Question 651209
your sequence represents {{{pentagonal}}} numbers are given by {{{n(3n-1)/2}}} with n=0, +/-1, +/-2, ...

in your case we will start with {{{n=1}}}

{{{n(3n-1)/2=1(3*1-1)/2=2/2=1}}}

{{{n(3n-1)/2=2(3*2-1)/2=5}}}

{{{n(3n-1)/2=3(3*3-1)/2=12}}}

{{{n(3n-1)/2=4(3*4-1)/2=22}}}