A332862 Array read by antidiagonals: T(m,n) = number of placements of zero or more dominoes on the m X n grid where no two empty squares are horizontally adjacent.
1, 1, 2, 2, 4, 3, 2, 11, 9, 5, 3, 25, 48, 25, 8, 4, 61, 172, 227, 64, 13, 5, 146, 731, 1427, 1054, 169, 21, 7, 351, 2976, 10388, 11134, 4921, 441, 34, 9, 844, 12039, 72751, 140555, 88733, 22944, 1156, 55, 12, 2028, 49401, 510779, 1693116, 1932067, 701926, 107017, 3025, 89
Offset: 1
Examples
Table starts: =================================================================== m\n| 1 2 3 4 5 6 7 ---|--------------------------------------------------------------- 1 | 1 1 2 2 3 4 5 ... 2 | 2 4 11 25 61 146 351 ... 3 | 3 9 48 172 731 2976 12039 ... 4 | 5 25 227 1427 10388 72751 510779 ... 5 | 8 64 1054 11134 140555 1693116 20414525 ... 6 | 13 169 4921 88733 1932067 40008789 831347033 ... 7 | 21 441 22944 701926 26425981 941088936 33656587715 ... 8 | 34 1156 107017 5567467 362036629 22168654178 1365206879940 ... ...
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..378
- Bjorn Huntemann, Svenja Huntemann, Neil A. McKay, SageMath code for Counting Domineering Positions
- Svenja Huntemann, Neil A. McKay, Counting Domineering Positions, arXiv:1909.12419 [math.CO], 2019.
Crossrefs
Programs
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PARI
\\ here R(n) is row 1 as vector. R(n)={Vec((1+x+x^2)/(1-x^2-x^3)+O(x*x^n))} F(b,r)={my(t=1); while(b, b=(b>>valuation(b,2))+1; my(s=valuation(b,2)); t*=r[s]; b>>=s+1); t} step(v,f)={vector(#v, t, my(i=t-1); sum(j=0, #v-1, if(!bitand(i,j), v[1+j]*(f[#v-bitor(i,j)]))))} T(m,n)={my(r=R(n), f=vector(2^n, i, F(i-1, r)), v=vector(2^n)); v[1]=1; for(k=2, m, v=step(v,f)); sum(j=0, #v-1, v[1+j]*f[#v-j])} {for(m=1, 8, for(n=1, 8, print1(T(m,n), ", ")); print)} \\ Andrew Howroyd, Feb 28 2020 -
Sage
# See Bjorn Huntemann, Svenja Huntemann, Neil A. McKay link.
Comments