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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
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refs;
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internal format)
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OFFSET
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1,1
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COMMENTS
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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
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REFERENCES
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Robbert Fokkink, Gerard Francis Ortega, Dan Rust, Corner the Empress, arXiv:2204.11805. See Table 2.
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LINKS
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FORMULA
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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.
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MAPLE
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See link.
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MATHEMATICA
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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}];
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PROG
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(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))))))}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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