander Adamchuk, Jan 25 2007
Complement of A135668.  Kieren MacMillan, Dec 16 2007
Numbers (sorted) on the main diagonal of a 2n X 2n spiral. For example, when n=2:
7...8...9...10
6...1...2...11
5...4...3...12
16..15..14..13  cf. A137928.  William A. Tedeschi, Feb 29 2008
a(n) = AlexanderPolynomial[n] defined as Det[Transpose[S]n S] where S is Seifert matrix {{1, 1}, {0, 1}}.  Artur Jasinski, Mar 31 2008
Starting (1, 3, 7, 13, 21,...) = binomial transform of [1, 2, 2, 0, 0, 0]; example: a(4) = 13 = (1, 3, 3, 1) dot (1, 2, 2, 0) = (1 + 6 + 6 + 0).  Gary W. Adamson, May 10 2008
Starting (1, 3, 7, 13,...) = triangle A158821 * [1, 2, 3,...].  Gary W. Adamson, Mar 28 2009
Starting with offset 1 = triangle A128229 * [1,2,3,...].  Gary W. Adamson, Mar 26 2009
a(n) = k such that floor(1/2 *(1 + sqrt(4*k3))) + k = (n^2+1). A000037(a(n)) = A002522(n) = n^2 + 1.  Jaroslav Krizek, Jun 21 2009
For n > 0: a(n) = A170950(A002522(n1)), A170950(a(n)) = A174114(n), A170949(a(n)) = A002522(n1).  Reinhard Zumkeller, Mar 08 2010
a(n) = A176271(n,1) for n > 0.  Reinhard Zumkeller, Apr 13 2010
a(n) == 3 (mod n+1).  Bruno Berselli, Jun 03 2010
From Emeric Deutsch, Sep 23 2010: (Start)
a(n) is also the Wiener index of the fan graph F(n). The fan graph F(n) is defined as the graph obtained by joining each node of an nnode path graph with an additional node. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of the graph F(n) is (1/2)t[(n1)(n2)t + 2(2n1)]. Example: a(2)=3 because the corresponding fan graph is a cycle on 3 nodes (a triangle), having distances 1, 1, and 1.
(End)
For all elements k = n^2  n + 1 of the sequence, sqrt(4*(k1)+1) is an integer because 4*(k1) + 1 = (2*n1)^2 is a perfect square. Building the intersection of this sequence with A000225, k may in addition be of the form k = 2^x  1, which happens only for k = 1, 3, 7, 31, and 8191. [Proof: Still 4*(k1)+1 = 2^(x+2)  7 must be a perfect square, which has the finite number of solutions provided by A060728: x = 1, 2, 3, 5, or 13.] In other words, the sequence A038198 defines all elements of the form 2^x  1 in this sequence. For example k = 31 = 6*6  6 + 1; sqrt((311)*4+1) = sqrt(121) = 11 = A038198(4).  Alzhekeyev Ascar M, Jun 01 2011
a(n) such that A002522(n1) * A002522(n) = A002522(a(n)) where A002522(n) = n^2 + 1.  Michel Lagneau, Feb 10 2012
Left edge of the triangle in A214661: a(n) = A214661(n, 1), for n > 0.  Reinhard Zumkeller, Jul 25 2012
a(n) = A215630(n, 1), for n > 0; a(n) = A215631(n1, 1), for n > 1.  Reinhard Zumkeller, Nov 11 2012
Sum_{n > 0} arccot(a(n)) = Pi/2.  Franz Vrabec, Dec 02 2012
If you draw a triangle with one side of unit length and one side of length n, with an angle of Pi/3 radians between them, then the length of the third side of the triangle will be the square root of a(n).  Elliott Line, Jan 24 2013
a(n) = A228643(n, 1), for n > 0.  Reinhard Zumkeller, Aug 29 2013
a(n+1) are the numbers j such that: j^2 = j + m + sqrt(j*m), with corresponding numbers m given by A100019(n). Also: sqrt(j*m) = A027444(n) = n * a(n+1).  Richard R. Forberg, Sep 03 2013
Let p(x) the interpolating polynomial of degree n1 passing through the n points (n,n) and (1,1), (2,1), ..., (n1,1). Then p(n+1) = a(n).  Giovanni Resta, Feb 09 2014
The number of square roots >= sqrt(n) and < n+1 (n >= 0) gives essentially the same sequence, 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133, 157, 183, 211, ... .  Michael G. Kaarhus, May 21 2014
