function p = nextn(n)
% Copyright (c) 2001-2009 by Ales Cerny
%--------------------------------------------------------------------------
% nextn(n) returns a highly composite number p >= n such that the divisors 
% of p are in the set {2, 3, 5, 7} and 7 is present only in 1st power if at
% all
%--------------------------------------------------------------------------

%************************************************************************%
% nextn.m - supplementary program to                            %
% Ales Cerny (2009) Mathematical Techniques in Finance (2nd ed.)         % 
% Princeton University Press http://press.princeton.edu/titles/9079.html %
%************************************************************************%
    
% This code is provided 'as-is', without any express or implied warranty.  
% 
% Permission is granted to anyone to use this code for any purpose, 
% subject to the following restrictions:
% 
% 1. The origin of this code must not be misrepresented; you must not
%    claim that you wrote the original code.
% 2. Modified code versions must be plainly marked as such, and must not 
%    be misrepresented as being the original code.
% 3. This notice may not be removed from any source distribution.

% NOTICE TO STUDENTS: To avoid accusations of plagiarism, if you use this 
% code or its modifications in assessed work you should prepend it with a 
% note stating: 
%   "This is the original/modified version of the code nextn.m by 
%    Ales Cerny (2009), Mathematical Techniques in Finance (2nd ed.),  
%    Princeton University Press. The original version is available from 
%    http://www.martingales.info/mtfweb2".
% A similar acknowledgement should appear prominently inside your written 
% report.


flag = 1;
if ((real(n)~=n) || (n < 1) ||(ceil(n)~=n));
    error('the argument must be a positive integer');
elseif (n <= 10);
    p = n; %#ok<NASGU>
    flag=0;
end
p = n;
while (flag == 1)    
      p0 = p;
      while (mod(p0,2) == 0)
          p0 = p0/2;
      end
            
      while (mod(p0,3) == 0)
          p0 = p0/3;
      end
           
      while (mod(p0,5) == 0)
          p0 = p0/5;
      end
             
      if ((p0 == 1)||(p0==7))
          flag = 0;
      else p=p+1;
      end
end