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 A088956 Triangle, read by rows, of coefficients of the hyperbinomial transform. 29

%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) = (n-k+1)^(n-k-1)*C(n,k).

%C Given a table in which the n-th row is the n-th 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 = x-x/( LambertW(-x)*(1+LambertW(-x)) ), e.g.f.: A055541(x) = x-x*LambertW(-x).

%C The m-th iteration of the hyperbinomial transform is given by the triangle of coefficients defined by T_m(n,k) = m*(n-k+m)^(n-k-1)*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*(n-k+m)^(n-k-1)*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*(n-k+m)^(n-k-1)*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) / (n-k+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.helms-net.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) = (n-k+1)^(n-k-1)*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^(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.

%F 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.

%F 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].

%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)^(n-1)*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)*(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^^(...).

%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,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

%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^(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 *)

%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

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Last modified March 29 21:32 EDT 2020. Contains 333117 sequences. (Running on oeis4.)