A287151
Array read by antidiagonals: T(m,n) = number of nonzero m X n binary arrays with all 1's connected.
Original entry on oeis.org
1, 3, 3, 6, 13, 6, 10, 40, 40, 10, 15, 108, 218, 108, 15, 21, 275, 1126, 1126, 275, 21, 28, 681, 5726, 11506, 5726, 681, 28, 36, 1664, 28992, 116166, 116166, 28992, 1664, 36, 45, 4040, 146642, 1168586, 2301877, 1168586, 146642, 4040, 45, 55, 9779, 741556, 11749134, 45280509, 45280509, 11749134, 741556, 9779, 55
Offset: 1
Table starts:
====================================================================
m\n| 1 2 3 4 5 6 7
---|----------------------------------------------------------------
1 | 1 3 6 10 15 21 28 ...
2 | 3 13 40 108 275 681 1664 ...
3 | 6 40 218 1126 5726 28992 146642 ...
4 | 10 108 1126 11506 116166 1168586 11749134 ...
5 | 15 275 5726 116166 2301877 45280509 889477656 ...
6 | 21 681 28992 1168586 45280509 1732082741 66037462454 ...
7 | 28 1664 146642 11749134 889477656 66037462454 4872949974666 ...
...
A203717
A Catalan triangle by rows.
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 1, 8, 4, 1, 1, 20, 15, 5, 1, 1, 50, 53, 21, 6, 1, 1, 126, 182, 84, 28, 7, 1, 1, 322, 616, 326, 120, 36, 8, 1, 1, 834, 2070, 1242, 495, 165, 45, 9, 1, 1, 2187, 6930, 4680, 1997, 715, 220, 55, 10, 1, 1, 5797, 23166, 17512, 7942, 3003, 1001, 286, 66, 11, 1
Offset: 1
First few rows of the array begin:
1,...1,...1,...1,...1,...;
1,...2,...4,...9,..21,...; = A001006
1,...2,...5,..13,..36,...; = A036765
1,...2,...5,..14,..41,...; = A036766
1,...2,...5,..14,..42,...; = A036767
... Taking finite differences of array terms starting from the top by columns, we obtain row terms of the triangle. First few rows of the triangle are:
1;
1, 1;
1, 3, 1;
1, 8, 4, 1;
1, 20, 15, 5, 1;
1, 50, 53, 21, 6, 1;
1, 126, 182, 84, 28, 7, 1;
1, 322, 616, 326, 120, 36, 8, 1;
1, 834, 2070, 1242, 495, 165, 45, 9, 1;
1, 2187, 6930, 4680, 1997, 715, 220, 55, 10, 1;
...
Example: Row 4 of the triangle = (1, 8, 4, 1) = the finite differences of (1, 9, 13, 14), column 4 of the array. Term (3,4) = 13 of the array is the upper left term of M^4, where M is an infinite square production matrix with four diagonals of 1's starting at (1,2), (1,1), (2,1), and (3,1); with the rest zeros.
-
b:= proc(n, t, k) option remember; `if`(n=0, 1, `if`(t>0,
add(b(j-1, k$2)*b(n-j, t-1, k), j=1..n), b(n-1, k$2)))
end:
T:= (n, k)-> b(n, k-1$2) -`if`(k=1, 0, b(n, k-2$2)):
seq(seq(T(n, k), k=1..n), n=1..14); # Alois P. Heinz, Jun 29 2014
# second Maple program:
b:= proc(u, o, k) option remember; `if`(u+o=0, 1,
add(b(u-j, o+j-1, k), j=1..min(1, u))+
add(b(u+j-1, o-j, k), j=1..min(k, o)))
end:
T:= (n, k)-> b(0, n, k)-`if`(k=0, 0, b(0, n, k-1)):
seq(seq(T(n, k), k=1..n), n=1..14); # Alois P. Heinz, Aug 28 2017
-
b[n_, t_, k_] := b[n, t, k] = If[n == 0, 1, If[t > 0, Sum[b[j-1, k, k]*b[n - j, t-1, k], {j, 1, n}], b[n-1, k, k]]]; T[n_, k_] := b[n, k-1, k-1] - If[k == 1, 0, b[n, k-2, k-2]]; Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 27 2016, after Alois P. Heinz *)
-
from sympy.core.cache import cacheit
@cacheit
def b(u, o, k): return 1 if u + o==0 else sum([b(u - j, o + j - 1, k) for j in range(1, min(1, u) + 1)]) + sum([b(u + j - 1, o - j, k) for j in range(1, min(k, o) + 1)])
def T(n, k): return b(0, n, k) - (0 if k==0 else b(0, n, k - 1))
for n in range(1, 16): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Aug 30 2017
A167630
Riordan array (1/(1-x),xm(x)) where m(x) is the g.f. of Motzkin numbers A001006.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 8, 4, 1, 1, 17, 20, 13, 5, 1, 1, 38, 50, 38, 19, 6, 1, 1, 89, 126, 107, 63, 26, 7, 1, 1, 216, 322, 296, 196, 96, 34, 8, 1, 1, 539, 834, 814, 588, 326, 138, 43, 9, 1, 1, 1374, 2187, 2236, 1728, 1052, 507, 190, 53, 10, 1
Offset: 0
Triangle begins:
1;
1, 1;
1, 2, 1;
1, 4, 3, 1;
1, 8, 8, 4, 1;
1, 17, 20, 13, 5, 1;
1, 38, 50, 38, 19, 6, 1;
...
