%I
%S 1,1,1,3,2,1,16,9,3,1,125,64,18,4,1,1296,625,160,30,5,1,16807,7776,
%T 1875,320,45,6,1,262144,117649,27216,4375,560,63,7,1,4782969,2097152,
%U 470596,72576,8750,896,84,8,1,100000000,43046721,9437184,1411788,163296,15750
%N Triangle, read by rows, of coefficients of the hyperbinomial transform.
%C The hyperbinomial transform of a sequence {b} is defined to be the sequence {d} given by d(n) = Sum_{k=0..n} T(n,k)*b(k)), where T(n,k) = (nk+1)^(nk1)*C(n,k).
%C Given a table in which the nth row is the nth binomial transform of the first row, then the hyperbinomial transform of any diagonal results in the next lower diagonal in the table.
%C The simplest example of a table of iterated binomial transforms is A009998, with a main diagonal of {1,2,9,64,625,...}; and the hyperbinomial transform of this diagonal gives the next lower diagonal, {1,3,16,125,1296,...}, since 1=(1)*1, 3=(1)*1+(1)*2, 16=(3)*1+(2)*2+(1)*9, 125=(16)*1+(9)*2+(3)*9+(1)*64, etc.
%C Another example: the hyperbinomial transform maps A065440 into A055541, since HYPERBINOMIAL([1,1,1,8,81,1024,15625]) = [1,2,6,36,320,3750,54432] where e.g.f.: A065440(x)+x = xx/( LambertW(x)*(1+LambertW(x)) ), e.g.f.: A055541(x) = xx*LambertW(x).
%C The mth iteration of the hyperbinomial transform is given by the triangle of coefficients defined by T_m(n,k) = m*(nk+m)^(nk1)*binomial(n,k).
%C Example: PARI code for T_m: {a=[1,1,1,8,81,1024,15625]; m=1; b=vector(length(a)); for(n=0,length(a)1, b[n+1]=sum(k=0,n, m*(nk+m)^(nk1)*binomial(n,k)*a[k+1]); print1(b[n+1],","))} RETURNS b=[1,2,6,36,320,3750,54432].
%C The INVERSE hyperbinomial transform is thus given by m=1: {a=[1,2,6,36,320,3750,54432]; m=1; b=vector(length(a)); for(n=0,length(a)1, b[n+1]=sum(k=0,n, m*(nk+m)^(nk1)*binomial(n,k)*a[k+1]); print1(b[n+1],","))} RETURNS b=[1,1,1,8,81,1024,15625].
%C Simply stated, the HYPERBINOMIAL transform is to LambertW(x)/x as the BINOMIAL transform is to exp(x).
%C Let A[n] be the set of all forests of labeled rooted trees on n nodes. Build a superset B[n] of A[n] by designating "some" (possibly all or none) of the isolated nodes in each forest. T(n,k) is the number of elements in B[n] with exactly k designated nodes. See A219034.  _Geoffrey Critzer_, Nov 10 2012
%C T(n,k) = A095890(n+1,k+1) * a007318(n,k) / (nk+1), 0 <= k <= n.  _Reinhard Zumkeller_, Jul 07 2013
%H T. D. Noe, <a href="/A088956/b088956.txt">Rows n=0..50 of triangle, flattened</a>
%H G. Helms, <a href="http://go.helmsnet.de/math/tetdocs/PascalMatrixTetrated.pdf">Pascalmatrix tetrated</a>
%H E. W. Weisstein, <a href="http://mathworld.wolfram.com/AbelPolynomial.html">Abel Polynomial</a>, From MathWorld  A Wolfram Web Resource.
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F T(n, k) = (nk+1)^(nk1)*C(n, k).
%F E.g.f.: LambertW(x)*exp(x*y)/x.  _Vladeta Jovovic_, Oct 27 2003
%F From Peter Bala, Sep 11 2012: (Start)
%F Let T(x) = Sum_{n >= 0} n^(n1)*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.
%F The matrix power (A088956)^r has the e.g.f. exp(r*T(x))*exp(t*x) with triangle entries given by r*(nk+r)^(nk1)*binomial(n,k) for n and k >= 0. See A215534 for the case r = 1.
%F Let A(n,x) = x*(x+n)^(n1) 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].
%F 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.
%F With T(x) the tree function as above, define E(x) = T(x)/x. Then A088956 = E(S) = sum {n>=0} (n+1)^(n1)*S^n/n!.
%F 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)*(A1)^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^^(...).
%F A089956 augmented with the sequence (x,x,x,...) on the first superdiagonal is the production matrix for the row polynomials of A105599.
%F (End)
%F Sum_{k = 0..n} T(n,nk)*(x  k  1)^(nk) = x^n. Setting x = n + 1 gives Sum_{k = 0..n} T(n,k)*k^k = (n + 1)^n.  _Peter Bala_, Feb 17 2017
%e Rows begin:
%e {1},
%e {1, 1},
%e {3, 2, 1},
%e {16, 9, 3, 1},
%e {125, 64, 18, 4, 1},
%e {1296, 625, 160, 30, 5, 1},
%e {16807, 7776, 1875, 320, 45, 6, 1},
%e {262144, 117649, 27216, 4375, 560, 63, 7, 1}, ...
%t nn=8; t=Sum[n^(n1)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 *)
%o (Haskell)
%o a088956 n k = a095890 (n + 1) (k + 1) * a007318' n k `div` (n  k + 1)
%o a088956_row n = map (a088956 n) [0..n]
%o a088956_tabl = map a088956_row [0..]
%o  _Reinhard Zumkeller_, Jul 07 2013
%Y Cf. A088957 (row sums), A000272 (first column), A009998, A105599, A132440, A215534 (matrix inverse), A215652.
%Y Cf. A227325 (central terms).
%K nonn,tabl,nice
%O 0,4
%A _Paul D. Hanna_, Oct 26 2003
