A = [4 8 14]
B = [1 2.5 5 11 17]
C = B(3:end) - A % take only the variables from 3 to end of vector
A more dynamic approach, assuming B always starts at 1 and A can be at different times
%% Assumng B always starts at 1
A(3:5) = [4 8 14]; % generate A with empty values in first 3 positions
B = [1 2.5 5 11 17];
[rowVal,colVal] = find(A); % find non-zero elements in A
newA = A(colVal); % only uses valid values of A
newB = B(colVal); % only uses same values that A has
C = newB - newA;
