STATUS
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48, 240, 560, 1008, 1584, 2288, 3120, 4080, 5168, 6384, 7728, 9200, 10800, 12528, 14384, 16368, 18480, 20720, 23088, 25584, 28208, 30960, 33840, 36848, 39984, 43248, 46640, 50160, 53808, 57584, 61488, 65520, 69680, 73968, 78384, 82928, 87600, 92400, 97328, 102384
From Vincenzo Librandi, Feb 09 2012: (Start)
G.f.: -16*x*(3 + 6*x - x^2)/(x - 1)^3. - _Vincenzo Librandi_, Feb 09 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 09 2012(End)
From Amiram Eldar, Mar 07 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/32.
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi-2)/64. (End)
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(MAGMAMagma) I:=[48, 240, 560]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 09 2012
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a(n) = 64*n^2 - 16.
The identity (8*n^2 - 1)^2 - (64*n^2 - 16)*(n)^2 = 1 can be written as A157914(n)^2 - a(n)*(n)^2 = 1. - _Vincenzo Librandi, _, Feb 09 2012
G.f.: -16*x*(3 + 6*x - x^2)/(x - 1)^3. - _Vincenzo Librandi, _, Feb 09 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _Vincenzo Librandi, _, Feb 09 2012
(MAGMA) I:=[48, 240, 560]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // _Vincenzo Librandi, _, Feb 09 2012
(PARI) for(n=1, 40, print1(64*n^2 - 16", ")); \\ _Vincenzo Librandi, _, Feb 09 2012
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<a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).