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A164934 Number of different ways to select 3 disjoint subsets from {1..n} with equal element sum.

%I A164934 #22 Feb 05 2015 04:08:21
%S A164934 0,0,0,0,1,3,8,22,63,157,502,1562,4688,15533,50953,165054,562376,
%T A164934 1911007,6467143,22447463,78021923,271410289,957082911,3384587525,
%U A164934 11998851674,42876440587,153684701645,552421854011,1995875594696,7231871165277,26274832876337
%N A164934 Number of different ways to select 3 disjoint subsets from {1..n} with equal element sum.
%C A164934 a(5) = 1, because {1,4}, {2,3}, {5} are disjoint subsets of {1..5} with element sum 5.
%C A164934 a(6) = 3: {1,4}, {2,3}, {5} have element sum 5, {1,5}, {2,4}, {6} have element sum 6, and {1,6}, {2,5}, {3,4} have element sum 7.
%H A164934 Alois P. Heinz and Vaclav Kotesovec, <a href="/A164934/b164934.txt">Table of n, a(n) for n = 1..104</a> (first 65 terms from Alois P. Heinz)
%F A164934 Conjecture: a(n) ~ 4^n / (Pi * sqrt(3) * n^3). - _Vaclav Kotesovec_, Oct 16 2014
%p A164934 b:= proc(n, k, i) option remember; local m;
%p A164934       m:= i*(i+1)/2;
%p A164934       if k>n then b(k, n, i)
%p A164934     elif k>=0 and n+k>m or k<0 and n-2*k>m then 0
%p A164934     elif [n, k, i] = [0, 0, 0] then 1
%p A164934     else b(n, k, i-1)+b(n+i, k+i, i-1)+b(n-i, k, i-1)+b(n, k-i, i-1)
%p A164934       fi
%p A164934     end:
%p A164934 a:= proc(n) option remember;
%p A164934       `if`(n>2, b(n, n, n-1)/2+ a(n-1), 0)
%p A164934     end:
%p A164934 seq(a(n), n=1..20);
%t A164934 b[n_, k_, i_] := b[n, k, i] = Module[{m = i*(i+1)/2}, Which[k>n , b[k, n, i], k >= 0 && n+k>m || k<0 && n-2*k > m, 0, {n, k, i} == {0, 0, 0}, 1, True, b[n, k, i-1] + b[n+i, k+i, i-1] + b[n-i, k, i-1] + b[n, k-i, i-1]]]; a[n_] := a[n] = If[n>2, b[n, n, n-1]/2 + a[n-1], 0]; Table[a[n], {n, 1, 20}] (* _Jean-François Alcover_, Feb 05 2015, after _Alois P. Heinz_ *)
%Y A164934 Column k=3 of A196231.
%Y A164934 Cf. A161943, A164949, A232534.
%K A164934 nonn
%O A164934 1,6
%A A164934 _Alois P. Heinz_, Aug 31 2009