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COMMENT
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Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers sum{k=0..n, (-1)^(n+k)n!/k!}. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry (pbarry(AT)wit.ie), Mar 27 2007
Comments from Tom Copeland (tcjpn(AT)msn.com), Nov 01 2007: (Start) T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = sum(j=0,...,n) Binom(n,j) * j! * x^(n-j) = sum(j=0,...,n) (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = sum(j=0,...,n) (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = sum(j=0,1,...) (xDx)^j A(x)
6) B(x) = sum(j=0,1,...) x^j * D^j * x^j A(x)
7) B(x) = sum(j=0,1,...) j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
Comments from Peter Bala (pbala(AT)toucansurf.com), Jul 10 2008 (Start): This array is the particular case P(1,1) of the generalised Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b)+3a*(a+b)*x+3a*x^2+x^3)*y^3/3!+ ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analogue of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >=2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3;...) [From Gary W. Adamson (qntmpkt(AT)yahoo.com), May 03 2009]
Contribution from Peter Luschny (peter(AT)luschny.de), Jun 01 2009: (Start)
If the first column is deleted and the triangle read from right to left resulting in
1|1,2|1,3,6|1,4,12,24|1,5,20,60,120|...,
then this triangle T'(m,k) (m>=0,m>=k>=0) has the definition
T'(m,k) = (-1)^k prod_{j=0..k-1} (j-m-1)
for n from -1 to 7 do print(seq(T'(n,n-k),k=-1..n)) od:
(Let G(m,k,p) = (-p)^k prod_{j=0..k-1} (j-m-1/p).
For G(m,k,2) see A112292 and for G(m,k,3) see A136214.) (End)
Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Oct 07 2009: (Start)
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - .. ) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n = >1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n).
(End)
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