For n>1: a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+1] X [n+1] chessboard. Specifically, this will be a lone queen of one color placed in any position on the perimeter of the board, facing an opponent's "army" of size a(n)1 == A002378(n1).  Bob Selcoe, Feb 07 2015
A256188(a(n)) = 1.  Reinhard Zumkeller, Mar 26 2015
a(n+1) is, for n >= 1, the number of points as well as the number of lines of a finite projective plane of order n (cf. Hughes and Piper, 1973, Theorem 3.5., pp. 7980). For n = 3, a(4) = 13, see the 'Finite example' in the Wikipedia link, section 2.3, for the pointline matrix.  Wolfdieter Lang, Nov 20 2015
Sum_{n>=0} 1/a(n) = 1 + Pi*tanh(Pi*sqrt(3)/2)/sqrt(3) = 2.79814728056269018... .  Vaclav Kotesovec, Apr 10 2016
Denominators of the solution to the generalization of the Feynman triangle problem. If each vertex of a triangle is joined to the point (1/p) along the opposite side (measured say clockwise), then the area of the inner triangle formed by these lines is equal to (p  2)^2/(p^2  p + 1) times the area of the original triangle, p>2. For example, when p = 3, the ratio of the areas is 1/7. The numerators of the ratio of the areas is given by A000290 with an offset of 2. [Cook & Wood, 2004].  Joe Marasco, Feb 20 2017
n^2 equal triangular tiles with side lengths 1 X 1 X 1 may be put together to form an n X n X n triangle. For n>=2 a(n1) is the number of different 2 X 2 X 2 triangles being contained.  Heinrich Ludwig, Mar 13 2017


FORMULA

G.f.: (12*x+3*x^2)/(1x)^3.  Simon Plouffe in his 1992 dissertation
a(n) = (n5)*a(n1) + (n2)*a(n2).
a(1n) = a(n).  Michael Somos, Sep 04 2006
a(n) = a(n1) + 2*(n1) = 2*a(n1)  a(n2) + 2 = 1+A002378(n1) = 2*A000124(n1)  1.  Henry Bottomley, Oct 02 2000 [Corrected by N. J. A. Sloane, Jul 18 2010]
a(n) = A000217(n) + A000217(n2) (sum of two triangular numbers).
x*(1+x^2)/(1x)^3 is g.f. for 0, 1, 3, 7, 13, ... a(n)=2*C(n, 2)+C(n1, 0). E.g.f.: (1+x^2)*exp(x).  Paul Barry, Mar 13 2003
a(n) = ceiling((n1/2)^2).  Benoit Cloitre, Apr 16 2003. [Hence the terms are about midway between successive squares and so (except for 1) are not squares.  N. J. A. Sloane, Nov 01 2005]
a(n) = 1 + sum_{j=0..n1} (2*j).  Xavier Acloque, Oct 08 2003
a(n) = floor(t(n^2)/t(n)), where t(n)= A000217(n).  Jon Perry, Feb 14 2004
a(n) = leftmost term in M^(n1) * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 0 1 2 / 0 0 1]. E.g., a(6) = 31 since M^5 * [1 1 1] = [31 11 1].  Gary W. Adamson, Nov 11 2004
a(n+1) = n^2 + n + 1. a(n+1)*a(n) = (n^61)/(n^21) = n^4+n^2+1 = a(n^2+1) (a product of two consecutive numbers from this sequence belongs to this sequence). (a(n+1)+a(n))/2 = n^2+1. (a(n+1)a(n))/2=n. a((a(n+1)+a(n))/2) = a(n+1)*a(n).  Alexander Adamchuk, Apr 13 2006
a(n+3) is the numerator of ((n + 1)! + (n  1)!)/ n!.  Artur Jasinski, Jan 09 2007
a(n) = A132111(n1, 1), for n>1.  Reinhard Zumkeller, Aug 10 2007
a(n) = Det[Transpose[{{1, 1}, {0, 1}}]  n {{1, 1}, {0, 1}}].  Artur Jasinski, Mar 31 2008
a(n) = 3*a(n1)3*a(n2)+a(n3), n>=3.  Jaume Oliver Lafont, Dec 02 2008
a(n) = (n1)^2 + (n1) + 1 = 111 read in base n1 (for n>2).  Jason Kimberley, Oct 18 2011
a(n) = sqrt(A058031(n)).  Richard R. Forberg, Sep 03 2013
G.f.: 1 / (1  x / (1  2*x / (1 + x / (1  2*x / (1 + x))))).  Michael Somos, Apr 03 2014
a(n) = A243201(n  1) / A003215(n  1), n > 0.  Mathew Englander, Jun 03 2014
For n>=2, a(n) = ceil(4/(sum(k = A000217(n1)... A000217(n)  1, 1/k))).  Richard R. Forberg, Aug 17 2014
a(n) = A101321(2,n1).  R. J. Mathar, Jul 28 2016
