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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.

%I A157933 #29 Oct 11 2023 03:55:58
%S A157933 1,3,3,7,10,7,15,25,25,15,31,56,66,56,31,63,119,154,154,119,63,127,
%T A157933 246,337,372,337,246,127,255,501,711,837,837,711,501,255,511,1012,
%U A157933 1468,1804,1930,1804,1468,1012,511,1023,2035,2992,3784,4246,4246,3784,2992,2035,1023
%N 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.
%C A157933 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 A157933 The first and last term in the (i+1)-th row is T(i,0) = 2^(i+1)-1.
%F A157933 The second and penultimate term is T(i,1) = T(i,0) + T(i-1,1).
%F A157933 G.f.: 1/((1-2*x)*(1-2*x*y)*(1-x-x*y)). - _Yu-Sheng Chang_, Sep 20 2023
%e A157933 To get the 3rd row of the triangle, consider the pyramid
%e A157933     f
%e A157933    d e
%e A157933   a b c
%e A157933 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 A157933 Triangle begins:
%e A157933    1,
%e A157933    3,   3;
%e A157933    7,  10,   7;
%e A157933   15,  25,  25,  15;
%e A157933   31,  56,  66,  56,  31;
%e A157933   63, 119, 154, 154, 119, 63;
%e A157933   ...
%Y A157933 Row sums give A001788(n+1).
%Y A157933 T(2n,n) gives A033504.
%K A157933 nonn,tabl
%O A157933 0,2
%A A157933 _M. F. Hasler_, Mar 16 2009