# (2,1)-Pascal triangle

In mathematics, the (2,1)-Pascal triangle (mirrored Lucas triangle)is a triangular array.

The rows of the (2,1)-Pascal triangle (sequence A029653 in the OEIS) are conventionally enumerated starting with row n = 0 at the top (the 0th row). The entries in each row are numbered from the left beginning with k = 0 and are usually staggered relative to the numbers in the adjacent rows.

The triangle is based on the Pascal's Triangle with the second line being (2,1) and the first cell of each row set to 2.

This construction is related to the binomial coefficients by Pascal's rule, with one of the terms being $2x+y$ .

$(2x+y)(x+y)^{n-1}$ ## Patterns and properties

(2,1)-Pascal triangle has many properties and contains many patterns of numbers. It can be seen as a sister of the Pascal's triangle, in the same way that a Lucas sequence is a sister sequence of the Fibonacci sequence.[citation needed]

### Rows

• Except the row n = 0, 1, The sum of the elements of a single row is twice the sum of the row preceding it. For example, row 1 has a value of 3, row 2 has a value of 6, row 3 has a value of 12, and so forth. This is because every item in a row produces two items in the next row: one left and one right. The sum of the elements of row n is equal to $3\times 2^{n-1}$ .(sequence A003945 in the OEIS) (sequence A007283 in the OEIS)
• The value of a row, if each entry is considered a decimal place (and numbers larger than 9 carried over accordingly) is a power of 11 multiplied by 21 ($21\times 11^{n-1}$ , for row n). Thus, in row 2, ⟨2, 3, 1⟩ becomes $21\times 11$ , while ⟨2, 9, 16, 14, 6, 1⟩ in row five becomes (after carrying) 307461, which is $21\times 11^{4}$ . This property is explained by setting x = 10 in the binomial expansion of (2x + 1)(x + 1)n−1, and adjusting values to the decimal system. But x can be chosen to allow rows to represent values in any base.
• In base 3: $231_{3}=7\times 4(28)$
• $(2,5,4,1)\rightarrow 11011_{3}=7\times 4^{2}(112)$
• In base 9: $231_{9}=19\times 10(190)$
•              $2541_{9}=19\times 10^{2}(1900)$
• $(2,9,16,14,6,1)\rightarrow 318561_{9}=19\times 10^{4}(190000)$
• Polarity: Yet another interesting pattern, when rows of Pascal's triangle are added and subtracted together sequentially, every row with a middle number, meaning rows that have an odd number of integers, they are always equal to 0. Example, row 4 is 2 7 9 5 1, so the formula would be 9 − (7 + 5) + (2 + 1) = 0, row 6 is 2 11 25 30 20 7 1, so the formula would be 30 − (25 + 20) + (11 + 7) − (2 + 1) = 0. So every even row of the Pascal triangle equals 0 when you take the middle number, then subtract the integers directly next to the center, then add the next integers, then subtract, so on and so forth until you reach the end of the row.
• Or we can say that when we take the first term of a row, then subtract the second term, then add the third term, then subtract, so on and so forth until you reach the end of the row, the result is always equal to 0.
• row 3: 2 − 3 + 1 = 0
• row 4: 2 − 5 + 4 − 1 = 0
• row 5: 2 − 7 + 9 − 5 + 1 = 0
• row 6: 2 − 9 + 16 − 14 + 6 − 1 = 0
• row 7: 2 − 11 + 25 − 30 + 20 − 7 + 1 = 0
• row 8: 2 − 13 + 36 − 55 + 50 − 27 + 8 − 1 = 0

### Diagonals

The diagonals of Pascal's triangle contain the figurate numbers of simplices:

### Overall patterns and properties

(2,1)-Pascal triangle overlaid on a grid gives the number of distinct paths to each square, assuming only rightward and downward movements are considered.
• The pattern obtained by coloring only the odd numbers in Pascal's triangle closely resembles the fractal called the Sierpinski triangle. This resemblance becomes more and more accurate as more rows are considered; in the limit, as the number of rows approaches infinity, the resulting pattern is the Sierpinski triangle, assuming a fixed perimeter. More generally, numbers could be colored differently according to whether or not they are multiples of 3, 4, etc.; this results in other similar patterns.
• Imagine each number in the triangle is a node in a grid which is connected to the adjacent numbers above and below it. Now for any node in the grid, count the number of paths there are in the grid (without backtracking) which connect this node to the top node (1) of the triangle. The answer is the Pascal number associated to that node.
• One property of the triangle is revealed if the rows are left-justified. In the triangle below, the diagonal coloured bands sum to successive Fibonacci numbers and Lucas numbers.
 1 2 1 2 3 1 2 5 4 1 2 7 9 5 1 2 9 16 14 6 1 2 11 25 30 20 7 1 2 13 36 55 50 27 8 1 2 15 49 91 105 77 35 9 1
 1 2 1 2 3 1 2 5 4 1 2 7 9 5 1 2 9 16 14 6 1 2 11 25 30 20 7 1 2 13 36 55 50 27 8 1 2 15 49 91 105 77 35 9 1
• This construction is also related to the expansion of $x^{n}+{\frac {1}{x^{n}}}$  , using $y=x+{\frac {1}{x}}$ .
• then
{\begin{aligned}x^{0}+{\frac {1}{x^{0}}}&=2\\[5pt]x^{1}+{\frac {1}{x^{1}}}&=y\\[5pt]x^{2}+{\frac {1}{x^{2}}}&=y^{2}-2\\[5pt]x^{3}+{\frac {1}{x^{3}}}&=y^{3}-3y\\[5pt]x^{4}+{\frac {1}{x^{4}}}&=y^{4}-4y^{2}+2\\[5pt]x^{5}+{\frac {1}{x^{5}}}&=y^{5}-5y^{3}+5y\end{aligned}}