<PRE><B>Question 155130</B>
&lt;p&gt;It can be any number you want.&lt;/p&gt;&lt;p&gt;The solution arises from the fact that anything times zero is just zero and also adding zero to something doesn&#39;t change it.&lt;/p&gt;&lt;p&gt;Now granted the formula will get a bit lengthy but it is a valid solution and it can be any number you want.&lt;/p&gt;&lt;p&gt;Lets start with a simple sequence of three random numbers between (1-10) from random.org I got 6,9,2,_ whats next, Ok back to random.org its came up with 5.&lt;/p&gt;&lt;p&gt;So our problem is to come up with a formula f(n) that produces f(1)=6, f(2)=9, f(3)=2, and f(4)=5&lt;/p&gt;&lt;p&gt;
f(1)=6
f(2)=9
f(3)=2
f(4)=5&lt;/p&gt;&lt;p&gt;Lets initially choose&lt;/p&gt;&lt;p&gt;f(n) = 6&lt;/p&gt;&lt;p&gt;f(1) = 6 check
f(2) = 6 fail&lt;/p&gt;&lt;p&gt;so lets refine it&lt;/p&gt;&lt;p&gt;f(n) = 6 + (n-1)(3/1)&lt;/p&gt;&lt;p&gt;f(1) = 6 check
f(2) = 9 check
f(3) = 12 fail&lt;/p&gt;&lt;p&gt;so we refine further&lt;/p&gt;&lt;p&gt;f(n) = 6 + (n-1)(3/1) + (n-1)(n-2)(-10/2)&lt;/p&gt;&lt;p&gt;f(1) = 6 check
f(2) = 9 check
f(3) = 2 check
f(4) = -15 fail&lt;/p&gt;&lt;p&gt;so we refine further&lt;/p&gt;&lt;p&gt;f(n) = 6 + (n-1)(3/1) + (n-1)(n-2)(-10/2) + (n-1)(n-2)(n-3)(10/3)&lt;/p&gt;&lt;p&gt;f(1) = 6 check
f(2) = 9 check
f(3) = 2 check
f(4) = 5 check&lt;/p&gt;&lt;p&gt;{{{ f(n) = 6 +( (n-1)3)/1  + ((n-1)(n-2)(-10))/2 + ((n-1)(n-2)(n-3)10)/3 }}}&lt;/p&gt;&lt;p&gt;So to answer the original question using the same methods the formula looks like this...&lt;/p&gt;&lt;p&gt;{{{ f(n)=9+(n-1)*64/1!+(n-1)*(n-2)*104/2!+(n-1)*(n-2)*(n-3)*48/3!+(n-1)*(n-2)*(n-3)*(n-4)*(0)/4!+(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*0/5!+(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(-2913)/6! }}}&lt;/p&gt;&lt;p&gt;f(7) = 0&lt;/p&gt;&lt;p&gt;simplified...&lt;/p&gt;&lt;p&gt;{{{ f(n)=9+(n-1)*64/1!+(n-1)*(n-2)*104/2!+(n-1)*(n-2)*(n-3)*48/3!+(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6)*(-2913)/6! }}}&lt;/p&gt;</PRE>