

A039799


For S a subset of [ n ] = {1,2,3,...n}, let B_S = {x+y : x,y in S, x<y}; then a(n) is maximal cardinality of B_S intersect B_{[ n ]S}.


1



0, 0, 0, 1, 1, 2, 3, 6, 6, 9, 10, 12, 14, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119
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OFFSET

1,6


COMMENTS

An easy upper bound is 2n7, since the sums in the intersection must come from {5,...,2n3}. 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 {2n7,...,2n3}. If this is true, then 2n11 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) = 2n11 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..j1), 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, i1, s), b(n, i1, 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, j1}], {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, i1, s], b[n, i1, s ~Union~ {i}]]]; a[n_] := b[n, n, {}]; Table[a[n], {n, 1, 15}] (* JeanFrançois Alcover, May 29 2015, after Alois P. Heinz *)


CROSSREFS

Sequence in context: A198516 A187326 A056907 * A185423 A144583 A155215
Adjacent sequences: A039796 A039797 A039798 * A039800 A039801 A039802


KEYWORD

hard,nonn,more


AUTHOR

Erich Friedman


EXTENSIONS

a(15)a(23) from Alois P. Heinz, Feb 28 2011
a(24)a(100) from Rob Pratt, Jul 14 2015


STATUS

approved



