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A185256 Stanley Sequence S(0,3). 26
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Given a finite increasing sequence V = [v_1, ..., v_k] containing no 3-term 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 3-term arithmetic progression.
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, E10.
LINKS
P. Erdos et al., Greedy algorithm, arithmetic progressions, subset sums and divisibility, Discrete Math., 200 (1999), 119-135.
J. L. Gerver and L. T. Ramsey, Sets of integers with no long arithmetic progressions generated by the greedy algorithm, Math. Comp., 33 (1979), 1353-1359.
R. A. Moy, On the growth of the counting function of Stanley sequences, arXiv:1101.0022 [math.NT], 2010-2012.
R. A. Moy, On the growth of the counting function of Stanley sequences, Discrete Math., 311 (2011), 560-562.
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 progression-free sets, Discrete Math., 306 (2006), 2131-2133.
EXAMPLE
After [0, 3, 4, 7, 9] the next term cannot be 10 or we would have the 3-term 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 n-th term:
for n from t1+1 to t1+M do # do 1
# Try i as next term:
for i from s2[n-1]+1 to mmm do # do 2
swi:=-1;
# Test against j-th term:
for j from 1 to n-2 do # do 3
p:=s2[n-j];
if 2*p-i < 0 then break; fi;
if chvec[2*p-i] = 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 n-th term *)
For[n = t1+1, n <= t1 + M, n++,
(* try i as next term *)
For[i = s2[[n-1]] + 1, i <= mmm, i++, swi = -1;
(* test against j-th term *)
For[j = 1, j <= n-2, 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] (* Jean-François 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)))>1||next(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
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Mar 19 2011
STATUS
approved

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Last modified March 18 22:24 EDT 2024. Contains 370951 sequences. (Running on oeis4.)