cp's OEIS Frontend

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.

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A266728 Stanley sequence S_5(0,3).

Original entry on oeis.org

0, 3, 4, 5, 6, 8, 9, 10, 11, 13, 14, 15, 16, 18, 19, 21, 25, 28, 29, 30, 31, 33, 34, 35, 36, 38, 39, 40, 41, 43, 44, 46, 50, 53, 54, 55, 56, 58, 59, 60, 61, 63, 64, 65, 66, 68, 69, 71, 75, 78, 79, 80, 81, 83, 84, 85, 86, 88, 89, 90, 91, 93, 94, 96, 125, 128, 129, 130
Offset: 0

Views

Author

N. J. A. Sloane, Jan 04 2016

Keywords

Comments

Lexicographic first increasing sequence with a(0) = 0, a(1) = 3 and for all n > 1, {a(0), ..., a(n)} does not contain 5 terms in arithmetic progression.

Links

Crossrefs

Cf. A005836, A188053, A266727.
Cf. A185256 = S_3(0,3) = S(0,3), A267650 = S_4(0,3).

Programs

  • PARI
    A266728(n,show=1,L=5,v=[0,3],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));v[n+1]} \\ M. F. Hasler, Jan 18 2016

Extensions

More terms from M. F. Hasler, Jan 18 2016

A267650 Stanley sequence S_4(0,3): lexicographic first increasing sequence with a(0) = 0, a(1) = 3 and no 4 terms in arithmetic progression.

Original entry on oeis.org

0, 3, 4, 5, 7, 8, 10, 11, 16, 17, 18, 20, 21, 27, 28, 29, 31, 32, 34, 35, 36, 53, 55, 56, 57, 60, 61, 62, 64, 67, 69, 75, 87, 91, 100, 101, 103, 104, 105, 108, 109, 110, 114, 116, 120, 125, 127, 128, 129, 132, 134, 135, 164, 168, 173, 174, 175, 177, 181, 182, 184, 188, 190
Offset: 0

Views

Author

M. F. Hasler, Jan 18 2016

Keywords

Comments

See A185256 for S(0,3) = S_3(0,3) and A266728 for S_5(0,3).

Links

Crossrefs

For other examples of Stanley Sequences see A005487, A005836, A187843, A188052, A188053, A188054, A188055, A188056, A188057, A266727, A266728.
See also A004793, A033160, A033163.

Programs

  • PARI
    a(n,show=1,L=4,v=[0,3], 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(1..n)] as return value, else just a(n). - M. F. Hasler, Jan 18 2016
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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