A065825 Smallest k such that n numbers may be picked in {1,...,k} with no three terms in arithmetic progression.

1, 2, 4, 5, 9, 11, 13, 14, 20, 24, 26, 30, 32, 36, 40, 41, 51, 54, 58, 63, 71, 74, 82, 84, 92, 95, 100, 104, 111, 114, 121, 122, 137, 145, 150, 157, 163, 165, 169, 174, 194, 204, 209

Offset: 1

Author

Ed Pegg Jr, Nov 23 2001

Keywords

• nonn
• hard
• nice
• more

Comments

"Sequences containing no 3-term arithmetic progressions" is another phrase people may be searching for. See also A003002.

Don Reble notes large gaps between a(4k) and a(4k+1).

Ed Pegg Jr conjectures the 2^k term always equals (3^k+1)/2 and calls these "unprogressive" sets. Jaroslaw Wroblewski (jwr(AT)math.uni.wroc.pl), Nov 04 2003, remarks that this conjecture is known to be false.

Further comments from Jaroslaw Wroblewski (jwr(AT)math.uni.wroc.pl), Nov 05 2003: log a(n) / log n tends to 1 was established in 1946 by Behrend. This was extended by me in the Math. Comp. paper. Using appropriately chosen intervals from B(4,9,4) and B(6,9,11) I have determined that log (2a(n)-1) / log n < log 3 / log 2 holds for n=60974 and for n=2^19 since a(60974) <= 19197041, a(524288) <= 515749566. See my web page for further bounds.

Bloom & Sisask prove that a(n) >> n (log n)^(1+c) for an absolute (small) constant c > 0. This improves the o(1) in Behrend's result that log a(n)/log n = 1 + o(1) to log log n/log n. - Charles R Greathouse IV, Aug 04 2020

Examples

			a(9) = 20 = 1 2 6 7 9 14 15 18 20
a(10) = 24 = 1 2 5 7 11 16 18 19 23 24
a(11) = 26 = 1 2 5 7 11 16 18 19 23 24 26
a(12) = 30 = 1 3 4 8 9 11 20 22 23 27 28 30 (unique)
a(13) = 32 = 1 2 4 8 9 11 19 22 23 26 28 31 32
a(14) = 36 = 1 2 4 8 9 13 21 23 26 27 30 32 35 36
a(15) = 40 = 1 2 4 5 10 11 13 14 28 29 31 32 37 38 40
a(16) = 41 = 1 2 4 5 10 11 13 14 28 29 31 32 37 38 40 41
a(17) = 51 = 1 2 4 5 10 13 14 17 31 35 37 38 40 46 47 50 51
a(18) = 54 = 1 2 5 6 12 14 15 17 21 31 38 39 42 43 49 51 52 54
a(19) = 58 = 1 2 5 6 12 14 15 17 21 31 38 39 42 43 49 51 52 54 58
a(20) = 63 = 1 2 5 7 11 16 18 19 24 26 38 39 42 44 48 53 55 56 61 63
a(21) = 71 = 1 2 5 7 10 17 20 22 26 31 41 46 48 49 53 54 63 64 68 69 71
a(22) = 74 = 1 2 7 9 10 14 20 22 23 25 29 46 50 52 53 55 61 65 66 68 73 74
a(23) = 82 = 1 2 4 8 9 11 19 22 23 26 28 31 49 57 59 62 63 66 68 71 78 81 82
a(24) = 84 = 1 3 4 8 9 16 18 21 22 25 30 37 48 55 60 63 64 67 69 76 77 81 82 84
a(25) = 92 = 1 2 6 8 9 13 19 21 22 27 28 39 58 62 64 67 68 71 73 81 83 86 87 90 92
a(26) = 95 = 1 2 4 5 10 11 22 23 25 26 31 32 55 56 64 65 67 68 76 77 82 83 91 92 94 95
a(27) = 100 = 1 3 6 7 10 12 20 22 25 26 29 31 35 62 66 68 71 72 75 77 85 87 90 91 94 96 100
		

References

• Donald E. Knuth, Satisfiability, Fascicle 6, volume 4 of The Art of Computer Programming. Addison-Wesley, 2015, pages 135 and 190, Problem 31.
• C. R. J. Singleton, "No Progress": Solution to Problem 2472, Journal of Recreational Mathematics, 30(4) 305 1999-2000.

