# Problem 61: Cyclical figurate numbers
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal
numbers are all figurate (polygonal) numbers and are generated by the
following formulae: Triangle   P3,n=n(n+1)/2   1, 3, 6, 10, 15, ...
Square   P4,n=n2   1, 4, 9, 16, 25, ... Pentagonal   P5,n=n(3n−1)/2   1,
5, 12, 22, 35, ... Hexagonal   P6,n=n(2n−1)   1, 6, 15, 28, 45, ...
Heptagonal   P7,n=n(5n−3)/2   1, 7, 18, 34, 55, ... Octagonal  
P8,n=n(3n−2)   1, 8, 21, 40, 65, ... The ordered set of three 4-digit
numbers: 8128, 2882, 8281, has three interesting properties. The set is
cyclic, in that the last two digits of each number is the first two
digits of the next number (including the last number with the first).
Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and
pentagonal (P5,44=2882), is represented by a different number in the
set. This is the only set of 4-digit numbers with this property. Find
the sum of the only ordered set of six cyclic 4-digit numbers for which
each polygonal type: triangle, square, pentagonal, hexagonal,
heptagonal, and octagonal, is represented by a different number in the
set.