A038522 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.

1, 1, 8, 20, 32, 52, 68, 76, 96, 96, 120, 120, 148, 148, 176, 176, 204, 204, 232, 232, 260, 260, 288, 288, 316, 316, 344, 344, 372, 372, 400, 400, 428, 428, 456, 456, 484, 484, 512, 512, 540, 540, 568, 568, 596, 596, 624, 624, 652, 652, 680, 680, 708, 708

Offset

0,3

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Andreas P. Hadjipolakis, Problem E2605, Am. Math. Monthly Vol. 83 (1976), no. 7 (Aug-Sept.), p. 566.

Roger Weitzenkamp, Solution to Problem E2605: Labels on a Chessboard, Am. Math. Monthly Vol. 84 (1977), p. 822.

Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Formula

a(n) = 28*floor(n/2) - 20 for n >= 10. - Andrew Howroyd, Feb 28 2020

From Stefano Spezia, Feb 29 2020: (Start)

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)).

a(n) = a(n-1) + a(n-2) - a(n-3) for n > 12. (End)

Example

On a 5 X 5 board, [ m(0),...,m(4) ]=[ 1,8,8,4,4 ], max=8, so a(2)=8.

Mathematica

LinearRecurrence[{1, 1, -1}, {1, 1, 8, 20, 32, 52, 68, 76, 96, 96, 120, 120, 148}, 60] (* Harvey P. Dale, Apr 15 2020 *)

Prog

(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

Crossrefs

Cf. A018842.

Sequence in context: A017617 A246309 A363518 * A267435 A348093 A186293

Adjacent sequences: A038519 A038520 A038521 * A038523 A038524 A038525

Keyword

easy,nonn,walk,nice

Author

Antreas P. Hatzipolakis (xpolakis(AT)hol.gr)

Extensions

Corrected and additional terms added by Andrew Howroyd, Feb 28 2020

Status

approved