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A241688 Number of Sidon subsets of {1,...,n} of size 4.

%I A241688 #24 Mar 28 2025 13:28:59
%S A241688 0,0,0,0,0,0,2,10,26,60,110,190,304,466,676,958,1312,1762,2310,2984,
%T A241688 3786,4750,5874,7196,8720,10484,12488,14780,17360,20276,23530,27174,
%U A241688 31210,35696,40630,46074,52032,58566,65676,73434,81840,90966,100814,111460,122906
%N A241688 Number of Sidon subsets of {1,...,n} of size 4.
%C A241688 A Sidon set is a set of natural numbers A={a_1,a_2,...}, finite or infinite, such that all pairwise sums a_i+a_j (i <= j) are different.
%H A241688 <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (2, 0, -1, 0, -2, 2, 0, 1, 0, -2, 1).
%F A241688 It appears to be the case that G.f.: 2*x^7*(1+3*x+3*x^2+5*x^3)/((1-x)^5*(1+x)^2*(1+x^2)*(1+x+x^2)), corrected by _Vaclav Kotesovec_, May 03 2014
%F A241688 a(n) ~ 1/24*n^4 (deduced from g.f.). - _Ralf Stephan_, Apr 29 2014
%F A241688 a(n) = a(n-11)+a(n-8)-a(n-3)+2*(a(n-6)+a(n-1)-a(n-10)-a(n-5)). - _Fung Lam_, May 02 2014
%F A241688 Explicit formula (derived from g.f.): a(n) = n^4/24 - 7*n^3/12 + 29*n^2/12 - 15*n/8 - floor(n/4) - 4/3*floor(n/3) + (n/2-9/4)*floor(n/2) - floor((n+1)/4) - 2/3*floor((n+1)/3). - _Vaclav Kotesovec_, May 03 2014
%e A241688 a(7)=2 since the only subsets of {1,...,7} satisfying the required conditions are {1,2,5,7} and {1,3,6,7}.
%t A241688 SidonQ[l__] := If[Length[Join[Plus @@@ Subsets[l, {2}], 2 l]] == Length[Union[Join[Plus @@@ Subsets[l, {2}], 2 l]]], True, False]
%t A241688 Table[Length@Select[Subsets[Range[n], {4}], SidonQ[#] &], {n, 1, 30}]
%Y A241688 Column k=4 of A381476.
%Y A241688 Cf. A054578.
%K A241688 nonn
%O A241688 1,7
%A A241688 _Carl Najafi_, Apr 27 2014