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%I #25 Aug 03 2014 14:01:40
%S 3,3,5,9,14,25,45,77,141,261,505,935,1849,3445,6865,12885,25741,48637,
%T 97241,184775,369513,705453,1410865,2704179,5408313,10400625,20801201,
%U 40116627,80233201,155117549,310235041,601080421,1202160781,2333606253,4667212441
%N Largest number of the form C(n,x) + C(n,y) + C(n,z) where x + y + z = n.
%C lim{n->infinity} a(n+1)/a(n)=2. Subset of A034703. From an idea of _Michael B. Porter_.
%C For n > 6, it appears that the solution is always x = n mod 2, y = z = floor(n/2). - _T. D. Noe_, Mar 05 2012
%H Paolo P. Lava, <a href="/A209083/b209083.txt">Table of n, a(n) for n = 0..360</a>
%e For n=5 [x,y,z] can be [0,0,5], [0,1,4], [0,2,3], [1,1,3] and [1,2,2].
%e C(5,0) + C(5,0) + C(5,5) = 1+1+1 = 3.
%e C(5,0) + C(5,1) + C(5,4) = 1+5+5 = 11.
%e C(5,0) + C(5,2) + C(5,3) = 1+10+10 =21.
%e C(5,1) + C(5,1) + C(5,3) = 5+5+10 = 20.
%e C(5,1) + C(5,2) + C(5,2) = 5+10+10 = 25.
%e Therefore 25 is in the sequence.
%p with(numtheory);
%p P:=proc(i)
%p local c,m,n,s,v;
%p v:=array[1..3];
%p for n from 3 to i do
%p s:=0; v[1]:=0; v[2]:=0; v[3]:=n;
%p while v[1]<=floor(n/3) do
%p while v[2]<=floor((n-v[1])/2) do
%p c:=0;
%p for m from 1 to 3 do c:=c+binomial(n,v[m]); od;
%p if c>s then s:=c; fi;
%p v[2]:=v[2]+1; v[3]:=v[3]-1;
%p od;
%p v[1]:=v[1]+1; v[2]:=v[1]; v[3]:=n-v[1]-v[2];
%p od;
%p print(s);
%p od;
%p end:
%p P(1000);
%t Table[Maximize[{Binomial[n, a] + Binomial[n, b] + Binomial[n, c], a + b + c == n, a >= 0, b >= 0, c >= 0, a <= n, b <= n, c <= n}, {a, b, c}, Integers][[1]], {n, 0, 30}] (* _T. D. Noe_, Mar 05 2012 *)
%o (PARI) A209083(n)={local(a,b,c,s);s=-1;for(a=0,n,for(b=0,n-a,c=n-a-b;s=max(s,binomial(n,a)+binomial(n,b)+binomial(n,c))));s}
%Y Cf. A034703.
%K nonn
%O 0,1
%A _Paolo P. Lava_, Mar 05 2012
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