OFFSET
0,5
COMMENTS
The e.g.f. has the form A(t)*exp(x*B(t)), where A(t) = 1 - t and B(t) = log(1 + t). Thus the row polynomials of this triangle form a Sheffer sequence for the pair (2 - exp(t), exp(t) - 1) (see Roman, p. 17).
Let x_(k) := x*(x-1)*...*(x-k+1) denote the k-th falling factorial polynomial. Define a sequence x_[n] of basis polynomials for the polynomial algebra C[x] by setting x_[0] = 1, and setting x_[n] = x_(n-1)*(x - 2*n + 1) for n >= 1. The sequence begins [1, x-1, x*(x-3), x*(x-1)*(x-5), x*(x-1)*(x-2)*(x-7), ...]. Then this is the triangle of connection constants for expressing the basis polynomials x_[n] as a linear combination of the monomial polynomials x^k, that is, x_[n] = Sum_{k = 0..n} T(n,k) x^k. An example is given below.
REFERENCES
S. Roman, The umbral calculus, Pure and Applied Mathematics 111, Academic Press Inc., New York, 1984. Reprinted by Dover in 2005.
LINKS
Eric Weisstein's World of Mathematics, Sheffer Sequence
FORMULA
T(n,k) = Stirling1(n,k) - n*Stirling1(n-1,k).
E.g.f.: (1 - t)*(1 + t)^x = 1 + (-1 + x)*t + (-3*x + x^2)*t^2/2! + (5*x - 6*x^2 + x^3)*t^3/3! + ....
E.g.f. for column k: (1/k!)*(1 - t)*(log(1 + t))^k.
The row polynomials R(n,x) satisfy the Sheffer identity R(n,x + y) = Sum_{k = 0..n} binomial(n,k)*y_(k)*R(n-k,x), where y_(k) is the falling factorial. As a particular case we have the identity R(n,x + 1) - R(n,x) = n*R(n-1,x) for n >= 1.
EXAMPLE
Triangle begins:
n\k| 0 1 2 3 4 5 6 7 8 9
===|==================================================================
0 | 1
1 | -1, 1;
2 | 0, -3, 1;
3 | 0, 5, -6, 1;
4 | 0, -14, 23, -10, 1;
5 | 0, 54, -105, 65, -15, 1;
6 | 0, -264, 574, -435, 145, -21, 1;
7 | 0, 1560, -3682, 3199, -1330, 280, -28, 1;
8 | 0, -10800, 27180, -26124, 12649, -3360, 490, -36, 1;
9 | 0, 85680, -227196, 236312, -128205, 40089, -7434, 798, -45, 1;
...
Connection constants. Row 4 = [0, -14 ,23, -10, 1]:
-14*x + 23*x^2 - 10*x^3 + x^4 = x*(x-1)*(x-2)*(x-7) = x_[4].
CROSSREFS
KEYWORD
AUTHOR
Peter Bala, Jul 11 2013
STATUS
approved