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%I #13 Jun 28 2021 18:42:19
%S 1,1,2,2,4,4,3,12,12,8,6,24,48,32,16,10,60,120,160,80,32,20,120,360,
%T 480,480,192,64,35,280,840,1680,1680,1344,448,128,70,560,2240,4480,
%U 6720,5376,3584,1024,256,126,1260,5040,13440,20160,24192,16128,9216,2304,512
%N Triangle read by rows: T(n,k) = 2^k*binomial(n,k)*binomial(n-k, floor((n-k)/2)), 0 <= k <= n.
%C Sum of terms in row n = binomial(2n+1,n) (A001700; see the Andreescu-Feng reference).
%D T. Andreescu and Z. Feng, 102 Combinatorial Problems (from the training of the USA IMO team), Birkhauser, Boston, 2003, Advanced problem # 15, pp. 11,61-63.
%H Harvey P. Dale, <a href="/A143358/b143358.txt">Table of n, a(n) for n = 0..1000</a>
%F E.g.f.: exp(2*x*y)*(BesselI(0,2*x) + BesselI(1,2*x)). - _Vladeta Jovovic_, Dec 02 2008
%e Triangle begins:
%e 1;
%e 1, 2;
%e 2, 4, 4;
%e 3, 12, 12, 8;
%e 6, 24, 48, 32, 16;
%e 10, 60, 120, 160, 80, 32;
%e ...
%p T:=proc(n,k) options operator, arrow: 2^k*binomial(n,k)*binomial(n-k,floor((1/2)*n-(1/2)*k)) end proc: for n from 0 to 9 do seq(T(n,k),k=0..n) end do; # yields sequence in triangular form
%t Flatten[Table[2^k Binomial[n,k]Binomial[n-k,Floor[(n-k)/2]],{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Jun 28 2021 *)
%Y Cf. A001700.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Aug 11 2008
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