0,3

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}.

Also surround numbers of an n X n square - Jason Earls (zevi_35711(AT)yahoo.com), Apr 16 2001

Also 8n + 8 is a square. - Cino Hilliard (hillcino368(AT)gmail.com), Dec 18 2003

The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2+...+2(n+1)^2-1 = (1/2)(2(n+1)^2-1-2n^2+1)(2(n+1)^2-1+2n^2)=(2n+1)^3. Eg 2+3+4+5+6+7 = 27 =3^3, then 8+9+10+..+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005

Sequence allows us to find X values of the equation: 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n)=2n(2*n^2 - 1). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 06 2007

Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2 [From Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008]

Equals row sums of triangle A143593 & binomial transform of [1, 6, 4, 0, 0, 0,...]. [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 26 2008]

Also, except (-1,1), a(n)=2*n^2+8n+7; fourth column of [A154685] [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 25 2009]

sqrt(a(n) + a(n+1) + 1) = 2n+1 [From Doug Bell (bell.doug(AT)gmail.com), Mar 09 2009]

Apart the first term which is -1 the number of units of a(n) belongs to a periodic sequence: 1, 7, 7, 1, 9.We conclude that a(n) and a(n+5) have the same number of units. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 05 2009]

Index entries for sequences related to linear recurrences with constant coefficients

G.f.: (-1+4x+x^2)/(1-x)^3.

a(n) = A119258(n+1,2) for n>0. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 11 2006

Starting with offset 1, equals binomial transform of [1, 6, 4, 0, 0, 0,...] [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 29 2008]

Contribution from Doug Bell (bell.doug(AT)gmail.com), Mar 08 2009: (Start)

a(0) = -1,

a(n) = sqrt(A001844(n)^2 - A069074(n-1)),

a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2) (End)

a(0) = 0^2-1*1 = -1, a(1) = 1^2-4*0 = 1, a(2) = 2^2-9*1 = 7 etc.

a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 29 2008]

lst={}; Do[AppendTo[lst, 2*n^2-1], {n, 0, 4^2}]; lst...or/and... lst={}; Do[AppendTo[lst, ChebyshevT[2, n]], {n, 0, 4^2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 11 2008]

1. Table[2*n^2 + 4*n + 1, {n, -1, 46}] (.) 2. lst = {}; Do[a = 2*n^2 + 4*n + 1; AppendTo[lst, a], {n, -1, 46}] lst [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 10 2009]

(PARI) a(n)=if(n<0, 0, 2*n^2-1)

Cf. A047875, A000105, A077585.

Cf. A005563, A046092, A001082, A002378, A036666, A062717, A028347, A087475, A000217.

Cf. A143593 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 26 2008]

Cf. A154685 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Jan 25 2009]

Sequence in context: A046118 A120092 A130284 this_sequence A024840 A024835 A144861

Adjacent sequences: A056217 A056218 A056219 this_sequence A056221 A056222 A056223

sign

N. J. A. Sloane (njas(AT)research.att.com), Aug 06 2000

Formula and additional comments from Henry Bottomley (se16(AT)btinternet.com), Dec 12 2000