OFFSET
1,6
COMMENTS
For any n > 2 and k > 2*n, we can show that the set {a_1+...+a_n: -floor((k-1)/2) <= a_1,...,a_n <= floor(k/2), and a_1^2,...,a_n^2 are pairwise distinct} contains exactly k*n-n^2+1 elements. Thus a(n) = 4*n - 15 for n > 8.
Conjecture 1: Let A_1,...,A_n (n > 3) be finite subsets of a field F with |A_i| > 2*n for all i = 1..n. Then the set {a_1+...+a_n: a_1,...,a_n belong to A_1,...,A_n respectively, and a_1^2,...,a_n^2 are pairwise distinct} contains at least min{p(F), |A_1|+...+|A_n|-n^2+1} elements, where p(F) is the positive infinity if F is of characteristic zero, and p(F) is p if the characteristic of F is a prime p.
By a 2002 paper of H. Pan and Z.-W. Sun, If A and B are finite subsets of a field F with 2 < |A| <= |B|, then the {a+b: a belongs to A, b belongs to B, and a^2 is not equal to b^2} contains at least min{p(F)-1, |A|+|B|-4}.
We also have the following conjecture extending Conjecture 1 to any group.
Conjecture 2: Let G be any group with |G| > 1, and let p(G) be the minimum of orders of all nonzero elements of G (which is the positive infinity if each nonzero element of G has infinite order). If A_1,...,A_n (n > 3) are finite subsets of G with |A_i| > 2*n for all i = 1..n. Then the set {a_1+...+a_n: a_1,...,a_n belong to A_1,...,A_n respectively, and there are no 1<= i < j <= n with a_i = a_j or a_i = -a_j} contains at least min{p(F), |A_1|+...+|A_n|-n^2+1} elements.
LINKS
Paul Balister and Jeffery Paul Wheeler, The Erdos-Heilbronn problem for finite groups, Acta Arith. 140 (2009), 105-118.
Hao Pan and Zhi-Wei Sun, A lower bound for |{a+b: a belongs to A, b belongs to B, and P(a,b) is nonzero}|, J. Combin. Theory Ser. A 100 (2002), 387-393.
Zhi-Wei Sun and Li-Lu Zhao, Linear extension of the Erdos-Heilbronn conjecture, J. Combin. Theory Ser. A 119 (2012), 364-381.
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
From Chai Wah Wu, Sep 27 2022: (Start)
a(n) = 2*a(n-1) - a(n-2) for n > 10.
G.f.: x^6*(-2*x^4 - 2*x^3 + 5*x^2 - x + 4)/(x - 1)^2. (End)
EXAMPLE
a(6) = 4 since {-floor(5/2),...,floor(6/2)} = {-2,-1,0,1,2,3} and the set {0+1+2+3, 0-1+2+3, 0+1-2+3, 0-1-2+3} = {6, 4, 2, 0} has cardinality 4.
MATHEMATICA
L={}; Do[V=Table[x, {x, -Floor[(n-1)/2], Floor[n/2]}]; tab={}; Do[a1=V[[g]]; a2=V[[h]]; a3=V[[i]]; a4=V[[j]];
If[Length[Union[{a1^2, a2^2, a3^2, a4^2}]]==4, tab=Append[tab, a1+a2+a3+a4]],
{g, 1, n}, {h, 1, g}, {i, 1, h}, {j, 1, i}]; L=Append[L, Length[Union[tab]]], {n, 1, 50}]; Print[L]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Sep 26 2022
STATUS
approved