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a(n) = 2*n - (-1)^n*(1+(n mod 2)).
1

%I #51 Oct 23 2022 22:56:29

%S 4,3,8,7,12,11,16,15,20,19,24,23,28,27,32,31,36,35,40,39,44,43,48,47,

%T 52,51,56,55,60,59,64,63,68,67,72,71,76,75,80,79,84,83,88,87,92,91,96,

%U 95,100,99,104,103,108,107,112,111,116,115,120,119,124,123,128,127,132,131,136,135,140,139,144,143,148,147,152,151,156,155,160,159

%N a(n) = 2*n - (-1)^n*(1+(n mod 2)).

%C Conjecture 1: Let G be an additive group with |G| > 1, and let p(G) be the minimum of those orders of nonzero elements of G (which is the positive infinity if each nonzero element of G has infinite order). Let n > 1 be an integer, and let A be a finite nonempty subset of G.

%C (i) For the set n'A = {a_1 + ... + a_n: a_1, ..., a_n are elements of A with a_i not equal to a_{i+1} for all 0 < i < n, and a_n not equal to a_1}, we have |n'A| >= min{p(G), n*|A|-a(n)}.

%C (ii) For the set n~A = {a_1 + ... + a_n: a_1, ..., a_n are elements of A with a_i not equal to a_{i+1} for all 0 < i < n}, we have |n~A| >= min{p(G), n*|A|-2n+1+(n mod 2)}.

%C When n > 1 and A = {0,...,k-1}, it is easy to see that |n'A| = n*|A|-a(n) and |n~A| = n*|A|-2n+1+(n mod 2).

%C Does the following extension of Conjecture 1 also hold?

%C Conjecture 2: Let G be an additive group with |G| > 1, and let p(G) be as in Conjecture 1. Let A_1, ..., A_n (n > 1) be finite subsets of G with |A_i| > 1 for all i = 1..n.

%C (i) For the set C = {a_1 + ... + a_n: a_1, ..., a_n are elements of A_1, ..., A_n respectively, with a_i not equal to a_{i+1} for all 0 < i < n, and a_n not equal to a_1}, we have |C| >= min{p(G), |A_1|+...+|A_n|-a(n)}.

%C (ii) For the set D = {a_1 + ... + a_n: a_1, ..., a_n are elements of A_1, ..., A_n respectively, with a_i not equal to a_{i+1} for all 0 < i < n}, we have |D| >= min{p(G), |A_1|+...+|A_n|-2n+1+(n mod 2)}.

%C Conjectures 1 and 2 were partially confirmed by Han Wang and Zhi-Wei Sun in their preprint arXiv:2210.12044. - _Zhi-Wei Sun_, Oct 23 2022

%H Zhi-Wei Sun, <a href="/A357130/b357130.txt">Table of n, a(n) for n = 1..10000</a>

%H Paul Balister and Jeffery Paul Wheeler, <a href="https://www.impan.pl/download/pdf/aa140-2-01">The Erdos-Heilbronn problem for finite groups</a>, Acta Arith. 140 (2009), 105-118.

%H Zhi-Wei Sun, <a href="https://www.impan.pl/download/pdf/aa99-1-4">Restricted sums of subsets of Z</a>, Acta Arith. 99 (2001), 41-60.

%H Zhi-Wei Sun and Li-Lu Zhao, <a href="https://doi.org/10.1016/j.jcta.2011.09.003">Linear extension of the Erdos-Heilbronn conjecture</a>, J. Combin. Theory Ser. A 119 (2012), 364-381.

%H Van Vu and Terence Tao, <a href="https://doi.org/10.1017/CBO9780511755149">Additive Combinatorics</a>, Cambridge Univ. Press, Cambridge, 2006.

%H Han Wang and Zhi-Wei Sun, <a href="http://arxiv.org/abs/2210.12044">On two new kinds of restricted sumsets</a>, arXiv:2210.12044 [math.NT], 2022.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(n) = 2*n+3*(n mod 2)-1. - _Chai Wah Wu_, Sep 14 2022

%F From _Stefano Spezia_, Sep 16 2022: (Start)

%F If we adopt a(0) = 0, then G.f.: x*(4 - x + x^2)/((1 - x)^2*(1 + x)).

%F a(n) = a(n-1) + a(n-2) - a(n-3) for n > 3. (End)

%F Sum_{n>=1} (-1)^n/a(n) = 3*log(2)/4 - Pi/8. - _Amiram Eldar_, Sep 27 2022

%e a(1) = 2*1 - (-1)*(1+ (1 mod 2)) = 2 + 2 = 4.

%e a(2) = 2*2 - (-1)^2*(1 + (2 mod 2)) = 4 - 1 = 3.

%p Suppose a(0) = -1. Then gf := (5*x - 1)/((x^2 - 1)*(x - 1)): ser := series(gf, x, 82):

%p seq(coeff(ser, x, n), n = 1..80); # _Peter Luschny_, Sep 17 2022

%t a[n_]:=a[n]=2n-(-1)^n*(1+Mod[n,2]);

%t Table[a[n],{n,1,80}]

%o (Python)

%o def A357130(n): return (n<<1)+3*(n&1)-1 # _Chai Wah Wu_, Sep 14 2022

%Y Cf. A005408, A005843, A168205.

%K nonn,easy

%O 1,1

%A _Zhi-Wei Sun_, Sep 13 2022