%I #94 Sep 13 2023 04:32:34
%S 0,16,48,96,160,240,336,448,576,720,880,1056,1248,1456,1680,1920,2176,
%T 2448,2736,3040,3360,3696,4048,4416,4800,5200,5616,6048,6496,6960,
%U 7440,7936,8448,8976,9520,10080,10656,11248,11856,12480,13120,13776
%N Total number of possible knight moves on an (n+2) X (n+2) chessboard, if the knight is placed anywhere.
%C 16 times the triangular numbers A000217.
%C Centered 16-gonal numbers A069129, minus 1. Also, sequence found by reading the segment (0, 16) together with the line from 16, in the direction 16, 48, ..., in the square spiral whose vertices are the triangular numbers A000217. - _Omar E. Pol_, Apr 26 2008, Nov 20 2008
%C For n >= 1, number of permutations of n+1 objects selected from 5 objects v, w, x, y, z with repetition allowed, containing n-1 v's. Examples: at n=1, n-1=0 (i.e., zero v's), and a(1)=16 because we have ww, wx, wy, wz, xw, xx, xy, xz, yw, yx, yy, yz, zw, zx, zy, zz; at n=2, n-1=1 (i.e., one v), and there are 3 permutations corresponding to each one in the n=1 case (e.g., the single v can be inserted in any of three places in the 2-object permutation xy, yielding vxy, xvy, and xyv), so a(2) = 3*a(1) = 3*16 = 48; at n=3, n-1=2 (i.e., two v's), and a(3) = C(4,2)*a(1) = 6*16 = 96; etc. - _Zerinvary Lajos_, Aug 07 2008 (this needs clarification, _Joerg Arndt_, Feb 23 2014)
%C Sequence found by reading the line from 0, in the direction 0, 16, ... and the same line from 0, in the direction 0, 48, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - _Omar E. Pol_, Oct 03 2011
%C For n > 0, a(n) is the area of the triangle with vertices at ((n-1)^2, n^2), ((n+1)^2, (n+2)^2), and ((n+3)^2, (n+2)^2). - _J. M. Bergot_, May 22 2014
%C For n > 0, a(n) is the number of self-intersecting points in star polygon {4*(n+1)/(2*n+1)}. - _Bui Quang Tuan_, Mar 28 2015
%C Equivalently: integers k such that k$ / (k/2)! and k$ / (k/2+1)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - _Bernard Schott_, Dec 02 2021
%H Vincenzo Librandi, <a href="/A035008/b035008.txt">Table of n, a(n) for n = 0..1000</a>
%H Omar E. Pol, <a href="http://polprimos.com">Determinacion geometrica de los numeros primos y perfectos</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StarPolygon.html">Star Polygon</a>.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 8*n*(n+1).
%F G.f.: 16*x/(1-x)^3.
%F a(n) = A069129(n+1) - 1. - _Omar E. Pol_, Apr 26 2008
%F a(n) = binomial(n+1,2)*4^2, n >= 0. - _Zerinvary Lajos_, Aug 07 2008
%F a(n) = 8*n^2 + 8*n = 16*A000217(n) = 8*A002378(n) = 4*A046092(n) = 2*A033996(n). - _Omar E. Pol_, Dec 12 2008
%F a(n) = a(n-1) + 16*n, with a(0)=0. - _Vincenzo Librandi_, Nov 17 2010
%F E.g.f.: 8*exp(x)*x*(2 + x). - _Stefano Spezia_, May 19 2021
%F From _Amiram Eldar_, Feb 22 2023: (Start)
%F Sum_{n>=1} 1/a(n) = 1/8.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/8.
%F Product_{n>=1} (1 - 1/a(n)) = -(8/Pi)*cos(sqrt(3/2)*Pi/2).
%F Product_{n>=1} (1 + 1/a(n)) = (8/Pi)*cos(Pi/(2*sqrt(2))). (End)
%e 3 X 3-Board: knight can be placed in 8 positions with 2 moves from each, so a(1) = 16.
%p seq(binomial(n+1,2)*4^2, n=0..33); # _Zerinvary Lajos_, Aug 07 2008
%t CoefficientList[Series[16 x/(1 - x)^3, {x, 0, 40}], x] (* _Vincenzo Librandi_, Feb 24 2014 *)
%t LinearRecurrence[{3,-3,1},{0,16,48},50] (* or *) 16*Accumulate[ Range[ 0,50]] (* _Harvey P. Dale_, Aug 05 2018 *)
%o (Magma) [8*n*(n+1): n in [0..50]]; // _Wesley Ivan Hurt_, May 22 2014
%o (PARI) a(n)=8*n*(n+1) \\ _Charles R Greathouse IV_, Sep 30 2015
%Y Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A002492 (Bishop) and A049450 (Pawn).
%Y Cf. A000217, A069129, A027468, A008586 A038231, A002378, A033996, A046092.
%Y Cf. A348692.
%Y Subsequence of A008586 and of A349081.
%K easy,nonn,nice
%O 0,2
%A Ulrich Schimke (ulrschimke(AT)aol.com), Dec 11 1999
%E More terms from _Erich Friedman_
%E Minor errors corrected and edited by _Johannes W. Meijer_, Feb 04 2010