Question 322462
<pre><b>
{{{a[1]=0}}},{{{a[2]=3}}},{{{a[3]=5}}},{{{a[4]=9}}},{{{a[5]=12}}},{{{a[6]=7}}}

That has 6 terms, and we don't see an obvious pattern for it.
1 less than 6 is 5, so we assume a 5th degree polynomial equation 
in n for the <i>n</i>th term {{{a[n]}}}. 

That is, we assume the 5th degree polynomial

{{{a[n]=An^5+Bn^4+Cn^3+Dn^2+En+F}}}

We substitute 1,2,3,4,5, and 6 for n, making 6 equations
in 6 unknowns:

{{{system(a[1]=A*1^5+B*1^4+C*1^3+D*1^2+E*1+F=0,
          a[2]=A*2^5+B*2^4+C*2^3+D*2^2+E*2+F=3, 
          a[3]=A*3^5+B*3^4+C*3^3+D*3^2+E*3+F=5,
          a[4]=A*4^5+B*4^4+C*4^3+D*4^2+E*4+F=9,
          a[5]=A*5^5+B*5^4+C*5^3+D*5^2+E*5+F=12,
          a[6]=A*6^5+B*6^4+C*6^3+D*6^2+E*6+F=7)}}}

Simplifying:

{{{system(A+B+C+D+E+F=0,
          32A+16B+8C+4D+2E+F=3, 
          243A+81B+27C+9D+3E+F=5,
          1024A+256B+64C+16D+4E+F=9,
          3125A+625B+125C+25D+5E+F=12,
          7776A+1296B+216C+36D+6E+F=7)}}}

Write the augmented matrix:

{{{(matrix(6,8,
             1,    1,  1, 1,1,1,"|",0,
            32,   16,  8, 4,2,1,"|",3, 
           243,   81, 27, 9,3,1,"|",5,
          1024,  256, 64,16,4,1,"|",9,
          3125,  625,125,25,5,1,"|",12,
          7776, 1296,216,36,6,1,"|",7))}}}

If you are using a TI-83 or TI-84, put that
in matrix [A], then in the 2nd MATRIX MATH
menu select rref([A]} and get

{{{(matrix(6,8,
             1,0,0,0,0,0,"|",.0166666667,
             0,1,0,0,0,0,"|",-.5, 
             0,0,1,0,0,0,"|",4.416666667,
             0,0,0,1,0,0,"|",-16,
             0,0,0,0,1,0,"|",27.06666667,
             0,0,0,0,0,1,"|",-15))}}}

You could do that without a calculator using row
operations, but it would take you a long time and
would be tedious 

Change all those decimals to fractions:

{{{(matrix(6,8,
             1,0,0,0,0,0,"|",1/60,
             0,1,0,0,0,0,"|",-1/2, 
             0,0,1,0,0,0,"|",53/12,
             0,0,0,1,0,0,"|",-16,
             0,0,0,0,1,0,"|",406/15,
             0,0,0,0,0,1,"|",-15))}}}

So that matrix gives us

{{{system(A=1/60,B=-1/2,C=53/12,D=-16,E=406/15,F=-15)}}}

So the equation that will produce 

{{{a[1]=0}}},{{{a[2]=3}}},{{{a[3]=5}}},{{{a[4]=9}}},{{{a[5]=12}}},{{{a[6]=7}}}

is found by substituting those values in

{{{a[n]=An^5+Bn^4+Cn^3+Dn^2+En+F}}}

{{{a[n]=(1/60)n^5-(1/2)n^4+(53/12)n^3-16n^2+(406/15)n-15}}} 

and if you like you can find as many terms as you like by substituting
n=7,8,9,10,11,12,...:

{{{a[7]=-15}}},{{{a[8]=-63}}},{{{a[9]=-144}}},{{{a[10]=-261}}},{{{a[11]=-411}}},{{{a[12]=-583}}}, etc.

Edwin</pre>