A227342 Expansion of (1 - t)*(1 + t)^x.

1, -1, 1, 0, -3, 1, 0, 5, -6, 1, 0, -14, 23, -10, 1, 0, 54, -105, 65, -15, 1, 0, -264, 574, -435, 145, -21, 1, 0, 1560, -3682, 3199, -1330, 280, -28, 1, 0, -10800, 27180, -26124, 12649, -3360, 490, -36, 1, 0, 85680, -227196, 236312, -128205, 40089, -7434, 798, -45, 1

Offset: 0

Author

Peter Bala, Jul 11 2013

Keywords

• sign
• tabl
• easy

Comments

Cf. A105793. Inverse array is A227343.

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.

Examples

			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].
		

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

Crossrefs

Cf. A048994, A105793, A227343 (matrix inverse).

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.