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A067728
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a(n) = 2*n^2 + 8*n.
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19
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10, 24, 42, 64, 90, 120, 154, 192, 234, 280, 330, 384, 442, 504, 570, 640, 714, 792, 874, 960, 1050, 1144, 1242, 1344, 1450, 1560, 1674, 1792, 1914, 2040, 2170, 2304, 2442, 2584, 2730, 2880, 3034, 3192, 3354, 3520, 3690, 3864, 4042, 4224, 4410, 4600, 4794
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OFFSET
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1,1
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COMMENTS
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Positive numbers k such that 8*(8 + k) is a perfect square.
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
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FORMULA
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a(n+1) = 2*n*n + 12*n + 10. - Frank Ellermann
a(n) = Sum_{k=0..n} Sum_{j=4..n} (j - k), n >= 4. - Zerinvary Lajos, May 11 2007
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Jul 08 2012
G.f.: 2*x*(5-3*x)/(1-x)^3. - Vincenzo Librandi, Jul 08 2012
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 25/96.
Sum_{n>=1} (-1)^(n+1)/a(n) = 7/96. (End)
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MATHEMATICA
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Select[ Range[10000], IntegerQ[ Sqrt[ 8(8 + # )]] & ]
CoefficientList[Series[2*(5-3*x)/(1-x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 08 2012 *)
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PROG
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(PARI) a(n)=2*n*(n+4) \\ Charles R Greathouse IV, Dec 07 2011
(Magma) [2*n*(n+4): n in [1..50]] // Vincenzo Librandi, Jul 08 2012
(Python)
def a(n): return (2*n + 8)*n
print([a(n) for n in range(1, 48)]) # Michael S. Branicky, Oct 24 2021
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CROSSREFS
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Cf. 7: A067727, 6: A067726, 5: A067724, 3: A067725.
Cf. A000217, A005563, A140091, A140681, A212331.
Sequence in context: A267431 A162817 A103573 * A352283 A058504 A250798
Adjacent sequences: A067725 A067726 A067727 * A067729 A067730 A067731
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KEYWORD
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nonn,easy
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AUTHOR
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Robert G. Wilson v, Feb 05 2002
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STATUS
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approved
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