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A111505 Right half of Pascal's triangle (A007318) with zeros.

%I A111505 #9 Sep 08 2013 13:30:49
%S A111505 1,0,1,0,2,1,0,0,3,1,0,0,6,4,1,0,0,0,10,5,1,0,0,0,20,15,6,1,0,0,0,0,
%T A111505 35,21,7,1,0,0,0,0,70,56,28,8,1,0,0,0,0,0,126,84,36,9,1,0,0,0,0,0,252,
%U A111505 210,120,45,10,1,0,0,0,0,0,0,462,330,165
%N A111505 Right half of Pascal's triangle (A007318) with zeros.
%C A111505 A034869 is the version without zeros.
%F A111505 Sum_{n, n>=k} T(n, k) = A001700(k).
%F A111505 Sum_{k =0..2*n} T(2*n, k) = A032443(n).
%F A111505 Sum_{k=0..2*n+1} T(2*n+1, k) = 4^n = A000302(n).
%F A111505 Sum_{k=0..2*n} T(2*n, k)^2 = A036910(n).
%F A111505 Sum_{k=0..2*n+1} T(2*n+1, k)^2 = C(4*n+1, 2*n) = A002458(n) . _Paul D. Hanna_
%e A111505 Triangle begins:
%e A111505 1;
%e A111505 0, 1;
%e A111505 0, 2, 1;
%e A111505 0, 0, 3, 1;
%e A111505 0, 0, 6, 4, 1;
%e A111505 0, 0, 0, 10, 5, 1;
%e A111505 0, 0, 0, 20, 15, 6, 1;
%e A111505 0, 0, 0, 0, 35, 21, 7, 1;
%e A111505 0, 0, 0, 0, 70, 56, 28, 8, 1;
%e A111505 0, 0, 0, 0, 0, 126, 84, 36, 9, 1;
%e A111505 0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1;
%e A111505 0, 0, 0, 0, 0, 0, 462, 330, 165, 55, 11, 1;
%e A111505 0, 0, 0, 0, 0, 0, 924, 792, 495, 220, 66, 12, 1;
%e A111505 0, 0, 0, 0, 0, 0, 0, 1716, 1287, 715, 286, 78, 13, 1;
%e A111505 0, 0, 0, 0, 0, 0, 0, 3432, 3003, 2002, 1001, 364, 91, 14, 1;
%Y A111505 Cf. A007318, A034869, A094527, A111418.
%K A111505 nonn,tabl
%O A111505 0,5
%A A111505 _Philippe Deléham_, Nov 16 2005