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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.
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%I #44 Oct 18 2022 03:00:39

%S 3,8,12,17,20,25,29,34,39,43,48,51,56,60,65,69,74,77,82,86,91,96,100,

%T 105,108,113,117,122,125,130,134,139,144,148,153,156,161,165,170,174,

%U 179,182,187,191,196,201,205,210,213,218,222,227,232,236,241,244,249

%N 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.

%C 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

%D Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 2.

%H N. J. A. Sloane, <a href="/A140103/b140103.txt">Table of n, a(n) for n=1..50000</a>, Sep 13 2016 (First 1001 terms from Reinhard Zumkeller)

%H F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, <a href="https://www.combinatorics.org/ojs/index.php/eljc/article/view/v27i1p52/8039">Queens in exile: non-attacking queens on infinite chess boards</a>, Electronic J. Combin., 27:1 (2020), #P1.52.

%H N. J. A. Sloane, <a href="/A140100/a140100.txt">Maple program for A140100, A140101, A140102, A140103</a>

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

%F 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.

%p See link.

%t 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]]]];

%t Do[x[n], {n, 1, nmax}];

%t yy + xx (* _Jean-François Alcover_, Aug 01 2018 *)

%o (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))))))}

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

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

%K nonn

%O 1,1

%A _Paul D. Hanna_, Jun 04 2008

%E Terms computed by _Reinhard Zumkeller_