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BINOMIAL COEFFICIENT AND THE GAUSSIAN The binomial coefficient is defined asC (n, k ) = n! k!(n − k )! and can be written out in the form of a Pascal Triangle starting at the zeroth row with element C(0,0)=1 and followed by the two numbers C(1, 0)=1 and C(1, 1)=1 in the first row. The second row produces the numbers [1, 2, 1] and the third row the four numbers [1, 3, 3, 1]. The row number is here given by the value of n, and the number of elements in a row will be n+1. From these first few results one can state that- C (n, k ) + C (n, k + 1) = C (n + 1, k + 1) The Pascal Triangle thus reads1 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 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 It is known that the numbers in the individual rows of this triangle also give the coefficients in the binomial expansion of (a+b)^n. Thus, for example, for n=4, we have (a + b) 4 = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4 One also notices that the binomial coefficients for fixed n are symmetric about their central maximum value. Plotting their normalized values at n=72 with -15<k<15 we get the graph- We have superimposed on these values the Gaussian exp(-k^2/36). The results are almost identical. Indeed as n is allowed to become large one has the result- exp(−2k 2 / n) ≈ [(n / 2)!]2 [(n / 2)!]2 C (n, k + n / 2) = n! (n / 2 + k )!(n / 2 − k )! with n>2k. At n=128 and k=8 one finds- exp(1) ≈ 128! 2 [64! C (128,72)] = 2.704150286… which is close to the correct value of 2.718281828.. An even better approximation occurs when k is near zero since the two sides of the equation are normalized at k=0. So keeping k=1 and k<<n, we find- N + 2 exp(1) ≈= N N This yields the approximation 2.71801005.. for N=10,000. Note that one can also use the Stirling Approximation for large n. It gives, for example, the estimaten +1 / 2 C (n, k ) ≈ 1 n [ ] 2π k k +1 / 2 (n − k ) n − k +1 / 2 for n >> 1 which yields the approximation C(1000,500)≈0.270355…·10^300 compared to the exact value of C(1000,5000)=0.270288…·10^300. Going back to the Pascal Triangle given above one notices the following2 ∑ C (2, k ) = 1 + 2 + 1 = 2 2 k =0 3 C (3, k ) = 1 + 3 + 3 + 1 = 23 ∑ k =0 4 ∑ C (4, k ) = 1 + 4 + 6 + 4 + 1 = 2 4 k =0 From this it follows that- n ∑ C ( n, k ) = 2 k =0 n . One also notices that if C(n,1) is prime then C(n,k)/C(n,1) is an integer provided 0<k<n. Furthermore a violation of this condition indicates that the number is composite . Thus one has thatC ( n, k ) ( N − 1)! 1 k −1 N is prime if T (k ) = C (n,1) = k!( N − k )! = ∏ ( N − n) k! n =1 is an integer for all values of k ranging from k=1 through k=N/2. Any violation will indicate the number N is composite. Consider the number N=177. On running the one line MAPLE programfor k from 1 to 60 do {k,(1/k!)*product((177-n),n=1..k-1)}od; one finds T(3),T(6),T(9),T(12),T(15),T(59) up to k=60 to be non-integer and hence the number is composite. An extra benefit is that one actually finds the factors to be the lowest values of k not multiples of a lower value. Thus k=3 and k=59 in this case, so that 177=3·59. Unfortunately this approach for determining the primeness of a large number will be quite cumbersome and is actually more time consuming than just looking at the ratio N/(2k+1) for 2k+1<sqrt(N). Running the one-linerfor k from 1 to 14 do {2*k+1,177/(2*k+1)}od; already produces an integer for k=1 yielding 177/3=57 so that 177=3(57). Going back to the Pascal Triangle we observe that the sequence of numbers lying on the vertical line dividing the triangle in half is1-2-6-20-70-252-924- … In terms of the C(n,k) coefficient this means the sequence goes asC(0,0)-C(2,1)-C(4,2)-C(6,3)-C(8,4)-…C(2n,n)… Adding the terms together through the 2m th row, we find the Sum(m) to bem m ( 2n)! 2 n = 0 n! Sum(m) = ∑ C (2n, n) = ∑ n =0 For m=6 this sum equals Sum(6)=1+2+6+20+70+252+924=1275. One can also ask for the value of the reciprocal series which converges to the finite value - 1 1 1 1 n!2 =1+ + + + + ... = 1.73639985871871507790979516836… ∑ 2 6 20 70 ( 2 )! n n=0 ∞ We can also work out the seriesn S (n) = ∑ [C (n, k )]2 k =0 It yields S(1)=2, S(2)=6, S(3)=20, and S(4)=70. From this one sees that S(n) sums to C(2n,n). Finally we point out that one can generalize the Pascal Triangle starting with m ones in the first row and then adding the m nearest values in the (n-1) row to find the element in the nth row. For m=3 this will lead to the triangle- 1 1 4 1 5 15 1 6 21 50 1 1 2 3 6 10 16 30 45 90 126 1 1 1 3 2 1 7 6 3 1 19 16 10 4 1 51 45 30 15 5 1 141 126 90 50 21 6 1 etc. with the generating formula- D(n, k ) = D(n − 1, k − 2) + D(n − 1, k − 1) + D(n − 1, k ) Thus D(5,5)=51=16+19+16 and D(6,7)=126=51+45+30. Note that for this m=3 case that the number of elements in the nth row is 2n+1 and they sum to 3^n. The elements in the nth row again have the character of a Gaussian distribution with very small discrepancies noticed near the tails as shown on the following graph- . Comparing this result and that for the classic Pascal triangle, we can conclude that- m ones in the first row of a modified Pascal Triangle will lead to a triangle whose elements in the nth row number [(m-1)n]+1 and add up to m^n. The element F(n,k) of such a modified Pascal Triangle will be given byF(n,k)= F(n-1,k-m+1)+F(n-1,k-m+2)+……..+F(n-1,k) Also one notes that there is a generating function for the coefficients F(n,k) for a given value of m . For example at m=3 the generating function is G=1+x+x^2 which has coefficients [1,1,1] and corresponds to the first row of D(1,k). G^2=1+2x+3x^2+2x^3+x^4 has the coefficients [1,2,3,2,1] and thus yields the coefficients D(2,k). For m=3 and n=6 we find12 (1 + x + x 2 ) 6 = ∑ D(6, k ) x k = 1 + 6 x + 21x 2 + 50 x 3 + 90 x 4 + 126 x 5 + 141x 6 k =0 + 126 x 7 + 90 x 8 + 50 x 9 + 21x10 + 6 x11 + x12 December 2009