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%I #43 Sep 06 2023 21:20:10
%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,18,57,155,351,714,1386,2495,4291,
%T 7043,11234,17270,26087,38105,54954,77453,107816,146881,198880,264222,
%U 348722,453447,586339,747970,950770,1193145,1492469,1848124,2280697,2787353,3400200,4110662,4959054,5937538
%N Number of ways to choose 4 disjoint subsets of {1..n} with the same sum and with respectively 1, 2, 3 and 4 distinct elements of {1..n}.
%e For n = 15, there are 6 ways to choose the four disjoint subsets:
%e 15 = 7 + 8 = 4 + 5 + 6 = 1 + 2 + 3 + 9,
%e 15 = 6 + 9 = 3 + 5 + 7 = 1 + 2 + 4 + 8,
%e 15 = 6 + 9 = 3 + 4 + 8 = 1 + 2 + 5 + 7,
%e 15 = 6 + 9 = 2 + 5 + 8 = 1 + 3 + 4 + 7,
%e 15 = 7 + 8 = 2 + 4 + 9 = 1 + 3 + 5 + 6,
%e 15 = 7 + 8 = 1 + 5 + 9 = 2 + 3 + 4 + 6, so a(15) = 6.
%Y Cf. A362717.
%K nonn,nice
%O 0,16
%A _Jean-Marc Rebert_, Jun 07 2023
%E More terms from _David A. Corneth_, Jun 08 2023