0,2

Can be obtained as sum of Smarandache mirror sequence terms. a(n+1)=a(n)+2(n+1) where a(1)=1 - Felice Russo (felice.russo(AT)katamail.com)

a(n) = A105728(n+2,n+1). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Apr 18 2005

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

Values of Fibonacci polynomial n^2-n-1 for n=2,3,4,5,... - Artur Jasinski (grafix(AT)csl.pl), Nov 19 2006

Row sums of triangle A135223 - Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 23 2007

Equals row sums of triangle A143596 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), 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). [From Aaron Dunigan AtLee (aaron(AT)duniganatlee.com), Feb 13 2009]

Contribution from Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 23 2009: (Start)

a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989...

Example: a(3) = 19 =(5 + .6180339...) * (3.381966...). Cf. next to leftmost

column in A162997 array. (End)

Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.

P. De Geest, World!Of Numbers

Not of form k + [ sqrt(k) ], k integer.

a(n)=sqrt( n(n+1)(n+2)(n+3) + 1 ). - Floor van Lamoen (fvlamoen(AT)hotmail.com), Oct 08 2001

(a(n))^2 = n(n+1)(n+2)(n+3) + 1 - Rainer Rosenthal (r.rosenthal(AT)web.de), Sep 04 2004

a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1 - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Sep 24 2004

a(n) = A109128(n+2, 2). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 20 2005

A127701 * [1, 2, 3,...] - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jan 24 2007

a(n) = 2*T(n) - 1, where T(n) = A000217 = the triangular series. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 15 2007

a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum(a(k): 0<=k<=n). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 20 2007

Binomial transform of [1, 4, 2, 0, 0, 0,...] = (1, 5, 11, 19,...) - Gary W. Adamson (qntmpkt(AT)yahoo.com), Sep 20 2007

G.f.: (1+2*x-x^2)/(1-x)^3. a(n)=3*a(n-1)-3*a(n-2)+a(n-3). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Jul 11 2009]

a(n)=a(n-1)+2n (with a(1)=1) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 08 2009]

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]+2*n od: seq(a[n], n=1..47); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 22 2008

with (combinat):seq(fibonacci(3, n)+n-2, n=1..50); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 07 2008

a:=n->sum(k, k=0..n):seq(a(n)+sum(k, k=2..n), n=1..47); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 10 2008

Table[n^2 - n - 1, {n, 2, 20}] - Artur Jasinski (grafix(AT)csl.pl), Nov 19 2006

Table[Numerator[((n + 1)! - (n - 1)!)/(n!)], {n, 1, 30}] - Artur Jasinski (grafix(AT)csl.pl), Jan 09 2007

s = 1; lst = {s}; Do[s += n + 3; AppendTo[lst, s], {n, 1, 100, 2}]; lst [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 11 2009]

sage: [i+(i+1)^2 for i in xrange(0, 48)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 03 2008

Complement of A028392. Third column of array A094954.

Cf. A000217, A002522, A062392, A127701, A135223.

Cf. A143596 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 26 2008]

Cf. a052905 [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 11 2009]

A162997 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jul 23 2009]

Adjacent sequences: A028384 A028385 A028386 this_sequence A028388 A028389 A028390

Sequence in context: A108151 A088059 A165900 this_sequence A110331 A106071 A073847

nonn,new

Patrick De Geest (pdg(AT)worldofnumbers.com)