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

0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 18, 19, 21, 22, 23, 24, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 44, 45, 46, 47, 49, 50, 52, 53, 54, 55, 57, 58, 59, 61, 62, 63, 64, 66, 67, 68, 70, 71, 72, 73, 75, 76, 78, 79, 80, 81, 83, 84, 85, 87, 88, 89, 90, 92, 93

Offset: 0

Author

Paul D. Hanna, Jun 04 2008

Keywords

• nonn

Links

• N. J. A. Sloane, Table of n, a(n) for n = 0..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. A140103 (complement); A140100, A140101; A058265.

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}];
Join[{0}, 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) = A140101(n) - A140100(n).

Theorem: the limit of A140103(n)/A140102(n) = t^2 = 3.38297576...

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.

Offset and initial term changed by N. J. A. Sloane, Oct 10 2016