A202952 A108235(n)-A202951(n).
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Links
- R. J. Nowakowski, Generalizations of the Langford-Skolem problem, M.S. Thesis, Dept. Math., Univ. Calgary, May 1975. [Scanned copy, with permission.]
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
For n = 3, the unique solution is 1 + 2 = 3. For n = 12, there are 8 solutions: 1 5 6 | 1 5 6 | 2 5 7 | 1 6 7 2 8 10 | 3 7 10 | 3 6 9 | 4 5 9 4 7 11 | 2 9 11 | 1 10 11 | 3 8 11 3 9 12 | 4 8 12 | 4 8 12 | 2 10 12 --------+---------+---------+-------- 2 4 6 | 2 6 8 | 3 4 7 | 3 5 8 1 9 10 | 4 5 9 | 1 8 9 | 2 7 9 3 8 11 | 3 7 10 | 5 6 11 | 4 6 10 5 7 12 | 1 11 12 | 2 10 12 | 1 11 12
nxyz(v,t)=local(n,r,x2); r=0; if(t==0,return(1)); for(i3=3*t,#v, n=v[i3]; for(i1=1,i3-2, x2=n-v[i1]; if(x2<=v[i1],break); for(i2=i1+1,i3-1, if(v[i2]>=x2, if(v[i2]==x2, r+=nxyz(vector(i3-3,k,v[if(kFranklin T. Adams-Watters
Examples from _Alois P. Heinz_, Feb 12 2010: A002848(7) = 3: [1, 3, 4], [2, 5, 7] [1, 5, 6], [3, 4, 7] [2, 3, 5], [1, 6, 7] A002848(8) = 7: [1, 3, 4], [2, 6, 8] [1, 4, 5], [2, 6, 8] [1, 6, 7], [3, 5, 8] [2, 3, 5], [1, 7, 8] [2, 4, 6], [1, 7, 8] [2, 4, 6], [3, 5, 8] [3, 4, 7], [2, 6, 8] A002848(10) = 12: [1, 4, 5], [2, 6, 8], [3, 7, 10] [1, 4, 5], [3, 6, 9], [2, 8, 10] [1, 5, 6], [3, 4, 7], [2, 8, 10] [1, 6, 7], [4, 5, 9], [2, 8, 10] [1, 7, 8], [2, 3, 5], [4, 6, 10] [1, 8, 9], [2, 3, 5], [4, 6, 10] [1, 8, 9], [2, 4, 6], [3, 7, 10] [1, 8, 9], [2, 5, 7], [4, 6, 10] [2, 4, 6], [3, 5, 8], [1, 9, 10] [2, 6, 8], [3, 4, 7], [1, 9, 10] [2, 6, 8], [4, 5, 9], [3, 7, 10] [2, 7, 9], [3, 5, 8], [4, 6, 10] See A002849 for further examples.
a(1) = 1: {1,2,3,4}. a(2) = 4: {1,2,7,8}, {3,4,5,6}; {1,3,6,8}, {2,4,5,7}; {1,4,5,8}, {2,3,6,7}; {1,4,6,7}, {2,3,5,8}.
b:= proc() option remember; local i, j, t, m; m:= args[nargs]; if args[1]=0 then `if`(nargs=2, 1, b(args[t] $t=2..nargs)) elif args[1]<1 then 0 else add(`if`(args[j]`if`(n=0, 1, b(((8*n+2)+4/97) $n, 4*n)/n!): seq(a(n), n=0..6);
b[l_] := b[l] = Module[{nl = Length[l], k = l[[-1]], m = l[[-2]]}, Which[l[[1]] == 0, If[nl == 3, 1, b[l[[2 ;; nl]]]], l[[1]] < 1, 0, True, Sum[If[l[[j]] < m, 0, b[Join[Sort[Table[l[[i]] - If[i == j, m + 1/97, 0], {i, 1, nl - 2}]], {m - 1, k}]]], {j, 1, nl - 2}]]]; a[n_] := If[n == 0, 1, b[Join[Array[8*n + 2 + 4/97& , n], {4*n, 4}]]/n!]; Table[a[n], {n, 0, 6}] (* Jean-François Alcover, Jun 03 2018, adapted from Maple *)
a(0) = 1: the empty permutation. a(2) = 1: 221121. a(3) = 1: 223321131. a(5) = 3: 223325534411514, 225523344531141, 552244253341131. a(6) = 6: 221121665544336543, 225523366534411614, 225526633544361141, 446611415563322532, 552266253344631141, 665544336543221121.
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