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%I #29 Oct 11 2023 03:55:58
%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 1468,1804,1930,1804,1468,1012,511,1023,2035,2992,3784,4246,4246,3784,2992,2035,1023
%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).
%F G.f.: 1/((1-2*x)*(1-2*x*y)*(1-x-x*y)). - _Yu-Sheng Chang_, Sep 20 2023
%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).
%e Triangle begins:
%e 1,
%e 3, 3;
%e 7, 10, 7;
%e 15, 25, 25, 15;
%e 31, 56, 66, 56, 31;
%e 63, 119, 154, 154, 119, 63;
%e ...
%Y Row sums give A001788(n+1).
%Y T(2n,n) gives A033504.
%K nonn,tabl
%O 0,2
%A _M. F. Hasler_, Mar 16 2009
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