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T:= proc(n, k) option remember; `if`(k=0, 1,
`if`(k>n, 0, T(n-1, k-1)+T(n-1, k)+T(n-1, k+1)))
end:
seq(seq(T(n, k), k=0..n), n=0..12); # Alois P. Heinz, Apr 20 2018
-
T[, 0] = T[n, n_] = 1;
T[n_, k_] /; 0, ] = 0;
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 09 2019 *)
A379838
Triangle read by rows: T(n,k) is the total number of humps with height k in all Motzkin paths of order n, n >= 2 and 1 <= k <= n/2.
Original entry on oeis.org
1, 3, 8, 1, 20, 5, 50, 19, 1, 126, 63, 7, 322, 196, 34, 1, 834, 588, 138, 9, 2187, 1728, 507, 53, 1, 5797, 5016, 1749, 253, 11, 15510, 14454, 5786, 1067, 76, 1, 41834, 41470, 18590, 4147, 416, 13, 113633, 118690, 58487, 15223, 1976, 103, 1, 310571, 339274, 181181, 53599, 8528, 635, 15
Offset: 2
Triangle begins:
[2] 1;
[3] 3;
[4] 8, 1;
[5] 20, 5;
[6] 50, 19, 1;
[7] 126, 63, 7;
[8] 322, 196, 34, 1;
[9] 834, 588, 138, 9;
[10] 2187, 1728, 507, 53, 1;
...
-
def A379838_triangel(dim):
M = matrix(ZZ, dim, dim)
for n in (2..dim+1):
for k in (1..math.floor(n/2)+1):
for i in range(n-2*k+1):
if ((n-i)%2)==0:
M[n-2,k-1]=M[n-2, k-1]+(4*k)/(n-i+2*k)*binomial(n,i)*binomial(n-i-1,(n-i)/2+k-1)
return M
A378947
Number of row states in an automaton for the enumeration of the number of fixed polyominoes with bounding box of width n.
Original entry on oeis.org
1, 2, 6, 16, 40, 99, 247, 625, 1605, 4178, 11006, 29292, 78652, 212812, 579672, 1588242, 4374282, 12103404, 33628824, 93786966, 262450878, 736710357, 2073834417, 5853011847, 16558618507, 46949351272, 133390812252, 379708642286, 1082797114046, 3092894319075, 8848275403639
Offset: 0
-
a:= proc(n) option remember; `if`(n<3, [1, 2, 6][n+1],
((3*n^2+2*n-12)*a(n-1)+(n^2-13*n+15)*a(n-2)
-3*(n-3)*(n-1)*a(n-3))/((n-2)*(n+3)))
end:
seq(a(n), n=0..30); # Alois P. Heinz, Dec 20 2024
-
a[n_] := a[n] = If[n < 3, {1, 2, 6}[[n+1]],
((3*n^2 + 2*n - 12)*a[n-1] + (n^2 - 13*n + 15)*a[n-2]
- 3*(n-3)*(n-1)*a[n-3])/((n-2)*(n+3))];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 26 2025, after Alois P. Heinz *)
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b(n) = (1 + (hammingweight(bitxor(n, n>>1)))) >> 1;
C(n) = binomial(2*n, n)/(n+1);
a(n) = 1 + sum(m=1, 2^n-1, C(b(m)) * 2^((m % 2)==0) * 2^(m<2^(n-1))); \\ Michel Marcus, Dec 12 2024
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a(n) = {1 + sum(k=1, (n+1)\2, (binomial(n+1, 2*k)+2*binomial(n,2*k)+binomial(n-1,2*k))*binomial(2*k, k)/(k+1))} \\ Andrew Howroyd, Dec 17 2024
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