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Number of squares on infinite half chessboard at <=n knight moves from a fixed point on the edge.
+10
7
1, 5, 23, 60, 110, 172, 248, 338, 442, 560, 692, 838, 998, 1172, 1360, 1562, 1778, 2008, 2252, 2510, 2782, 3068, 3368, 3682, 4010, 4352, 4708, 5078, 5462, 5860, 6272, 6698, 7138, 7592, 8060, 8542, 9038, 9548, 10072, 10610, 11162, 11728, 12308, 12902, 13510
FORMULA
a(n) = 7*n^2 - n + 2, for n>3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>6. G.f.: -(2*x^6 -x^5 -6*x^4 +5*x^3 +11*x^2 +2*x +1) / (x -1)^3. - Colin Barker, Jul 14 2013
EXAMPLE
5 squares are reachable after 1 move, from these you can reach 18 new squares more, so a(1)=5 and a(2)=23.
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 5, 23, 60, 110, 172, 248}, 50] (* Paolo Xausa, Jul 17 2024 *)
Number of squares on infinite half chessboard at <=n knight moves from a fixed point on the diagonal.
+10
4
1, 5, 23, 57, 109, 169, 246, 334, 439, 555, 688, 832, 993, 1165, 1354, 1554, 1771, 1999, 2244, 2500, 2773, 3057, 3358, 3670, 3999, 4339, 4696, 5064, 5449, 5845, 6258, 6682, 7123, 7575, 8044, 8524, 9021, 9529, 10054, 10590, 11143, 11707, 12288, 12880, 13489
FORMULA
a(n) = (1/4) [28n^2 - 6n + 9 + 3(-1)^n], for n>3.
G.f.: -(3*x^7-x^6-8*x^5+4*x^4+13*x^3+13*x^2+3*x+1) / ((x-1)^3*(x+1)). - Colin Barker, Jul 14 2013
EXAMPLE
5 squares are reachable after 1 move, from these you can reach 18 new squares more, so a(1)=5, a(2)=23.
Number of squares on infinite quarter chessboard at <=n knight moves from the corner.
+10
4
1, 3, 12, 32, 59, 91, 130, 176, 229, 289, 356, 430, 511, 599, 694, 796, 905, 1021, 1144, 1274, 1411, 1555, 1706, 1864, 2029, 2201, 2380, 2566, 2759, 2959, 3166, 3380, 3601, 3829, 4064, 4306, 4555, 4811, 5074, 5344, 5621, 5905, 6196, 6494, 6799, 7111, 7430
FORMULA
a(n) = (1/2) * (7*n^2 + n + 2), for n>3.
G.f.: -(2*x^6-2*x^5-4*x^4+4*x^3+6*x^2+1) / (x-1)^3. - Colin Barker, Jul 15 2013
EXAMPLE
3 squares are reachable after 1 move, from these you can reach 8 new squares more, so a(1)=3, a(2)=12.
The number of distinct positions on an infinite chessboard reachable by the (2,3)-leaper in <= n moves.
+10
4
1, 9, 41, 129, 321, 625, 997, 1413, 1885, 2425, 3033, 3709, 4453, 5265, 6145, 7093, 8109, 9193, 10345, 11565, 12853, 14209, 15633, 17125, 18685, 20313, 22009, 23773, 25605, 27505, 29473, 31509, 33613, 35785, 38025, 40333, 42709, 45153, 47665, 50245, 52893
FORMULA
a(n) = 34*n^2 + 30*n + 9 for n >= 6.
G.f.: (1 + x)*(1 + 5*x + 12*x^2 + 20*x^3 + 28*x^4 - 20*x^5 - 24*x^6 + 12*x^8) / (1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>9. (End)
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 9, 41, 129, 321, 625, 997, 1413, 1885, 2425}, 50] (* Paolo Xausa, Mar 17 2024 *)
PROG
(PARI) Vec((1 + x)*(1 + 5*x + 12*x^2 + 20*x^3 + 28*x^4 - 20*x^5 - 24*x^6 + 12*x^8) / (1 - x)^3 + O(x^40)) \\ Colin Barker, Jan 07 2018
Number of squares on infinite chessboard at n knight's moves from center.
+10
3
1, 8, 32, 68, 96, 120, 148, 176, 204, 232, 260, 288, 316, 344, 372, 400, 428, 456, 484, 512, 540, 568, 596, 624, 652, 680, 708, 736, 764, 792, 820, 848, 876, 904, 932, 960, 988, 1016, 1044, 1072, 1100, 1128, 1156
FORMULA
a(n) = 28*n-20, n >= 5.
G.f.: (1 + 5*x + 12*x^2 - 8*x^4 + 4*x^5)*(1+x)/(1-x)^2.
