|
%I #29 Apr 15 2020 13:28:13
%S 1,1,8,20,32,52,68,76,96,96,120,120,148,148,176,176,204,204,232,232,
%T 260,260,288,288,316,316,344,344,372,372,400,400,428,428,456,456,484,
%U 484,512,512,540,540,568,568,596,596,624,624,652,652,680,680,708,708
%N On a (2n+1) X (2n+1) board, let m(i) be the number of squares that are i knight's moves from center; sequence gives max m(i) for i >= 0.
%H Colin Barker, <a href="/A038522/b038522.txt">Table of n, a(n) for n = 0..1000</a>
%H Andreas P. Hadjipolakis, <a href="https://www.jstor.org/stable/2319360">Problem E2605</a>, Am. Math. Monthly Vol. 83 (1976), no. 7 (Aug-Sept.), p. 566.
%H Roger Weitzenkamp, <a href="https://www.jstor.org/stable/2322069">Solution to Problem E2605: Labels on a Chessboard</a>, Am. Math. Monthly Vol. 84 (1977), p. 822.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).
%F a(n) = 28*floor(n/2) - 20 for n >= 10. - _Andrew Howroyd_, Feb 28 2020
%F From _Stefano Spezia_, Feb 29 2020: (Start)
%F G.f.: (1 + 6*x^2 + 12*x^3 + 5*x^4 + 8*x^5 + 4*x^6 - 12*x^7 + 4*x^8 - 8*x^9 + 4*x^10 + 4*x^12)/((1 - x)^2*(1 + x)).
%F a(n) = a(n-1) + a(n-2) - a(n-3) for n > 12. (End)
%e On a 5 X 5 board, [ m(0),...,m(4) ]=[ 1,8,8,4,4 ], max=8, so a(2)=8.
%t LinearRecurrence[{1,1,-1},{1,1,8,20,32,52,68,76,96,96,120,120,148},60] (* _Harvey P. Dale_, Apr 15 2020 *)
%o (PARI) Vec((1 + x^2)*(1 + 5*x^2 + 12*x^3 - 4*x^5 + 4*x^6 - 8*x^7 + 4*x^10) / ((1 - x)^2*(1 + x)) + O(x^50)) \\ _Colin Barker_, Mar 16 2020
%Y Cf. A018842.
%K easy,nonn,walk,nice
%O 0,3
%A Antreas P. Hatzipolakis (xpolakis(AT)hol.gr)
%E Corrected and additional terms added by _Andrew Howroyd_, Feb 28 2020
|
