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A028560
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a(n) = n*(n + 6), also numbers j such that 9*(9 + j) is a perfect square.
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42
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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
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OFFSET
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0,2
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COMMENTS
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Nonnegative X values of solutions to 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, Nov 12 2007
Number of units of a(n) belongs to a periodic sequence: 0, 7, 6, 7, 0, 5, 2, 1, 2, 5. - Mohamed Bouhamida, Sep 04 2009
a(m) where m is a positive integer are the only positive integer values of t for which the Binet-de Moivre Formula of the recurrence b(n)=6*b(n-1)+t*b(n-2) with b(0)=0 and b(1)=1 has a root which is a square. In particular, sqrt(6^2+4*t) is an integer since 6^2+4*t=6^2+4*a(m)=(2*m+6)^2. Thus, the charcteristic roots are k1=6+m and k2=-m. - Felix P. Muga II, Mar 27 2014
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LINKS
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FORMULA
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a(n) = (n+3)^2 - 3^2 = n*(n+6).
G.f.: x*(7-5*x)/(1-x)^3.
Sum_{n>=1} 1/a(n) = 49/120 = 0.4083333... - R. J. Mathar, Mar 22 2011
Sum_{n>=1} (-1)^(n+1)/a(n) = 37/360. - Amiram Eldar, Nov 04 2020
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MAPLE
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MATHEMATICA
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Table[n(n + 6), {n, 0, 65}] (* or *) Select[ Range[0, 5000], IntegerQ[ Sqrt[9(9 + #)]]& ]
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PROG
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(Sage) [lucas_number2(2, n, 4-n) for n in range(2, 49)] # Zerinvary Lajos, Mar 19 2009
(Haskell)
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CROSSREFS
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a(n-3), n>=4, third column (used for the Paschen series of the hydrogen atom) of triangle A120070.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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