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A028560
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n(n + 6), also numbers such that 9(9 + n) is a perfect square.
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26
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0, 7, 16, 27, 40, 55, 72, 91, 112, 135, 160, 187, 216, 247, 280, 315, 352, 391, 432, 475, 520, 567, 616, 667, 720, 775, 832, 891, 952, 1015, 1080, 1147, 1216, 1287, 1360, 1435, 1512, 1591, 1672, 1755, 1840, 1927, 2016, 2107, 2200, 2295, 2392
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| 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]
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LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (3,-3,1).
P. De Geest, Palindromic Quasipronics of the form n(n+x)
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FORMULA
| a(n) = (n+3)^2 -3^2 = n*(n+6).
G.f.: x*(7-5*x)/(1-x)^3.
a(n)=2*n+a(n-1)+5. [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Aug 05 2010]
sum_{n>=1} 1/a(n) = 49/120 = 0.4083333... - R. J. Mathar, Mar 22 2011
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MATHEMATICA
| Table[n(n + 6), {n, 0, 65}] or Select[ Range[0, 5000], IntegerQ[ Sqrt[9(9 + #)]] & ]
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PROG
| (Other) SAGE: [lucas_number2(2, n, 4-n) for n in xrange(2, 49)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Mar 19 2009]
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CROSSREFS
| 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 * A190530 A133694 A024627
Adjacent sequences: A028557 A028558 A028559 * A028561 A028562 A028563
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KEYWORD
| nonn,easy
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AUTHOR
| Patrick De Geest (pdg(AT)worldofnumbers.com)
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EXTENSIONS
| Edited by Robert G. Wilson v (rgwv(AT)rgwv.com), Feb 06 2002
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