A176388 - OEIS

%I #2 Mar 30 2012 17:34:40

%S 1,1,1,1,2,1,1,4,4,1,1,3,5,3,1,1,11,21,21,11,1,1,6,13,16,13,6,1,1,34,

%T 76,106,106,76,34,1,1,15,33,50,56,50,33,15,1,1,112,258,402,493,493,

%U 402,258,112,1,1,40,91,146,188,204,188,146,91,40,1

%N A symmetrical triangle:t(n,m)=Floor[(n!/Floor[n/2]!^2)*(Exp[ -(m - n/2)^2/( 2*((n + 1)/4)^2)] - Exp[ -(n/2)^2/(2*((n + 1)/4)^2)]) + 1]

%C The sequence is an approximate adjusted normal probability distribution made integer by the Floor[] operation.

%C Row sums are:

%C {1, 2, 4, 10, 13, 66, 56, 434, 254, 2532, 1136,...}.

%F t(n,m)=Floor[(n!/Floor[n/2]!^2)*(Exp[ -(m - n/2)^2/( 2*((n + 1)/4)^2)] - Exp[ -(n/2)^2/(2*((n + 1)/4)^2)]) + 1]

%e {1},

%e {1, 1},

%e {1, 2, 1},

%e {1, 4, 4, 1},

%e {1, 3, 5, 3, 1},

%e {1, 11, 21, 21, 11, 1},

%e {1, 6, 13, 16, 13, 6, 1},

%e {1, 34, 76, 106, 106, 76, 34, 1},

%e {1, 15, 33, 50, 56, 50, 33, 15, 1},

%e {1, 112, 258, 402, 493, 493, 402, 258, 112, 1},

%e {1, 40, 91, 146, 188, 204, 188, 146, 91, 40, 1}

%t t0[n_, m_] = Floor[(n!/Floor[n/2]!^2)*(Exp[ -(m - n/2)^2/(2*((n + 1)/4)^2)] - Exp[ -(n/2)^2/(2*((n + 1)/4)^2)]) + 1];

%t Table[Table[t0[n, m], {m, 0, n}], {n, 0, 10}];

%t Flatten[%]

%K nonn,tabl,uned

%O 0,5

%A _Roger L. Bagula_, Apr 16 2010