Question 1048127
An arithmetic sequence might work.
i=1 gives 3, and i=4 gives 17.


Let d be the common difference to successive terms.
The difference from i=1 to i=4 is 3d and this takes the sequence from element 3 to element 17.


{{{3+3d=17}}}


{{{3d=17-3}}}

{{{3d=14}}}

{{{d=14/3}}}

Therefore, an arithmetic sequence can be  {{{3+(n-1)(14/3)}}}; but other patterns may also be possible, but maybe giving other elements different values for indices not ( i=1 and i=4 ).