A088956 Triangle, read by rows, of coefficients of the hyperbinomial transform.
1, 1, 1, 3, 2, 1, 16, 9, 3, 1, 125, 64, 18, 4, 1, 1296, 625, 160, 30, 5, 1, 16807, 7776, 1875, 320, 45, 6, 1, 262144, 117649, 27216, 4375, 560, 63, 7, 1, 4782969, 2097152, 470596, 72576, 8750, 896, 84, 8, 1, 100000000, 43046721, 9437184, 1411788, 163296, 15750
Offset: 0
Examples
Rows begin:
{1},
{1, 1},
{3, 2, 1},
{16, 9, 3, 1},
{125, 64, 18, 4, 1},
{1296, 625, 160, 30, 5, 1},
{16807, 7776, 1875, 320, 45, 6, 1},
{262144, 117649, 27216, 4375, 560, 63, 7, 1}, ...
Links
- T. D. Noe, Rows n=0..50 of triangle, flattened
- T. Copeland, Composition, Conjugation, and the Umbral Calculus-Part I, 2021.
- G. Helms, Pascalmatrix tetrated
- E. W. Weisstein, Abel Polynomial, From MathWorld - A Wolfram Web Resource.
- Index entries for triangles and arrays related to Pascal's triangle
Crossrefs
Programs
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Haskell
a088956 n k = a095890 (n + 1) (k + 1) * a007318' n k `div` (n - k + 1) a088956_row n = map (a088956 n) [0..n] a088956_tabl = map a088956_row [0..] -- Reinhard Zumkeller, Jul 07 2013
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Mathematica
nn=8; t=Sum[n^(n-1)x^n/n!, {n,1,nn}]; Range[0,nn]! CoefficientList[Series[Exp[t+y x] ,{x,0,nn}], {x,y}] //Grid (* Geoffrey Critzer, Nov 10 2012 *)
Formula
T(n, k) = (n-k+1)^(n-k-1)*C(n, k).
E.g.f.: -LambertW(-x)*exp(x*y)/x. - Vladeta Jovovic, Oct 27 2003
From Peter Bala, Sep 11 2012: (Start)
Let T(x) = Sum_{n >= 0} n^(n-1)*x^n/n! denote the tree function of A000169. The e.g.f. is (T(x)/x)*exp(t*x) = exp(T(x))*exp(t*x) = 1 + (1 + t)*x + (3 + 2*t + t^2)*x^2/2! + .... Hence the triangle is the exponential Riordan array [T(x)/x,x] belonging to the exponential Appell group.
The matrix power (A088956)^r has the e.g.f. exp(r*T(x))*exp(t*x) with triangle entries given by r*(n-k+r)^(n-k-1)*binomial(n,k) for n and k >= 0. See A215534 for the case r = -1.
Let A(n,x) = x*(x+n)^(n-1) be an Abel polynomial. The present triangle is the triangle of connection constants expressing A(n,x+1) as a linear combination of the basis polynomials A(k,x), 0 <= k <= n. For example, A(4,x+1) = 125*A(0,x) + 64*A(1,x) + 18*A(2,x) + 4*A(3,x) + A(4,x) gives row 4 as [125,64,18,4,1].
Let S be the array with the sequence [1,2,3,...] on the main subdiagonal and zeros elsewhere. S is the infinitesimal generator for Pascal's triangle (see A132440). Then the infinitesimal generator for this triangle is S*A088956; that is, A088956 = Exp(S*A088956), where Exp is the matrix exponential.
With T(x) the tree function as above, define E(x) = T(x)/x. Then A088956 = E(S) = Sum_{n>=0} (n+1)^(n-1)*S^n/n!.
For commuting lower unit triangular matrices A and B, we define A raised to the matrix power B, denoted A^^B, to be the matrix Exp(B*log(A)), where the matrix logarithm Log(A) is defined as Sum_{n >= 1} (-1)^(n+1)*(A-1)^n/n. Let P denote Pascal's triangle A007318. Then the present triangle, call it X, solves the matrix equation P^^X = X . See A215652 for the solution to X^^P = P. Furthermore, if we denote the inverse of X by Y then X^^Y = P. As an infinite tower of matrix powers, A088956 = P^^(P^^(P^^(...))).
A088956 augmented with the sequence (x,x,x,...) on the first superdiagonal is the production matrix for the row polynomials of A105599.
(End)
Sum_{k = 0..n} T(n,n-k)*(x - k - 1)^(n-k) = x^n. Setting x = n + 1 gives Sum_{k = 0..n} T(n,k)*k^k = (n + 1)^n. - Peter Bala, Feb 17 2017
As lower triangular matrices, this entry, T, equals unsigned A137542 * A007318 * signed A059297. The Pascal matrix is sandwiched between a pair of inverse matrices, so this entry is conjugate to the Pascal matrix, allowing convergent analytic expressions of T, say f(T), to be computed as f(A007318) sandwiched between the inverse pair. - Tom Copeland, Dec 06 2021
Comments