A227343 Matrix inverse of triangle A227342.

1, 1, 1, 3, 3, 1, 13, 13, 6, 1, 75, 75, 37, 10, 1, 541, 541, 270, 85, 15, 1, 4683, 4683, 2341, 770, 170, 21, 1, 47293, 47293, 23646, 7861, 1890, 308, 28, 1, 545835, 545835, 272917, 90930, 22491, 4158, 518, 36, 1, 7087261, 7087261, 3543630, 1181125, 294525, 57351, 8400, 822, 45, 1

Offset: 0

Author

Peter Bala, Jul 11 2013

Keywords

• nonn
• easy
• tabl

Comments

The e.g.f. has the form A(t)*exp(x*B(t)), where A(t) = 1/(2 - exp(t)) and B(t) = exp(t) - 1. Thus the row polynomials of this triangle form a Sheffer sequence for the pair (1 - t, log(1 + t)) (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 monomial polynomials x^n as a linear combination of the basis 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
= = = = = = = = = = = = = = = = =
0 |   1
1 |   1    1
2 |   3    3    1
3 |  13   13    6    1
4 |  75   75   37   10    1
5 | 541  541  270   85   15    1
...
Connection constants. Row 4 = [75,75,37,10,1]: Thus
75 + 75*(x - 1) + 37*x*(x - 3) + 10*x*(x - 1)*(x - 5)+ 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

A000670 (columns 1 and 2), A048993, A059099 (row sums), A105794, A227342 (matrix inverse).

Programs

Mathematica

T[n_, k_] := n!/k! SeriesCoefficient[Series[1/(2 - Exp[t]) (Exp[t] - 1)^k, {t, 0, n}], n]
Flatten[Table[T[n, k], {n, 0, 12}, {k, 0, n}]]
U[n_, k_] := n!/k! SeriesCoefficient[Series[1/(1 - t^2) (t/Log[1 + t])^(n + 1), {t, 0, n - k}], n - k]
Flatten[Table[U[n, k], {n, 0, 8}, {k, 0, n}]] (* Emanuele Munarini, Dec 21 2016 *)

Formula

E.g.f.: 1/(2 - exp(t))*exp(x*(exp(t) - 1)) = 1 + (1 + x)*t + (3 + 3*x + x^2)*t^2/2! + (13 + 13*x + 6*x^2 + x^3)*t^3/3! + ....

Recurrence equation: T(n,0) = A000670(n), and for k >= 1, T(n,k) = 1/k*sum {i = 1..n} binomial(n,i)*T(n-i,k-1).

The row polynomials R(n,x) satisfy the Sheffer identity R(n,x + y) = sum {k = 0..n} binomial(n,k)*Bell(k,y)*R(n-k,x), where Bell(k,y) is the Bell or exponential polynomial (row polynomials of A048993).

The row polynomials also satisfy d/dx(R(n,x)) = sum {k = 0..n-1} binomial(n,k)*R(k,x).

Row sums A059099. Column 1 and column 2 = A000670. 1 + 2*column 3 = A000670 (apart from the first two terms).

From Emanuele Munarini, Dec 21 2016: (Start)

T(n,k) = (n!/k!)*[t^n](exp(t)-1)^k/(2-exp(t)).

T(n,k) = (n!/k!)*[t^(n-k)](t/log(1+t))^(n+1)/(1-t^2). (End)