A274105 Triangle read by rows: T(n,k) = number of configurations of k nonattacking bishops on the black squares of an n X n chessboard (0 <= k <= n - [n>1]).
1, 1, 1, 1, 2, 1, 5, 4, 1, 8, 14, 4, 1, 13, 46, 46, 8, 1, 18, 98, 184, 100, 8, 1, 25, 206, 674, 836, 308, 16, 1, 32, 356, 1704, 3532, 2816, 632, 16, 1, 41, 612, 4196, 13756, 20476, 11896, 1912, 32, 1, 50, 940, 8480, 38932, 89256, 93800, 37600, 3856, 32, 1, 61, 1440, 16940, 106772, 361780, 629336, 506600, 154256, 11600, 64
Offset: 0
Examples
Triangle begins: 1; 1, 1; 1, 2; 1, 5, 4; 1, 8, 14, 4; 1, 13, 46, 46, 8; 1, 18, 98, 184, 100, 8; 1, 25, 206, 674, 836, 308, 16; 1, 32, 356, 1704, 3532, 2816, 632, 16; 1, 41, 612, 4196, 13756, 20476, 11896, 1912, 32; 1, 50, 940, 8480, 38932, 89256, 93800, 37600, 3856, 32; 1, 61, 1440, 16940, 106772, 361780, 629336, 506600, 154256, 11600, 64; ... Corresponding independence polynomials: 1, (empty graph) 1+x, (K_1) 1+2*x, (P_2 = K_2) 1+5*x+4*x^2, (butterfly graph) 1+8*x+14*x^2+4*x^3, ...
Links
- Irving Kaplansky and John Riordan, The problem of the rooks and its applications, Duke Mathematical Journal 13.2 (1946): 259-268. See Section 9.
- Irving Kaplansky and John Riordan, The problem of the rooks and its applications, in Combinatorics, Duke Mathematical Journal, 13.2 (1946): 259-268. See Section 9. [Annotated scanned copy]
- Eder G. Santos, Counting non-attacking chess pieces placements: Bishops and Anassas. arXiv:2411.16492 [math.CO], 2024. (considered as white board).
- Eric Weisstein's World of Mathematics, Black Bishop Graph
- Eric Weisstein's World of Mathematics, Independence Polynomial
Crossrefs
Programs
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Maple
with(combinat); with(gfun); T:=n->add(stirling2(n+1,n+1-k)*x^k, k=0..n); # bishops on black squares bish:=proc(n) local m,k,i,j,t1,t2; global T; if n<2 then return [1$(n+1)] fi; if (n mod 2) = 0 then m:=n/2; t1:=add(binomial(m,k)*T(2*m-1-k)*x^k, k=0..m); else m:=(n-1)/2; t1:=add(binomial(m+1,k)*T(2*m-k)*x^k, k=0..m+1); fi; seriestolist(series(t1,x,2*n+1)); end; for n from 0 to 12 do lprint(bish(n)); od: # second Maple program: T:= (n,k)-> add(binomial(ceil(n/2),j)*Stirling2(n-j,n-k),j=0..k): seq(seq(T(n,k), k=0..n-`if`(n>1,1,0)), n=0..11); # Alois P. Heinz, Dec 01 2024
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Mathematica
CoefficientList[Table[Sum[x^n Binomial[Ceiling[n/2], k] BellB[n - k, 1/x], {k, 0, Ceiling[n/2]}], {n, 10}], x] (* Eric W. Weisstein, Jun 26 2017 *) -
SageMath
def stirling2_negativek(n, k): if k < 0: return 0 else: return stirling_number2(n, k) print([sum([binomial(ceil(n/2), l)*stirling2_negativek(n-l, n-k) for l in [0..k]]) for n in [0..10] for k in [0..n-1+kronecker_delta(n,1)+kronecker_delta(n,0)]]) # Eder G. Santos, Dec 01 2024
Formula
From Eder G. Santos, Dec 01 2024: (Start)
T(n,k) = Sum_{j=0..k} binomial(ceiling(n/2),j) * Stirling2(n-j,n-k).
T(n,k) = T(n-1,k) + (n-k+A000035(n)) * T(n-1,k-1), T(n,0) = 1, T(0,k) = delta(k,0). (End)
Extensions
T(0,0) prepended by Eder G. Santos, Dec 01 2024
Comments