Kamis, 05 Maret 2009

BINOMIAL COEFFFICIENT

Binomial Coefficient

The binomial coefficient (n; k) is the number of ways of picking k unordered outcomes from n possibilities, also known as a combination or combinatorial number. The symbols _nC_k and (n; k) are used to denote a binomial coefficient, and are sometimes read as "n choose k." (n; k) therefore gives the number of k-subsets possible out of a set of n distinct items. For example, The 2-subsets of {1,2,3,4} are the six pairs {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, and {3,4}, so (4; 2)=6.

The value of the binomial coefficient is given explicitly by

 _nC_k=(n; k)=(n!)/((n-k)!k!),
(1)

where z! denotes a factorial. Writing the factorial as a gamma function z!=Gamma(z+1) allows the binomial coefficient to be generalized to non-integer arguments, including complex n and k. The binomial coefficient is implemented in Mathematica as Binomial[n, k].

The binomial coefficients form the rows of Pascal's triangle, and the number of lattice paths from the origin (0,0) to a point (a,b) is the binomial coefficient (a+b; a) (Hilton and Pedersen 1991).

BinomialCoefficient

Plotting the binomial coefficient

 (x; y)=(x!)/(y!(x-y)!)
(2)

in the xy-plane (Fowler 1996) gives the beautiful plot shown above, which has a very complicated graph for negative x and y and is therefore difficult to render using standard plotting programs.

For a positive integer n, the binomial theorem gives

 (x+a)^n=sum_(k=0)^n(n; k)x^ka^(n-k).
(3)

The finite difference analog of this identity is known as the Chu-Vandermonde identity. A similar formula holds for negative integers,

 (x+a)^(-n)=sum_(k=0)^infty(-n; k)x^ka^(-n-k).
(4)

There are a number of elegant binomial sums.

The binomial coefficients satisfy the identities

(n; 0)=(n; n)=1
(5)
(n; k)=(n; n-k)
(6)
=(-1)^k(k-n-1; k)
(7)
(n; k+1)=(n; k)(n-k)/(k+1)
(8)
(n+1; k)=(n; k)+(n; k-1).
(9)

The product of binomial coefficients is given by

 product_(k=0)^n(n; k)=(H^2(n))/((n!)^(n+1)),
(10)

where H(n) is a hyperfactorial and n! is a factorial.

As shown by Kummer in 1852, if p^k is the largest power of a prime p that divides (m+n; m), where m and n are nonnegative integers, then k is the number of carries that occur when m is added to n in base p (Graham et al. 1989, Exercise 5.36, p. 245; Ribenboim 1989; Vardi 1991, p. 68). Kummer's result can also be stated in the form that the exponent of a prime p dividing (n; m) is given by the number of integers =0" border="0" height="14" width="28"> for which

frac(n/p^j), " border="0" height="21" width="148">
(11)

where frac(x) denotes the fractional part of x. This inequality may be reduced to the study of the exponential sums sum_(n)Lambda(n)e(x/n), where Lambda(n) is the Mangoldt function. Estimates of these sums are given by Jutila (1973, 1974), but recent improvements have been made by Granville and Ramare (1996).

R. W. Gosper showed that

 f(n)=(n-1; 1/2(n-1))=(-1)^((n-1)/2) (mod n)
(12)

for all primes, and conjectured that it holds only for primes. This was disproved when Skiena (1990) found it also holds for the composite number n=3×11×179. Vardi (1991, p. 63) subsequently showed that n=p^2 is a solution whenever p is a Wieferich prime and that if n=p^k with 3" border="0" height="14" width="28"> is a solution, then so is n=p^(k-1). This allowed him to show that the only solutions for composite n<1.3×10^7 are 5907, 1093^2, and 3511^2, where 1093 and 3511 are Wieferich primes.

