A336351 Number of cyclic arrangements of S = {1,2,...,6n - 3} such that any three neighbors can be reordered in an arithmetic progression.
1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 7, 9, 12, 16, 22, 30, 41, 55, 74, 100
Offset: 1
Examples
The cycle with n=1 is {1,2,3}.
The cycle with n=2 is {1,3,2,4,6,8,7,9,5}.
Links
- Fausto A. C. Cariboni, Complete solutions for a(1)-a(21)
Crossrefs
Cf. A334625.
Programs
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PARI
getval(va, pos, typ) = {my(vr = vecsort([va[pos-2], va[pos-1]])); if (typ==0, return (2*vr[1] - vr[2])); if (typ==1, return ((vr[1]+vr[2])/2)); if (typ==2, return (2*vr[2] - vr[1])); error("typ is wrong");} okval(val, n) = {if ((val < 0) || (val > n) || (denominator(val) != 1), return (0)); return (1);} ok3(v3) = my(vs=vecsort(v3)); (vs[3]-vs[2] == vs[2]-vs[1]); oklast(va, n, val) = ok3([va[n-1], val, va[1]]) && ok3([val, va[1], va[2]]); a(n) = {my(va = vector(n), vp = vector(n)); va[1] = 1; my(nb = 0); for (k=2, n, va[2] = k; my(end = 0, pos = 3, vp = vector(n)); while (! end, my(val = getval(va, pos, vp[pos])); if (okval(val, n) && ! vecsearch(vecsort(va), val), my(ok = 1); if (pos == n, ok = oklast(va, n, val)); if (ok, va[pos] = val; if (pos == n, nb++; vp[pos]++; , pos++;); , vp[pos]++;);, vp[pos]++;); while((pos>=3) && (vp[pos] == 3) && !end, if (pos == 3, end=1, vp[pos] = 0; va[pos] = 0; pos --);););); nb;} for (n=1, 10, print1(a(6*n-3)/2, ", ")) \\ Michel Marcus, Aug 02 2020
Extensions
a(21) from Fausto A. C. Cariboni, Aug 22 2020
Comments