A259697 Triangle read by rows: T(n,k) = number of rook patterns on n X n board where bottom rook is in column k.
1, 1, 1, 1, 2, 2, 1, 4, 5, 5, 1, 9, 12, 15, 15, 1, 24, 32, 42, 52, 52, 1, 76, 99, 129, 166, 203, 203, 1, 279, 354, 451, 575, 726, 877, 877, 1, 1156, 1434, 1786, 2232, 2792, 3466, 4140, 4140, 1, 5296, 6451, 7883, 9664, 11881, 14621, 17884, 21147, 21147
Offset: 0
Examples
Triangle begins: 1, 1, 1, 1, 2, 2, 1, 4, 5, 5, 1, 9, 12, 15, 15, 1, 24, 32, 42, 52, 52, 1, 76, 99, 129, 166, 203, 203, ... From _Andrew Howroyd_, Jun 13 2017: (Start) For n=3, the four solutions with the bottom rook in column 1 are: . . . x . . . x x . . . x . . x . . . . . . . . For n=3, the five solutions with the bottom rook in column 2 are: . . x . x . . x . . . . x . x . x . . x . . x . . . . . . . (End)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1274
- H. W. Becker, Rooks and rhymes, Math. Mag. 22 (1948/49), 23-26. See Table V.
- H. W. Becker, Rooks and rhymes, Math. Mag., 22 (1948/49), 23-26. [Annotated scanned copy]
Crossrefs
Programs
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Mathematica
a11971[n_, k_] := Sum[Binomial[k, i]*BellB[n - k + i], {i, 0, k}]; T[, 0] = 1; T[n, k_] := Sum[a11971[i - 1, k - 1], {i, k, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 07 2018, after Andrew Howroyd *)
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PARI
A(N)={my(T=matrix(N,N),U=matrix(N,N));T[1,1]=1;U[1,1]=1; for(n=2,N,for(k=1,n, T[n,k]=if(k==1,T[n-1,n-1],T[n-1,k-1]+T[n,k-1]); U[n,k]=T[n,k]+U[n-1,k]));U} {my(T=A(10));for(n=0,length(T),for(k=0,n,print1(if(k==0,1,T[n,k]),", "));print)} \\ Andrew Howroyd, Jun 13 2017
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Python
from sympy import bell, binomial def a011971(n, k): return sum([binomial(k, i)*bell(n - k + i) for i in range(k + 1)]) def T(n, k): return 1 if k==0 else sum([a011971(i - 1, k - 1) for i in range(k, n + 1)]) for n in range(10): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Jun 17 2017
Formula
T(n,0) = 1, T(n,k) = Sum_{i=k..n} A011971(i-1,k-1) for k>0. - Andrew Howroyd, Jun 13 2017
Extensions
Terms a(28) and beyond from Andrew Howroyd, Jun 13 2017
Comments