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For example, the third triangular number is (3 × 2 =) 6, the seventh is (7 × 4 =) 28, the 31st is (31 × 16 =) 496, and the 127th is (127 × 64 =) 8128. The final digit of a triangular number is 0, 1, 3, 5, 6, or 8, and thus such numbers never end in 2, 4, 7, or 9. A final 3 must be preceded by a 0 or 5; a final 8 must be preceded by a 2 or 7.
This gnomonic technique also provides a mathematical proof that the sum of the first n odd numbers is n 2; the figure illustrates 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64 = 8 2. There is a similar gnomon with centered hexagonal numbers adding up to make cubes of each integer number.
the number of vertices in a 6-cube, the fourth dodecagonal number, [8] and the seventh centered triangular number. [9] Since it is possible to find sequences of 65 consecutive integers (intervals of length 64) such that each inner member shares a factor with either the first or the last member, 64 is the seventh Erdős–Woods number. [10]
The triangular number sequence is the representation of the numbers in the form of equilateral triangle arranged in a series or sequence. These numbers are in a sequence of 1, 3, 6, 10, 15, 21, 28, 36, 45, and so on.
8 2 = 64 9 2 = 81. 10 2 = 100 11 2 = 121 12 2 ... Since all triangular numbers have an odd factor, ... An analogous pattern applies for the last 3 digits around ...
The difference of the n-th and the (n+1)-th consecutive centered k-gonal numbers is k(2n+1). The n-th centered k-gonal number is equal to the n-th regular k-gonal number plus (n-1) 2. Just as is the case with regular polygonal numbers, the first centered k-gonal number is 1. Thus, for any k, 1 is both k-gonal and centered k-gonal.
Every hexagonal number is a triangular number, but only every other triangular number (the 1st, 3rd, 5th, 7th, etc.) is a hexagonal number. Like a triangular number, the digital root in base 10 of a hexagonal number can only be 1, 3, 6, or 9. The digital root pattern, repeating every nine terms, is "1 6 6 1 9 3 1 3 9".
A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. From Gulley (2010).The n th coloured region shows n squares of dimension n by n (the rectangle is 1 evenly divided square), hence the area of the n th region is n times n × n.