A293087
T(n,k) = Number of n X k 0..1 arrays with the number of 1's horizontally or antidiagonally adjacent to some 0 two less than the number of 0's adjacent to some 1.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 5, 1, 0, 0, 23, 53, 10, 0, 1, 124, 486, 498, 60, 0, 3, 518, 4091, 8108, 4013, 304, 0, 10, 2103, 32070, 125848, 126903, 31689, 1420, 0, 24, 8304, 246852, 1900201, 3786544, 1955375, 246743, 6312, 0, 60, 32561, 1897509, 28709764
Offset: 1
Some solutions for n=4, k=4
..0..0..1..0. .0..0..0..0. .0..0..0..0. .0..0..1..1. .1..1..0..0
..0..0..0..0. .1..0..0..0. .0..0..0..0. .0..1..1..1. .1..1..0..0
..1..1..0..1. .1..0..0..1. .1..1..0..1. .0..0..0..0. .0..0..0..1
..0..0..1..1. .0..0..0..0. .0..0..1..0. .0..1..0..0. .0..0..1..0
A293100
T(n,k) = Number of n X k 0..1 arrays with the number of 1's horizontally, diagonally or antidiagonally adjacent to some 0 two less than the number of 0's adjacent to some 1.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 9, 5, 0, 0, 42, 48, 30, 0, 1, 150, 567, 554, 126, 0, 3, 596, 3619, 8088, 3334, 520, 0, 10, 2220, 29952, 114318, 115870, 30837, 2088, 0, 24, 8376, 205195, 1676740, 3125547, 1710994, 202363, 8192, 0, 60, 31959, 1668655, 24773296
Offset: 1
Some solutions for n=4, k=4
..0..0..0..0. .0..0..0..1. .0..1..1..0. .1..1..0..1. .1..1..0..0
..1..0..0..0. .1..1..0..0. .1..0..0..1. .0..1..0..1. .0..1..1..0
..1..1..0..1. .1..0..1..0. .1..0..1..0. .0..0..1..0. .1..1..0..0
..0..0..1..0. .0..1..0..1. .0..1..0..0. .0..0..0..1. .1..0..0..0
A289651
T(n,k) = Number of n X k 0..1 arrays with the number of 1's horizontally, antidiagonally or vertically adjacent to some 0 two less than the number of 0's adjacent to some 1.
Original entry on oeis.org
0, 0, 0, 0, 2, 0, 0, 11, 11, 0, 0, 44, 61, 44, 0, 1, 170, 607, 607, 170, 1, 3, 642, 3160, 8554, 3160, 642, 3, 10, 2421, 31271, 121162, 121162, 31271, 2421, 10, 24, 9142, 183390, 1756800, 2662095, 1756800, 183390, 9142, 24, 60, 34572, 1714758, 25760875
Offset: 1
Some solutions for n=4, k=4
..1..0..0..0. .1..0..0..0. .0..1..1..0. .0..1..0..0. .1..0..0..1
..1..1..1..0. .0..1..0..1. .0..0..1..0. .1..1..0..1. .0..1..0..1
..1..0..1..0. .0..1..1..0. .0..1..1..1. .0..1..0..1. .0..1..1..0
..0..0..1..0. .0..1..0..1. .0..1..0..0. .0..1..0..0. .0..0..1..0
A286979
T(n,k) = Number of n X k 0..1 arrays with the number of 1's horizontally or vertically adjacent to some 0 two less than the number of 0's adjacent to some 1.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 5, 5, 0, 0, 30, 50, 30, 0, 1, 126, 538, 538, 126, 1, 3, 520, 3932, 8270, 3932, 520, 3, 10, 2088, 32253, 123706, 123706, 32253, 2088, 10, 24, 8192, 238532, 1853532, 3584756, 1853532, 238532, 8192, 24, 60, 32083, 1875212, 27938670
Offset: 1
Some solutions for n=4, k=4
..0..1..1..1. .1..0..0..1. .0..0..0..0. .1..0..0..1. .0..1..1..1
..0..0..0..1. .1..0..0..1. .0..1..1..0. .0..1..0..1. .0..0..1..0
..1..0..0..0. .0..0..1..1. .0..1..1..0. .0..0..1..1. .0..1..0..1
..1..0..0..0. .1..0..0..0. .1..0..0..1. .1..0..0..0. .0..1..0..0
A307796
Number T(n,k) of binary words of length n such that k is the difference of numbers of occurrences of subword 101 and subword 010; triangle T(n,k), n>=0, -floor(n/3)<=k<=floor(n/3), read by rows.
Original entry on oeis.org
1, 2, 4, 1, 6, 1, 2, 12, 2, 6, 20, 6, 1, 12, 38, 12, 1, 3, 28, 66, 28, 3, 10, 56, 124, 56, 10, 1, 24, 119, 224, 119, 24, 1, 4, 60, 236, 424, 236, 60, 4, 15, 134, 481, 788, 481, 134, 15, 1, 42, 304, 950, 1502, 950, 304, 42, 1, 5, 114, 656, 1902, 2838, 1902, 656, 114, 5
Offset: 0
T(8,2) = 10: 01101101, 10101101, 10110101, 10110110, 10110111, 10111011, 10111101, 11011011, 11011101, 11101101.
T(8,-2) = 10: 00010010, 00100010, 00100100, 01000010, 01000100, 01001000, 01001001, 01001010, 01010010, 10010010.
T(9,3) = 1: 101101101.
T(9,-3) = 1: 010010010.
Triangle T(n,k) begins:
: 1 ;
: 2 ;
: 4 ;
: 1, 6, 1 ;
: 2, 12, 2 ;
: 6, 20, 6 ;
: 1, 12, 38, 12, 1 ;
: 3, 28, 66, 28, 3 ;
: 10, 56, 124, 56, 10 ;
: 1, 24, 119, 224, 119, 24, 1 ;
: 4, 60, 236, 424, 236, 60, 4 ;
: 15, 134, 481, 788, 481, 134, 15 ;
: 1, 42, 304, 950, 1502, 950, 304, 42, 1 ;
T(3n-4,n-2) gives
A000217 for n >= 3.
-
b:= proc(n, t, h) option remember; `if`(n=0, 1, expand(
`if`(h=3, 1/x, 1)*b(n-1, [1, 3, 1][t], 2)+
`if`(t=3, x, 1)*b(n-1, 2, [1, 3, 1][h])))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=-iquo(n, 3)..iquo(n, 3)))(b(n, 1$2)):
seq(T(n), n=0..15);
-
b[n_, t_, h_] := b[n, t, h] = If[n == 0, 1, Expand[If[h == 3, 1/x, 1]* b[n-1, {1, 3, 1}[[t]], 2] + If[t == 3, x, 1]*b[n-1, 2, {1, 3, 1}[[h]]]]];
T[n_] := Table[Coefficient[#, x, i], {i, -Quotient[n, 3], Quotient[n, 3]}]& @ b[n, 1, 1];
Table[T[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, May 08 2019, after Alois P. Heinz *)
Showing 1-5 of 5 results.
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