0,2

Sequence allows us to find X values of the equation: X + (X + 3)^2 + (X + 6)^3 = Y^2. To prove that X = n^2 + 6n: Y^2 = X + (X + 3)^2 + (X + 6)^3 = X^3 + 19*X^2 + 115X + 225 = (X + 9)(X^2 + 10X + 25) = (X + 9)*(X + 5)^2 it means: (X + 9) must be a perfect square, so X = k^2 - 9 with k>=3. we can put: k = n + 3, which gives: X = n^2 + 6n and Y = (n + 3)(n^2 + 6n + 5). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 12 2007

Number of units of a(n) belongs to a periodic sequence: 0, 7, 6, 7, 0, 5, 2, 1, 2, 5. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009]

Apart the first term, a(n)=2*n+a(n-1)+5 (with a(1)=7) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 08 2009]

Index entries for sequences related to linear recurrences with constant coefficients

P. De Geest, Palindromic Quasipronics of the form n(n+x)

a(n) = (n+3)^2 -3^2 = n*(n+6*n), n>=0.

G.f.: x*(7-5*x)/(1-x)^3.

a(n)=2*n+a(n-1)+3 (with a(1)=0) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 29 2009]

For n=2, a(2)=2*2+0+3=7; n=3, a(3)=2*3+7+3=16; n=4, a(4)=2*4+16+3=27 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 29 2009]

a:=n->sum(n, j=7..n): seq(a(n), n=6..52); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Feb 12 2007

a:=n->sum(1+sum(1, k=1..n), k=6..n):seq(a(n), n=5...51); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 11 2008

Table[n(n + 6), {n, 0, 65}] or Select[ Range[0, 5000], IntegerQ[ Sqrt[9(9 + #)]] & ]

(Other) SAGE: [lucas_number2(2, n, 4-n) for n in xrange(2, 49)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 19 2009]

a(n-3), n>=4, third column (used for the Paschen series of the hydrogen atom) of triangle A120070.

Cf. A005563.

Sequence in context: A017245 A052221 A119461 this_sequence A133694 A024627 A140511

Adjacent sequences: A028557 A028558 A028559 this_sequence A028561 A028562 A028563

nonn,new

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

Edited by Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 06 2002