A266727 Stanley sequence S(0,1,7).

0, 1, 7, 8, 10, 11, 17, 18, 30, 31, 37, 38, 40, 41, 47, 48, 90, 91, 97, 98, 100, 101, 107, 108, 120, 121, 127, 128, 130, 131, 137, 138, 270, 271, 277, 278, 280, 281, 287, 288, 300, 301, 307, 308, 310, 311, 317, 318, 360, 361, 367, 368, 370, 371, 377, 378, 390

Offset: 0

Author

N. J. A. Sloane, Jan 04 2016

Keywords

• nonn

Comments

Lexicographic first increasing sequence starting (0, 1, 7, ...) and such that no 3 terms are in arithmetic progression. - M. F. Hasler, Jan 18 2016

Examples

			a(3) = 8 because no 3 terms among (0,1,7,8) are in AP. a(4) is not 9 because (7,8,9) would be an AP, which is excluded. a(4) = 10 works, a(5) = 11 also.
For a(6), 12 is forbidden because of (8,10,12), 13 because of (7,10,13), 14 because of (8,11,14), 15 because of (7,11,15), 16 because of (0,8,16), thus a(6) = 17 is the smallest possible choice. - _M. F. Hasler_, Jan 18 2016
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 0..8191
• R. A. Moy and D. Rolnick, Novel structures in Stanley sequences, arXiv:1502.06013 [math.CO], 2015.
• R. A. Moy and D. Rolnick, Novel structures in Stanley sequences, Discrete Math., 339 (2016), 689-698.
• A. M. Odlyzko and R. P. Stanley, Some curious sequences constructed with the greedy algorithm, 1978.

Crossrefs

Cf. A005836, A188053, A266728.

Programs

Mathematica

s = {0, 1, 7};
next[] := For[k = Last[s]+1; AppendTo[s, 0];, True, k++, s[[-1]] = k; sp = SequencePosition[s, {_, a_, _, b_, _, c_, _} /; b-a == c-b, 1]; If[sp == {}, Print[k]; Break[]]];
Do[next[], {60}];
s (* Jean-François Alcover, Oct 04 2018 *)

PARI

A266727(n,show=1,L=3,v=[0,1,7],D=v->v[2..-1]-v[1..-2])={while(#v<=n, show&&print1(v[#v]","); v=concat(v,v[#v]); while(v[#v]++, forvec(i=vector(L,j,[if(j1||next(2),2);break)); if(type(show)=="t_VEC",v,v[n+1])} \\ 2nd (optional) arg: zero = silent, nonzero = verbose, vector (e.g. [] or [1]) = get the whole list [a(0..n)] as return value, else just a(n). - M. F. Hasler, Jan 18 2016

Python

A266727_list = [0,1,7]
for i in range(1000):
    n, flag = A266727_list[-1]+1, False
    while True:
        for j in range(i+2,0,-1):
            m = 2*A266727_list[j]-n
            if m in A266727_list:
                break
            if m < A266727_list[0]:
                flag = True
                break
        else:
            A266727_list.append(n)
            break
        if flag:
            A266727_list.append(n)
            break
        n += 1 # Chai Wah Wu, Jan 05 2016