

A185256


Stanley Sequence S(0,3).


25



0, 3, 4, 7, 9, 12, 13, 16, 27, 30, 31, 34, 36, 39, 40, 43, 81, 84, 85, 88, 90, 93, 94, 97, 108, 111, 112, 115, 117, 120, 121, 124, 243, 246, 247, 250, 252, 255, 256, 259, 270, 273, 274, 277, 279, 282, 283, 286, 324, 327, 328, 331, 333, 336, 337, 340, 351, 354, 355, 358, 360, 363
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OFFSET

1,2


COMMENTS

Given a finite increasing sequence V = [v_1, ..., v_k] containing no 3term arithmetic progression, the Stanley Sequence S(V) is obtained by repeatedly appending the smallest term that is greater than the previous term and such that the new sequence also contains no 3term arithmetic progression.


REFERENCES

R. K. Guy, Unsolved Problems in Number Theory, E10.


LINKS

Alois P. Heinz, Table of n, a(n) for n = 1..2048
P. Erdos et al., Greedy algorithm, arithmetic progressions, subset sums and divisibility, Discrete Math., 200 (1999), 119135.
J. L. Gerver and L. T. Ramsey, Sets of integers with no long arithmetic progressions generated by the greedy algorithm, Math. Comp., 33 (1979), 13531359.
R. A. Moy, On the growth of the counting function of Stanley sequences, arXiv:1101.0022 [math.NT], 20102012.
R. A. Moy, On the growth of the counting function of Stanley sequences, Discrete Math., 311 (2011), 560562.
A. M. Odlyzko and R. P. Stanley, Some curious sequences constructed with the greedy algorithm, 1978.
S. Savchev and F. Chen, A note on maximal progressionfree sets, Discrete Math., 306 (2006), 21312133.
Index entries for nonaveraging sequences


EXAMPLE

After [0, 3, 4, 7, 9] the next term cannot be 10 or we would have the 3term A.P. 4,7,10; it cannot be 11 because of 7,9,11; but 12 is OK.


MAPLE

# Stanley Sequences, Discrete Math. vol. 311 (2011), see p. 560
ss:=proc(s1, M) local n, chvec, swi, p, s2, i, j, t1, mmm; t1:=nops(s1); mmm:=1000;
s2:=Array(1..t1+M, s1); chvec:=Array(0..mmm);
for i from 1 to t1 do chvec[s2[i]]:=1; od;
# Get nth term:
for n from t1+1 to t1+M do # do 1
# Try i as next term:
for i from s2[n1]+1 to mmm do # do 2
swi:=1;
# Test against jth term:
for j from 1 to n2 do # do 3
p:=s2[nj];
if 2*pi < 0 then break; fi;
if chvec[2*pi] = 1 then swi:=1; break; fi;
od; # od 3
if swi=1 then s2[n]:=i; chvec[i]:=1; break; fi;
od; # od 2
if swi=1 then ERROR("Error, no solution at n = ", n); fi;
od; # od 1;
[seq(s2[i], i=1..t1+M)];
end;
ss([0, 3], 80);


MATHEMATICA

ss[s1_, M_] := Module[{n, chvec, swi, p, s2, i, j, t1, mmm}, t1 = Length[s1]; mmm = 1000; s2 = Table[s1, {t1 + M}] // Flatten; chvec = Array[0&, mmm];
For[i = 1, i <= t1, i++, chvec[[s2[[i]] ]] = 1];
(* get nth term *)
For[n = t1+1, n <= t1 + M, n++,
(* try i as next term *)
For[i = s2[[n1]] + 1, i <= mmm, i++, swi = 1;
(* test against jth term *)
For[j = 1, j <= n2, j++, p = s2[[n  j]]; If[2*p  i < 0, Break[] ];
If[chvec[[2*p  i]] == 1, swi = 1; Break[] ] ];
If[swi == 1, s2[[n]] = i; chvec[[i]] = 1; Break[] ] ];
If[swi == 1, Print["Error, no solution at n = ", n] ] ];
Table[s2[[i]], {i, 1, t1 + M}] ];
ss[{0, 3}, 80] (* JeanFrançois Alcover, Sep 10 2013, translated from Maple *)


PROG

(PARI) A185256(n, show=1, L=3, 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(j<L, j, #v), #v]), #Set(D(vecextract(v, i)))>1next(2), 2); break)); if(type(show)=="t_VEC", v, v[n])} \\ 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


CROSSREFS

For other examples of Stanley Sequences see A005487, A005836, A187843, A188052, A188053, A188054, A188055, A188056, A188057.
See also A004793, A033160, A033163.
Sequence in context: A329963 A034022 A198772 * A070992 A246514 A060142
Adjacent sequences: A185253 A185254 A185255 * A185257 A185258 A185259


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Mar 19 2011


STATUS

approved



