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a(n) = 5*n^2 - 1.
5

%I #24 Sep 08 2022 08:45:32

%S 4,19,44,79,124,179,244,319,404,499,604,719,844,979,1124,1279,1444,

%T 1619,1804,1999,2204,2419,2644,2879,3124,3379,3644,3919,4204,4499,

%U 4804,5119,5444,5779,6124,6479,6844,7219,7604,7999,8404,8819,9244,9679,10124

%N a(n) = 5*n^2 - 1.

%C For k != 0, the quintic polynomials of the form x^5 + 5*(5*k^2-1)*x + 4*(5*k^2-1) have Galois group A5 (order 60) over rational numbers.

%H Vincenzo Librandi, <a href="/A134538/b134538.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: x*(-4-7*x+x^2)/(-1+x)^3. - _R. J. Mathar_, Nov 14 2007

%F From _Amiram Eldar_, Feb 04 2021: (Start)

%F Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(5))*cot(Pi/sqrt(5)))/2.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(5))*csc(Pi/sqrt(5)) - 1)/2.

%F Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(5))*csc(Pi/sqrt(5)).

%F Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(5))*sin(sqrt(2/5)*Pi)/sqrt(2). (End)

%t Table[5n^2 - 1, {n, 1, 50}]

%t CoefficientList[Series[(4+7*x-x^2)/(1-x)^3,{x,0,50}],x] (* _Vincenzo Librandi_, Jul 09 2012 *)

%o (Magma) [5*n^2-1: n in [1..50]]; // _Vincenzo Librandi_, Jul 09 2012

%o (PARI) a(n)=5*n^2-1 \\ _Charles R Greathouse IV_, Jul 01 2013

%K nonn,easy

%O 1,1

%A _Artur Jasinski_, Oct 30 2007