A140103 Term-by-term sums of A140101 and A140100; also, equals the complement of A140102, which is the term-by-term differences of A140101 and A140100, where A140101 is the complement of A140100.

3, 8, 12, 17, 20, 25, 29, 34, 39, 43, 48, 51, 56, 60, 65, 69, 74, 77, 82, 86, 91, 96, 100, 105, 108, 113, 117, 122, 125, 130, 134, 139, 144, 148, 153, 156, 161, 165, 170, 174, 179, 182, 187, 191, 196, 201, 205, 210, 213, 218, 222, 227, 232, 236, 241, 244, 249

Offset: 1

Author

Paul D. Hanna, Jun 04 2008

Keywords

• nonn

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

Crossrefs

Cf. A140102 (complement); A140100, A140101; A058265, A276800.

For first differences of A140100, A140101, A140102, A140103 see A305392, A305374, A305393, A305394.

Programs

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 *)

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))))))}

Formula

a(n) = A140100(n) + A140101(n).

Conjecture: the limit of A140103(n)/A140102(n) = t^2 = 3.38297576... (cf. A276800) where the limit of A140101(n)/A140100(n) = t = 1.839286755.. and t = tribonacci constant satisfies: t^3 = 1 + t + t^2.

Extensions

Terms computed by Reinhard Zumkeller