Consider the binomial coefficients f(n)=(2n-1; n), the first few of which are 1, 3, 10, 35, 126, ... (Sloane's A001700). The generating function is

 1/2[1/(sqrt(1-4x))-1]=x+3x^2+10x^3+35x^4+....
(13)

These numbers are squarefree only for n=2, 3, 4, 6, 9, 10, 12, 36, ... (Sloane's A046097), with no others known. It turns out that f(n) is divisible by 4 unless n belongs to a 2-automatic set S_2, which happens to be the set of numbers whose binary representations contain at most two 1s: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 16, 17, 18, ... (Sloane's A048645). Similarly, f(n) is divisible by 9 unless n belongs to a 3-automatic set S_3, consisting of numbers n for which the representation of 2n in ternary consists entirely of 0s and 2s (except possibly for a pair of adjacent 1s). The initial elements of S_3 are 1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 18, 19, 21, 22, 27, ... (Sloane's A051382). If f(n) is squarefree, then n must belong to S=S_2 intersection S_3. It is very probable that S is finite, but no proof is known. Now, squares larger than 4 and 9 might also divide f(n), but by eliminating these two alone, the only possible n for n<=2^(64) are 1, 2, 3, 4, 6, 9, 10, 12, 18, 33, 34, 36, 40, 64, 66, 192, 256, 264, 272, 513, 514, 516, 576 768, 1026, 1056, 2304, 16392, 65664, 81920, 532480, and 545259520. All of these but the last have been checked, establishing that there are no other n such that f(n) is squarefree for n<=545259520.

Erdős showed that the binomial coefficient (n; k) with 3<=k<=n/2 is a power of an integer for the single case (50; 3)=140^2 (Le Lionnais 1983, p. 48). Binomial coefficients T_(n-1)=(n; 2) are squares a^2 when a^2 is a triangular number, which occur for a=1, 6, 35, 204, 1189, 6930, ... (Sloane's A001109). These values of a have the corresponding values n=2, 9, 50, 289, 1682, 9801, ... (Sloane's A052436).

The binomial coefficients (n; |_n/2_|) are called central binomial coefficients, where |_x_| is the floor function, although the subset of coefficients (2n; n) is sometimes also given this name. Erdős and Graham (1980, p. 71) conjectured that the central binomial coefficient (2n; n) is never squarefree for 4" border="0" height="14" width="29">, and this is sometimes known as the Erdős squarefree conjecture. Sárkőzy's theorem (Sárkőzy 1985) provides a partial solution which states that the binomial coefficient (2n; n) is never squarefree for all sufficiently large =n_0" border="0" height="16" width="35"> (Vardi 1991). Granville and Ramare (1996) proved that the only squarefree values are n=2 and 4. Sander (1992) subsequently showed that (2n+/-d; n) are also never squarefree for sufficiently large n as long as d is not "too big."

For p, q, and r distinct primes, then the function (◇) satisfies

 f(pqr)f(p)f(q)f(r)=f(pq)f(pr)f(qr) (mod pqr)
(14)

(Vardi 1991, p. 66).

Most binomial coefficients (n; k) with =2k" border="0" height="14" width="38"> have a prime factor p<=n/k, and Lacampagne et al. (1993) conjecture that this inequality is true for all 17.125k" border="0" height="14" width="70">, or more strongly that any such binomial coefficient has least prime factor p<=n/k or p<=17 with the exceptions (62; 6), (959; 56), (474; 66), (284; 28) for which p=19, 19, 23, 29 (Guy 1994, p. 84).

The binomial coefficient (m; n) (mod 2) can be computed using the XOR operation n XOR m, making Pascal's triangle mod 2 very easy to construct.

Sondow (2005) and Sondow and Zudilin (2006) noted the inequality

 1/(4rm)[((r+1)^(r+1))/(r^r)]^m<((r+1)m; m)<[((r+1)^(r+1))/(r^r)]^m
(15)

for m a positive integer and =1" border="0" height="14" width="27"> a real number.

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