Calculating binary progression using for loop

How to correct folowing- I am using for loop.

%Find open nozzles from binary data
%Binary no=sum(2^nozzle no.) --- e.g. 2^2+2^5+2^6=100
%We need to reverse calculate nozzle nos from binary pattern.
% e.g. When we enter 100, output should be 2, 5, 6.
clc
N=1:12;
for i = 1:12
    a(i)=2^i;
end

T=table(N(:),a(:),'VariableNames', {'Nozzle_number', 'binarycode'});
T1=table(0,1,'VariableNames', {'Nozzle_number','binarycode'});
Tout=[T1;T]

numin=1064;

for j=1:N  
    n(j)=floor(log(j)/log(2));
    j=j-(2^n(j));
end

T2=table(j(:),n(:),'VariableNames',{'binary','Nozzle_numbers'})

Expected answer is table of nozzle no. in this case (1064)- 10, 5, 3 (i.e. 2^10+2^5+2^3)

[KALYAN ACHARJYA]

"Expected answer is table of nozzle no. in this case (1064)- 10, 5, 3 (i.e. 2^10+2^5+2^3)"

Can you elaboate?

Or what exactly you are looking for? What are the inputs you have?

[Vikas Salunke]

Input is a number (numin) in example it is 1064.

=1064

I am looking to calculate all power of two required to form this no. in a binary series.

Answer 1

[Guillaume]

"Input is a number (numin), I am looking to calculate all power of two required to form this no. in a binary series"

Easily done:

numin = 1064
find(fliplr(dec2bin(numin)) == '1') - 1  

[Stephen23]

+1 slight variation:

>> find(fliplr(dec2bin(1064))-'0')-1
ans =
  3  5  10

[Vikas Salunke]

This helps. Thanks a lot.

Answer 2

[KALYAN ACHARJYA]

num=1064;
j=1;r=[];
bin_num=str2num(dec2bin(num));
num_array=num2str(bin_num)-'0';
for i=length(num_array):-1:1
    if num_array(j)==1
    r(j)=(i-1);
    end
    j=j+1;
end
disp('The 2 power are');
disp(nonzeros(r));

Result:

The 2 power are
    10
     5
     3

Please note, three might be more easier way

[Vikas Salunke]

Thanks.