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%I #31 Aug 02 2018 09:38:05
%S 1,1,2,2,3,2,4,3,5,4,6,4,7,6,8,6,9,6,10,7,11,8,12,10,13,9,14,10,15,10,
%T 16,12,17,14,18,12,19,13,20,14,21,14,22,18,23,16,24,16,25,18,26,18,27,
%U 22,28,19,29,20,30,20,31,21,32,26,33,22,34,24,35,24
%N Maximal size of sum-free set in additive group of integers mod n.
%D Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, Manuscript, May 2017. See Table in Section 1.6.1.
%D A. P. Street, Counting non-isomorphic sum-free sets, in Proc. First Australian Conf. Combinatorial Math., Univ. Newcastle, 1972, pp. 141-143.
%H Reinhard Zumkeller, <a href="/A211316/b211316.txt">Table of n, a(n) for n = 2..10000</a>
%H Bela Bajnok, <a href="https://arxiv.org/abs/1705.07444">Additive Combinatorics: A Menu of Research Problems</a>, arXiv:1705.07444 [math.NT], May 2017. See Table in Section 1.6.1.
%H Ben Green and Imre Z. Ruzsa, <a href="http://arxiv.org/abs/math/0307142">Sum-free sets in abelian groups</a>, arXiv:math/0307142 [math.CO], 2004.
%F If n is divisible by a prime == 2 mod 3 then a(n) = n(p+1)/(3p) where p is the smallest such prime divisor; otherwise if n is divisible by 3 then a(n) = n/3; otherwise all prime divisors of n are == 1 mod 3 and a(n) = (n-1)/3.
%F In particular, a(2n) = n (cf. A211317).
%t a[n_] := Module[{f = FactorInteger[n][[All, 1]]}, For[i = 1, i <= Length[f], i++, If[Mod[f[[i]], 3]==2, Return[n*(f[[i]] + 1)/3/f[[i]]]]]; If[Mod[n, 3] == 1, n-1, n]/3]
%t Table[a[n], {n, 2, 100}] (* _Jean-François Alcover_, Aug 02 2018, from PARI *)
%o (Haskell)
%o a211316 n | not $ null ps = n * (head ps + 1) `div` (3 * head ps)
%o | m == 0 = n'
%o | otherwise = (n - 1) `div` 3
%o where ps = [p | p <- a027748_row n, mod p 3 == 2]
%o (n',m) = divMod n 3
%o -- _Reinhard Zumkeller_, Apr 25 2012
%o (PARI) a(n)=my(f=factor(n)[,1]); for(i=1,#f, if(f[i]%3==2, return(n*(f[i]+1)/3/f[i]))); if(n%3, n-1, n)/3 \\ _Charles R Greathouse IV_, Sep 02 2015
%Y Bisection: A211317. Cf. A007865, A027748, A003627.
%K nonn
%O 2,3
%A _N. J. A. Sloane_, Apr 24 2012
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