Pascal's triangle is a number triangle with numbers arranged in staggered rows such that
a_(nr)=(n!)/(r!(n-r)!)=(n; r),
(1)
where (n; r) is a binomial coefficient. The triangle was studied by B. Pascal, although it had been described centuries earlier by Chinese mathematician Yanghui (about 500 years earlier, in fact) and the Persian astronomer-poet Omar Khayyám. It is therefore known as the Yanghui triangle in China. Starting with n=0, the triangle is
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1
(2)
(Sloane's A007318). Pascal's formula shows that each subsequent row is obtained by adding the two entries diagonally above,
(n; r)=(n!)/((n-r)!r!)=(n-1; r)+(n-1; r-1).
(3)
Binary plot for Pascal's triangle
The plot above shows the binary representations for the first 255 (top figure) and 511 (bottom figure) terms of a flattened Pascal's triangle.
The first number after the 1 in each row divides all other numbers in that row iff it is a prime.
The sums P_n of the number of odd entries in the first n rows of Pascal's triangle for n=0, 1, ... are 0, 1, 3, 5, 9, 11, 15, 19, 27, 29, 33, 37, 45, 49, ... (Sloane's A006046). It is then true that
0.812...<P_nn^(-theta)<=1
(4)
(Harborth 1976, Le Lionnais 1983), with equality for n a power of 2, and the power of n given by the constant
theta=(ln3)/(ln2)=log_23=1.58496250072115...
(5)
(Sloane's A020857). The sequence of cumulative counts of odd entries has some amazing properties, and the minimum possible value beta=0.812... (Sloane's A077464) is known as the Stolarsky-Harborth constant.
Pascal's triangle contains the figurate numbers along its diagonals, as can be seen from the identity
sum_(i=1)^(n)(i; j) = (n+1)/(j+1)(n; j)
(6)
= (n+1; j+1).
(7)
In addition, the sum of the elements of the ith row is
sum_(j=0)^i(i; j)=2^i,
(8)
so the sum of the first k rows (i.e., rows 0 to k-1) is the Mersenne number
sum_(i=0)^(k-1)2^i=2^k-1.
(9)
FibonacciShallowDiags
The "shallow diagonals" of Pascal's triangle sum to Fibonacci numbers, i.e.,
1 = 1
(10)
1 = 1
(11)
2 = 1+1
(12)
3 = 2+1
(13)
5 = 1+3+1
(14)
8 = 3+4+1
(15)
and, in general,
sum_(k=0)^(|_n/2_|)(n-k; k)=F_(n+1).
(16)
The numbers of times that the numbers 2, 3, 4, ... occur in Pascal's triangle are given by 1, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 4, ... (Sloane's A003016; Ogilvy 1972, p. 96; Comtet 1974, p. 93; Singmaster 1971). Similarly, the numbers of rows in which the numbers 2, 3, 4, ... occur are 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, ... (Sloane's A059233).
By row 210, the numbers
120 = (10; 3)=(10; 7)=(16; 2)=(16; 14)=(120; 1)=(120; 119)
(17)
210 = (10; 4)=(10; 6)=(21; 2)=(21; 19)=(210; 1)=(210; 209)
(18)
3003 = (14; 6)=(14; 8)=(15; 5)=(15; 10)=(78; 2)=(78; 76)
(19)
have appeared six times, more than any other number (excluding 1). By row 1540,
1540=(22; 3)=(22; 19)=(56; 2)=(56; 54)=(1540; 1) =(1540; 1539)
(20)
has now occurred six times, by row 3003,
3003=(14; 6)=(14; 8)=(15; 5)=(15; 10)=(78; 2) =(78; 76)=(3003; 1)=(3003; 3002)
(21)
has now occurred 8 times, and by row 7140, 7140 has appeared six times as well. In fact, the numbers that occur five or more times in Pascal's triangle are 1, 120, 210, 1540, 3003, 7140, 11628, 24310, ... (Sloane's A003015), with no others up to 33×10^(16).
It is known that there are infinitely many numbers that occur at least 6 times in Pascal's triangle, namely the solutions to
r=(n; m-1)=(n-1; m)
(22)
given by
m = F_(2k-1)F_(2k)
(23)
n = F_(2k)F_(2k+1),
(24)
where F_i is the ith Fibonacci number (Singmaster 1975). The first few such values of r for k=1, 2, ... are 1, 3003, 61218182743304701891431482520, ... (Sloane's A090162).
There is an unexpected connection between Pascal's triangle and the Delannoy numbers via Cholesky decomposition (G. Helms, pers. comm., Aug. 29, 2005).
Pascal's triangle (mod 2) turns out to be equivalent to the Sierpi?ski sieve (Wolfram 1984; Crandall and Pomerance 2001; Borwein and Bailey 2003, pp. 46-47). Guy (1990) gives several other unexpected properties of Pascal's triangle.
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