Links

• Tanbir Ahmed, Janusz Dybizbanski and Hunter Snevily, Unique Sequences Containing No k-Term Arithmetic Progressions, Electronic Journal of Combinatorics, 20(4), 2013, #P29.
• Cyriac Antony, Jacob Antony, D. Laavanya, and S. Devi Yamini, Graceful Coloring is Computationally Hard, Algorithms and Discrete Applied Mathematics, Eds. Daya Gaur, Rogers Mathew, Proc. 11th Int'l Conf. (CALDAM 2025) Lect. Notes Comp. Sci. (LNCS) Vol. 15536, Springer, 13-24. See also arXiv:2407.02179, [math.CO], 2024. See pp. 1-2.
• F. Behrend, On sets of integers which contain no three terms in an arithmetic progression, Proc. Nat. Acad. Sci. USA, v. 32, 1946, pp. 331-332.
• Thomas F. Bloom and Olof Sisask, Breaking the logarithmic barrier in Roth's theorem on arithmetic progressions, arXiv:2007.03528 [math.NT], 2020.
• Janusz Dybizbanski, Sequences containing no 3-term arithmetic progressions, The Electronic Journal of Combinatorics, 19, no. 2 (2012), #P15.
• Erica Klarreich, Landmark Math Proof Clears Hurdle in Top Erdős Conjecture, Quanta Magazine, Aug 03 2020.
• Tom Sanders, On Roth's theorem on progressions, Annals of Mathematics 174:1 (2011), pp. 619-636; arXiv version, arXiv:1011.0104 [math.CA], 2010-2011.
• Samuel S. Wagstaff, Jr., On k-free sequences of integers, Math. Comp., 26 (1972), 767-771.
• Wikipedia, Salem-Spencer set
• J. Wroblewski, A Nonaveraging Set of Integers with a Large Sum of Reciprocals, Math. Comput. 43, 261-262, 1984.
• J. Wroblewski, Nonaveraging Sets
• Index entries related to non-averaging sequences

Crossrefs

Cf. A003002 (three-free sequences), A003003, A003004, A003005, A003278, A005047, A225745.

Programs

Mathematica

ThreeAPFree[n_, k_, a_] := Module[{d, j},
   For[d = 1, d < k/2, d ++,
    For[j = 1, j <= n - 2, j++,
     If[MemberQ[a, a[[j]] + d] && MemberQ[a, a[[j]] + 2 d],
      Return[False]]]];
   Return[True]];
A065825[n_] := Module[{k, x, a},
  k = n;
  While[True,
   x = Subsets[Range[k], {n}];
   For[i = 1, i <= Length[x], i++,
    a = x[[i]];
    If[a[[1]] != 1 || a[[n]] != k, Continue[]];
    If[ThreeAPFree[n, k, a], Return[k]]];
   k++]]
Table[A065825[n], {n, 1, 10}]  (* Robert Price, Mar 11 2019 *)

PARI

\\ brute force
has3AP(v)=for(i=1,#v-2,for(j=i+2,#v,my(t=(v[i]+v[j])/2);if(denominator(t)==1 && setsearch(v,t),return([v[i],t,v[j]]))));0
a(n)=for(k=n,oo,forvec(u=vector(n,i,if(i==1,[1,1],i==k,[k,k],[2,k-1])),if(has3AP(u)==0, /* print(u); */ return(u[n])),2)) \\ Charles R Greathouse IV, Aug 04 2020

Formula

a(n) = A005047(n) + 1. - Rob Pratt, Jul 09 2015

Extensions

a(19) found by Guenter Stertenbrink in response to an A003278-based puzzle on www.mathpuzzle.com

More terms from Don Reble, Nov 25 2001

a(28)-a(32) from William Rex Marshall, Mar 24 2002

a(33) from William Rex Marshall, Nov 15 2003

a(34) from William Rex Marshall, Jan 24 2004

a(35)-a(36) (found by Gavin Theobold in 2004) communicated by William Rex Marshall, Mar 10 2007

a(37)-a(41) (from Wroblewski's web page) added by Joerg Arndt, Apr 25 2012

a(42)-a(43) from Fausto A. C. Cariboni, Sep 02 2018