login
a(n) = 4*n^2 + 24*n + 8.
5

%I #27 Nov 22 2023 22:17:07

%S 36,72,116,168,228,296,372,456,548,648,756,872,996,1128,1268,1416,

%T 1572,1736,1908,2088,2276,2472,2676,2888,3108,3336,3572,3816,4068,

%U 4328,4596,4872,5156,5448,5748,6056,6372,6696,7028,7368,7716,8072,8436,8808,9188

%N a(n) = 4*n^2 + 24*n + 8.

%C 2*(fifth subdiagonal of triangle A144562).

%C Sequence gives values x of solutions (x, y) to the Diophantine equation x^3+28*x^2 = y^2. For a more comprehensive list of solutions see A155135.

%C For n >= 3, a(n - 1) is the number of checkmate positions with white queen and white king against black king on an n X n board. Reason: The black king can only be on the edge. There are 4*(4*n + 1) checkmate positions where the black king is in the corner, 4*(2*n + 4) checkmate positions where the black king is immediately adjacent to the corner square, and there are 4*(n - 4)*(n + 2) checkmate positions where the black king is on another edge square. That's a total of 4*n^2 + 16*n - 12 = a(n - 1) checkmate positions. - _Felix Huber_, Oct 29 2023

%H Vincenzo Librandi, <a href="/A153642/b153642.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 a(n) = A155135(2n+8) = A155136(2n+7).

%F a(n) = 4*A028881(n+3).

%F G.f.: 4*(3 - x)*(3 - 2*x)/(1-x)^3.

%F a(n)= 3*a(n-1) - 3*a(n-2) + a(n-3).

%F E.g.f.: 4*(-2 + (2 + 7*x + x^2)*exp(x)). - _G. C. Greubel_, Aug 23 2016

%F From _Amiram Eldar_, Mar 02 2023: (Start)

%F Sum_{n>=1} 1/a(n) = 1/56 - cot(sqrt(7)*Pi)*Pi/(8*sqrt(7)).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 31/168 - cosec(sqrt(7)*Pi)*Pi/(8*sqrt(7)). (End)

%t LinearRecurrence[{3, -3, 1}, {36, 72, 116}, 50] (* _Vincenzo Librandi_, Feb 25 2012 *)

%o (Magma) [ 4*(n+3)^2-28: n in [1..45] ];

%o (PARI) a(n)=4*n*(n+6)+8 \\ _Charles R Greathouse IV_, Dec 28 2011

%Y Cf. A028881, A144562, A155135, A155136,

%K nonn,easy,less

%O 1,1

%A _Vincenzo Librandi_, Dec 30 2008

%E Edited and extended by _Klaus Brockhaus_, Jan 21 2009