MAPLE
(1 + 5*x + 12*x^2 - 8*x^4 + 4*x^5)*(1+x)/(1-x)^2; seq(coeff(series(%, x, n+1), x, n), n=0..50);
MATHEMATICA
CoefficientList[Series[(1+5x+12x^2-8x^4+4x^5)(1+x)/(1-x)^2, {x, 0, 50}], x] (* or *) Join[{1, 8, 32, 68, 96}, LinearRecurrence[{2, -1}, {120, 148}, 46]] (* Harvey P. Dale, Jul 05 2011 *)
Number of squares on infinite octant of chessboard at <=n knight moves from the corner. The octant includes the diagonal.
+10
3
1, 2, 5, 13, 31, 49, 70, 93, 121, 151, 186, 223, 265, 309, 358, 409, 465, 523, 586, 651, 721, 793, 870, 949, 1033, 1119, 1210, 1303, 1401, 1501, 1606, 1713, 1825, 1939, 2058, 2179, 2305, 2433, 2566, 2701, 2841, 2983, 3130, 3279, 3433, 3589, 3750, 3913, 4081
FORMULA
a(n) = (1/8) * [14n^2 + 8n + 5 + 3(-1)^n], for n>4.
G.f.: -(2*x^8+2*x^7-7*x^6-5*x^5+8*x^4+5*x^3+x^2+1) / ((x-1)^3*(x+1)). - Colin Barker, Jul 14 2013
EXAMPLE
2 squares are reachable after 1 move, from these you can reach 3 new squares more, so a(1)=2, a(2)=5.
The number of distinct positions on an infinite chessboard reachable by the (3,4)-leaper in <= n moves.
+10
2
1, 9, 41, 129, 321, 681, 1289, 2121, 3081, 4121, 5233, 6445, 7777, 9233, 10813, 12517, 14345, 16297, 18373, 20573, 22897, 25345, 27917, 30613, 33433, 36377, 39445, 42637, 45953, 49393, 52957, 56645, 60457, 64393, 68453, 72637, 76945, 81377, 85933, 90613, 95417
FORMULA
Conjecture: a(n) = 62*n^2 + 30*n - 55 for n >= 10.
G.f.: (1 + 6*x + 17*x^2 + 32*x^3 + 48*x^4 + 64*x^5 + 80*x^6 - 24*x^7 - 96*x^8 - 48*x^9 - 8*x^10 + 28*x^11 + 20*x^12 + 4*x^13) / (1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>13.
(End)
Number of squares on infinite chessboard that a knight can reach in n moves from a fixed square.
+10
1
1, 8, 33, 76, 129, 196, 277, 372, 481, 604, 741, 892, 1057, 1236, 1429, 1636, 1857, 2092, 2341, 2604, 2881, 3172, 3477, 3796, 4129, 4476, 4837, 5212, 5601, 6004, 6421, 6852, 7297, 7756, 8229, 8716, 9217, 9732, 10261, 10804, 11361, 11932, 12517, 13116, 13729, 14356, 14997, 15652
REFERENCES
M. Petkovic, Mathematics and Chess, Dover Publications (2003), Problem 3.11.
FORMULA
a(n) = -3 + 4*n + 7*n^2 + 4*sign((n-2)*(n-1)).
G.f.: (1 + 5*x + 12*x^2 - 8*x^4 + 4*x^5)/(1 - x)^3.
E.g.f.: exp(x)*(1 + 11*x + 7*x^2) - 2*x*(x + 2). - Stefano Spezia, Jul 27 2022
EXAMPLE
a(2)=33 because knight in 2 moves from square (0,0) can reach one of the following squares: {{0,0}, {-4,-2}, {-4,0}, {-4,2}, {-3,-3}, {-3,-1}, {-3,1}, {-3,3}, {-2,-4}, {-2,0}, {-2,4}, {-1,-3}, {-1,-1}, {-1,1}, {-1,3}, {0,-4}, {0,-2}, {0,2}, {0,4}, {1,-3}, {1,-1}, {1,1}, {1,3}, {2,-4}, {2,0}, {2,4}, {3,-3}, {3,-1}, {3,1}, {3,3}, {4,-2}, {4,0}, {4,2}}.
MATHEMATICA
Table[ -3 + 4*n + 7*n^2 + 4*Sign[(n - 2)(n - 1)], {n, 0, 100}]
CoefficientList[Series[(1+5*x+12*x^2-8*x^4+4*x^5)/(1-x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 09 2012 *) Join[{1, 8, 33}, LinearRecurrence[{3, -3, 1}, {76, 129, 196}, 50]] (* Harvey P. Dale, Dec 05 2014 *)
PROG
(Magma) I:=[1, 8, 33, 76, 129, 196, 277]; [n le 7 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]: // Vincenzo Librandi, Jul 09 2012
AUTHOR
Anton Chupin (chupin(AT)icmm.ru), May 14 2006
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