"It looks there is a pattern that can help, but I am not able to find it."
Should you the one who should define how is the storage and derive appropriate indexing rather than "find it"?
It looks like if you index some vector with length 4^N is iquivalent to indexing an N-dimensional object (tensor?) where each dimention has 4 length:
T(4^N,4^N) ~ TT(4,4,4, ...) % 2N = N+N times on the RHS.
The indexing of the LHS array T and RHS multi-dimensional TT can be converted back and forth with sub2ind and ind2sub function.
T(i,j) ~ TT(i1,i2,...,iN,j1,j2,...,jN);
where
i = sub2ind([4,4,...4], i1,i2,...,iN);
[i1,i2,...,iN] = ind2sub([4,4,...4], i);
and similar for j.
If this is the case might be indexing in multi-dimensional array TT will be clearer for you (with the same kind of conversion for P).
Note that all the "..." of the above formal description can be effectively implemented in MATLAB syntax using cell/comma list of length N.
% As illustration for first dimmension of T only
% second dimension is supposed = 1
N = 6;
szT = [4^N,1];
% Generate random array
T = rand(szT);
% multi-dimensional
szTT = 4*ones(1,N); % [4 4 4 4 4 4]
TT = reshape(T, szTT);
% T = reshape(TT, szT)
Tindex = 2020 % example test linear index
ic = cell(1,N);
[ic{:}] = ind2sub(szTT,Tindex); % szTT is [4 4 4 4 4 4]
i = [ic{:}]; % [4 1 3 4 4 2] is this example
% Equivalent method to get i without using cell
% i = flip(dec2base(Tindex-1,4,N)-'0') + 1
% Get the element 2020 from T
Ti = T(Tindex)
% Get the same elemnt from TT using multi indexing (comma-list) array i
ic = num2cell(i);
TTi = TT(ic{:}) % TT(4,1,3,4,4,2) equal to T(2020)
% Check
isequal(Ti,TTi)
Note that back conversion from index vector i to linear index Tindex can be carried out like this
ic = num2cell(i);
Tindex = sub2ind(szTT,ic{:})
% or equivalently
Tindex = polyval(flip(i)-1,4)+1
% or
Tindex = base2dec(flip(i)+'0'-1,4)+1
My guess is if my formalism is suitable and once you sort out the equivalence, the tedious code can be written as few neat for-loop and array operations.
