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Number of distinct values v satisfying v = sum of elements in S = product of elements in P for any partition of {1,...,n} into two sets S and P.
2

%I #12 Dec 07 2020 08:12:32

%S 1,0,1,0,1,1,1,1,1,2,1,2,1,3,3,2,3,1,2,3,3,2,2,4,3,5,3,2,3,3,4,4,5,1,

%T 3,2,4,4,6,3,3,2,3,4,9,3,4,9,4,3,5,4,4,4,6,6,5,5,4,7,4,8,6,4,7,3,6,5,

%U 3,4,6,5,4,6,6,5,7,4,6,9,7,6,6,8,4,7,5

%N Number of distinct values v satisfying v = sum of elements in S = product of elements in P for any partition of {1,...,n} into two sets S and P.

%H Alois P. Heinz, <a href="/A213237/b213237.txt">Table of n, a(n) for n = 1..1000</a>

%e a(1) = 1: S={1}, P={}, v=1.

%e a(2) = 0: no partition of {1,2} satisfies the condition.

%e a(3) = 1: S={1,2}, P={3}, v=3.

%e a(10) = 2: three partitions of {1,2,...,10} into S and P satisfy v = Sum_{i:S} i = Product_{k:P} k but there are only two distinct values v: S={2,3,5,6,7,8,9}, P={1,4,10}, v=40; S={4,5,6,8,9,10}, P={1,2,3,7}, v=42; S={1,2,3,4,5,8,9,10}, P={6,7}, v=42.

%p b:= proc(n, s, p)

%p `if`(s=p, {s}, `if`(n<1, {}, {b(n-1, s, p)[],

%p `if`(s-n<p*n, {}, b(n-1, s-n, p*n))[]}))

%p end:

%p a:= n-> nops(b(n, n*(n+1)/2, 1)):

%p seq(a(n), n=1..100);

%t b[n_, s_, p_] := b[n, s, p] = If[s == p, {s}, If[n < 1, {}, Union[b[n - 1, s, p], If[s - n < p n, {}, b[n - 1, s - n, p n]]]]];

%t a[n_] := Length[b[n, n(n+1)/2, 1]];

%t Array[a, 100] (* _Jean-François Alcover_, Dec 07 2020, after _Alois P. Heinz_ *)

%Y The values v are in A213238.

%Y Cf. A178830, A207852.

%K nonn

%O 1,10

%A _Alois P. Heinz_, Jun 07 2012