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A157933 Triangle T[i,j] such that sum_{j=0...i} T[i,j]*x[i,j]/2^i = sum_{k=0...i, j=0...k} x[k,j], if x[k-1,j]=(x[k,j]+x[k,j+1])/2 0

%I

%S 1,3,3,7,10,7,15,25,25,15,31,56,66,56,31,63,119,154,154,119,63,127,

%T 246,337,372,337,246,127,255,501,711,837,837,711,501,255,511,1012,

%U 1468,1804,1930,1804,1468,1012,511

%N Triangle T[i,j] such that sum_{j=0...i} T[i,j]*x[i,j]/2^i = sum_{k=0...i, j=0...k} x[k,j], if x[k-1,j]=(x[k,j]+x[k,j+1])/2

%C Rows and columns are numbered starting with 0. Consider a pyramid (triangle) where each element is the mean value of the two elements below. Then the sum of all elements is expressed as linear combination of the elements at the base. This sequence gives the coefficients times the necessary power of 2.

%F The first and last term in the (i+1)-th row is T[i,0] = 2^(i+1)-1.

%F The second and penultimate term is T[i,1] = T[i,0] + T[i-1,1].

%e To get the 3rd row of the triangle, consider the pyramid

%e __f

%e _d e

%e a b c

%e where d=(a+b)/2, e=(b+c)/2, f=(d+e)/2. Then a+b+c+d+e+f=(7a+10b+7c)/2^2, which yields the row (7,10,7).

%K nonn,tabl

%O 0,2

%A _M. F. Hasler_, Mar 16 2009

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