0
$\begingroup$

I have come across a tree of numbers that represents a probability distribution and was wondering if anybody can identify it.

$$ 1 \\ 1\;\;\; 1 \\ 1\;\;\; 3\;\;\; 1 \\ 1\;\;\; 6\;\;\; 7\;\;\; 1 \\ 1\;\;\; 10\;\; 25\;\; 15\;\;\; 1\\ 1\;\;\; 15\;\;\; 65\;\;\; 90 \;\;\; 31\;\;\; 1\\ 1\;\;\; 21\;\;\; 140\;\;\; 350\;\;\; 301\;\;\; 63\;\;\; 1 \\ \vdots $$

The distribution appears skewed to the right looking at the numbers. The sequence for the left and right second diagonals are simple (i.e. left 2nd diagonal is +n-1 where n is row number starting with unity).

If it at all helps, this sequence comes from the differential operator that I have been studying

$$ T_n(x) = \left(x\partial_x \right)^n $$

by expanding out the operator.

  • 0
    Sum the numbers horizontally & search OEIS http://oeis.org/search?q=1%2C2%2C5%2C15%2C52&sort=&language=english&go=Search ... They are the Bell numbers ... your grid is a known grading of them ... – 2017-02-28

1 Answers 1

1

Check out the "Online Encyclopedia of Integer Sequences" (oeis.org). You can type in your numbers, and it will try to match it up.

This triangle gives the Stirling numbers of 2nd kind, $S(n,n-k+1)$.