A236266 Lexicographically earliest sequence of nonnegative integers such that no three points (i,a(i)), (j,a(j)), (n,a(n)) are collinear.
0, 0, 1, 1, 4, 3, 8, 2, 2, 5, 7, 4, 5, 8, 16, 3, 7, 14, 12, 23, 16, 12, 25, 31, 13, 6, 11, 28, 11, 17, 9, 9, 22, 34, 6, 15, 13, 29, 23, 22, 29, 45, 26, 19, 51, 14, 24, 39, 28, 39, 18, 37, 57, 17, 38, 41, 15, 68, 32, 24, 66, 42, 10, 50, 27, 10, 53, 72, 25, 26
Offset: 0
A174085 Number of permutations of length n with no consecutive triples i,...i+r,...i+2r for all positive and negative r, and for all equal spacings d.
1, 1, 2, 4, 18, 72, 396, 2328, 17050, 131764, 1199368, 11379524, 123012492, 1386127700, 17450444866, 227152227940
Offset: 0
Comments
Here we count both the sequence 1,2,3 (r=1) as a progression in 1,2,3,0,4,5, (note d=1) and in 1,0,2,4,3,5 (here, d=2).
Number of permutations of 1..n with no 2-dimensional arithmetic progression of length 3: that is, no three points (i,p(i)), (j,p(j)) and (k,p(k)) such that j-i = k-j and p(j)-p(i) = p(k)-p(j). - David Bevan, Jun 16 2021
Examples
a(3) = 4; 123 and 321 each contain a 3-term arithmetic progression. Since the only possibilities for progressions for n=4 are d=1 and r=1 and -1, we get the same term as A095816(4).
Crossrefs
Formula
a(n) >= A003407(n) with equality only for n in {0, 1, 2, 3}.
Extensions
a(0)-a(3) and a(10)-a(13) from David Bevan, Jun 16 2021
a(14)-a(15) from Bert Dobbelaere, May 18 2025
A255708 No three points (i,a(i)), (j,a(j)), (k,a(k)) are collinear, for n = 0,1,2,... the value of a(n) is chosen to be m or -m (in this order) for the smallest m>=0 satisfying the condition.
0, 0, 1, 1, -1, -1, 4, 2, 2, -3, -5, -2, -7, -2, 5, 3, 3, -5, -4, -4, 6, 5, -6, -3, -10, 11, -6, 4, 18, 11, 19, 7, 12, 12, 6, -13, 19, 7, -10, -7, -9, -14, 13, 23, -28, -8, -14, 9, 8, -22, -9, -8, 23, -11, 15, 22, 13, 8, -21, -13, -26, 9, -12, -12, -11, 40, 21
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..20000
Programs
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Maple
a:= proc(n) option remember; local i, j, k, t, ok; for t from 0 do for k in [t, -t] do ok:=true; for j from n-1 to 1 by -1 while ok do for i from j-1 to 0 by -1 while ok do ok:= (n-j)*(a(j)-a(i))<>(j-i)*(k-a(j)) od od; if ok then return k fi od od end: seq(a(n), n=0..60);
A255709 No three points (i,a(i)), (j,a(j)), (k,a(k)) are collinear and all values distinct, for n = 0,1,2,... the value of a(n) is chosen to be m or -m (in this order) for the smallest m>=0 satisfying the condition.
0, 1, -1, 2, 3, -2, -5, -3, 4, -6, 6, -7, -4, 5, 12, 16, 7, 8, -10, -8, 9, 19, 14, -12, -14, -9, 21, 10, -11, -15, 17, 15, -19, 13, -22, -13, -16, -24, 11, 18, 22, -18, 25, 23, -17, 24, 40, -21, -38, 20, -29, 36, -30, -20, 32, -34, 26, 43, -23, 37, -26, 33
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..20000
Programs
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Maple
b:= proc() true end: a:= proc(n) option remember; local i, j, k, t, ok; for t from 0 do for k in [t, -t] do ok:=b(k); for j from n-1 to 1 by -1 while ok do for i from j-1 to 0 by -1 while ok do ok:= (n-j)*(a(j)-a(i))<>(j-i)*(k-a(j)) od od; if ok then b(k):=false; return k fi od od end: seq(a(n), n=0..60);
Comments
Examples
For n=4 the value of a(n) cannot be less than 4 because otherwise we would have a set of three collinear points, {(0,0),(1,0),(4,0)} or {(2,1),(3,1),(4,1)} or {(0,0),(2,1),(4,2)} or {(1,0),(2,1),(4,3)}. Thus a(4) = 4 is the first value that is in accordance with the constraints.Links
Crossrefs
Programs
Maple
a:= proc(n) option remember; local i, j, k, ok; for k from 0 do ok:=true; for j from n-1 to 1 by -1 while ok do for i from j-1 to 0 by -1 while ok do ok:= (n-j)*(a(j)-a(i))<>(j-i)*(k-a(j)) od od; if ok then return k fi od end: seq(a(n), n=0..60);Mathematica
a[0] = a[1] = 0; a[n_] := a[n] = Module[{i, j, k, ok}, For[k = 0, True, k++, ok = True; For[j = n-1, ok && j >= 1, j--, For[i = j-1, ok && i >= 0, i--, ok = (n-j)*(a[j]-a[i]) != (j-i)*(k-a[j])]]; If[ok, Return[k]]]]; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Jun 16 2018, after Alois P. Heinz *)Formula