OFFSET
1,1
COMMENTS
Conjecture: a(n) = A003145(n) + n. This is the most direct connection between the Greedy Queens sequence and the tribonacci word that I know. - Michel Dekking, Mar 19 2019. [My notes show that I made this conjecture on Jul 20 2018. There are many similar conjectures relating the two problems. For example A140100 = A003145(n)-A003144(n), A140101(n) = A003146(n)-A003145(n), a(n) = A003146(n)-A003144(n). - N. J. A. Sloane, Mar 19 2019] All these conjectures are now theorems - see the Dekking et al. paper. - N. J. A. Sloane, Jul 22 2019
REFERENCES
Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 2.
LINKS
N. J. A. Sloane, Table of n, a(n) for n=1..50000, Sep 13 2016 (First 1001 terms from Reinhard Zumkeller)
F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), #P1.52.
N. J. A. Sloane, Maple program for A140100, A140101, A140102, A140103
FORMULA
MAPLE
See link.
MATHEMATICA
nmax = 100; y[0] = 0; x[1] = 1; y[1] = 2; x[n_] := x[n] = For[yn = y[n-1] + 1, True, yn++, For[xn = x[n-1] + 1, xn < yn, xn++, xx = Array[x, n-1]; yy = Array[y, n-1]; If[FreeQ[xx, xn | yn] && FreeQ[yy, xn | yn] && FreeQ[yy - xx, yn - xn] && FreeQ[yy + xx, yn - xn], y[n] = yn; Return[xn]]]];
Do[x[n], {n, 1, nmax}];
yy + xx (* Jean-François Alcover, Aug 01 2018 *)
PROG
(PARI) {X=[1]; Y=[2]; D=[1]; S=[3]; print1(Y[1]-X[1]", "); for(n=1, 100, for(j=2, 2*n, if(setsearch(Set(concat(X, Y)), j)==0, Xt=concat(X, j); for(k=j+1, 3*n, if(setsearch(Set(concat(Xt, Y)), k)==0, if(setsearch(Set(concat(D, S)), k-j)==0, if(setsearch(Set(concat(D, S)), k+j)==0, X=Xt; Y=concat(Y, k); D=concat(D, k-j); S=concat(S, k+j); print1(Y[ #X]-X[ #Y]", "); break); break))))))}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 04 2008
EXTENSIONS
Terms computed by Reinhard Zumkeller
STATUS
approved