Find all possible triplets a,b,c that appear as adjacent coefficents in the same row of Pascal triangle and form an arithmetic sequence.

Let the coefficients  ⁿC_r-1 , ⁿC_r  and ⁿC_r+1 be in arithmetic progression,

so that:             ⁿC_r-1  +  ⁿC_r+1  =  2 ⁿC_r

Multiplying by (r + 1)! (n - r + 1)! / n! Gives

           r(r + 1)  +  (n - r + 1)(n - r)  =  2(r + 1)(n - r + 1)

                        4r²  +  n²  - 4nr  =  n + 2

                        (n  -  2r)²  =  n + 2

Let  n - 2r =  t,   so that             n  =  t² - 2        (1)

The parameter, t, must be an integer, and we need only consider t > 0 since this indicates that r < n/2  and the progression will have increasing values in the first half of Pascal's row. t < 0 will give the same values in reverse order in the second half of Pascal's row.

So, from 1:        r = (n - t)/2  =  (t² - t - 2)/2         (2)

and for t = 3, 4, 5, ....  (1) and (2) give the following results:

t           n          r           Sequence of coefficients

3          7          2          7, 21, 35
4          14         5          1001, 2002, 3003
5          23         9          490314, 817190, 1144066
6          34         14         927983760, 1391975640, 1855967520

etc.

Edited on July 14, 2011, 3:17 pm

Posted by Harry on 2011-07-14 15:14:17