%I #24 May 08 2024 02:15:57
%S 9,17,28,42,59,79,102,128,157,189,224,262,303,347,394,444,497,553,612,
%T 674,739,807,878,952,1029,1109,1192,1278,1367,1459,1554,1652,1753,
%U 1857,1964,2074,2187,2303,2422,2544,2669,2797,2928,3062,3199,3339,3482,3628,3777,3929,4084
%N Subset sums (see Comments).
%C For n > 2, take the set [3*(n-1)] and form three subsets all of which: a) have cardinality of n, b) have the same sum of elements, and c) share one element with the other subset and another element with the third subset. a(n) is the sum of the elements of each subset.
%C a(n) is the minimum value of the magic constant in a normal magic triangle of order n (see formula 5 in Trotter). - _Stefano Spezia_, Feb 18 2021
%D a(4) is mentioned in: Gary Gruber, "The World's 200 Hardest Brain Teasers", Sourcebooks, 2010, p. 55.
%H Colin Barker, <a href="/A285009/b285009.txt">Table of n, a(n) for n = 3..1000</a>
%H Terrel Trotter, <a href="https://www.trottermath.net/simpleops/magictri.html">Normal Magic Triangles of Order n</a>, Journal of Recreational Mathematics Vol. 5, No. 1, 1972, pp. 28-32.
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), for n > 5.
%F a(n) = (8 + (n-2)*(3*n+1))/2, for n > 2.
%F G.f.: x^3*(9 - 10*x + 4*x^2) / (1 - x)^3. - _Colin Barker_, Apr 08 2017
%F E.g.f.: (1/2)*exp(x)*(3*x^2 - 2*x + 6) - 2*x*(x + 1) - 3. - _Indranil Ghosh_, Apr 08 2017; corrected by _Ilya Gutkovskiy_, Apr 10 2017
%F a(n) = A005449(n-1) + 2. - _Hugo Pfoertner_, Feb 18 2021
%e For n = 3, the set is S = {1,2,3,4,5,6} and the subsets are S1 = {1,2,6}, S2 = {1,3,5} and S3 = {2,3,4}. Therefore, a(3) = 9.
%t Table[(8+(n-2)*(3 *n+1))/2,{n,3,53}]
%t Drop[CoefficientList[Series[x^3*(9 - 10*x + 4*x^2) / (1 - x)^3 , {x, 0, 60}], x], 3] (* _Indranil Ghosh_, Apr 08 2017 *)
%o (PARI) Vec(x^3*(9 - 10*x + 4*x^2) / (1 - x)^3 + O(x^60)) \\ _Colin Barker_, Apr 08 2017
%Y Cf. A005449.
%K easy,nonn
%O 3,1
%A _Ivan N. Ianakiev_, Apr 07 2017