OFFSET
1,6
COMMENTS
An easy upper bound is 2n-7, since the sums in the intersection must come from {5,...,2n-3}. It is easy to see that 5 and 6 cannot both appear as sums. It appears that at most three sums can come from {5,...,9} and at most three from {2n-7,...,2n-3}. If this is true, then 2n-11 is an upper bound for n >= 9. - Rob Pratt, Jul 14 2015
LINKS
Rob Pratt, Table of n, a(n) for n = 1..100
FORMULA
Conjecture: a(n) = 2n-11 for n >= 14. - Rob Pratt, Jul 14 2015
EXAMPLE
a(7) = 3 since we can divide [ 7 ] into S={1,5,6} and T={2,3,4,7} giving B_S={6,7,11} and B_T={5,6,7,9,10,11}, with intersection {6,7,11} of cardinality 3.
MAPLE
B:= proc(s) local l, i, j;
l:= [s[]];
{seq(seq(l[i]+l[j], i=1..j-1), j=2..nops(l))}
end:
b:= proc(n, i, s)
if i=0 then nops(B(s) intersect B({$1..n} minus s))
else max(b(n, i-1, s), b(n, i-1, s union {i}))
fi
end;
a:= n-> b(n, n, {}):
seq(a(n), n=1..15); # Alois P. Heinz, Feb 28 2011
MATHEMATICA
B[s_List] := B[s] = Module[{l=s}, Flatten @ Table[Table[l[[i]] + l[[j]], {i, 1, j-1}], {j, 2, Length[l]}]]; b[n_, i_, s_List] := b[n, i, s] = If[i == 0, Length[B[s] ~Intersection~ B[Range[n] ~Complement~ s]], Max[b[n, i-1, s], b[n, i-1, s ~Union~ {i}]]]; a[n_] := b[n, n, {}]; Table[a[n], {n, 1, 15}] (* Jean-François Alcover, May 29 2015, after Alois P. Heinz *)
CROSSREFS
KEYWORD
hard,nonn
AUTHOR
EXTENSIONS
a(15)-a(23) from Alois P. Heinz, Feb 28 2011
a(24)-a(100) from Rob Pratt, Jul 14 2015
STATUS
approved