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A093375
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Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.
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7
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1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 18, 8, 1, 6, 25, 40, 30, 10, 1, 7, 36, 75, 80, 45, 12, 1, 8, 49, 126, 175, 140, 63, 14, 1, 9, 64, 196, 336, 350, 224, 84, 16, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 11, 100, 405, 960, 1470, 1512, 1050, 480, 135, 20, 1, 12
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OFFSET
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1,2
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COMMENTS
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Number of n-long m-ary words avoiding the pattern 1-1'2'.
T(n,n+1) = Sum_{i=1..n} T(n,i).
Exponential Riordan array [(1+x)e^x, x] as a number triangle. - Paul Barry, Feb 17 2009
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A059298. (End)
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LINKS
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FORMULA
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Triangle = P*M, the binomial transform of the infinite bidiagonal matrix M with (1,1,1,...) in the main diagonal and (1,2,3,...) in the subdiagonal, and zeros elsewhere. P = Pascal's triangle as an infinite lower triangular matrix. - Gary W. Adamson, Nov 05 2006
E.g.f. for triangle: (1 + z)*exp((1 + x)*z) = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2/2! + ....
O.g.f. for triangle: (1 - x*z)/(1 - z - x*z)^2 = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2 + ....
The n-th row polynomial R(n,x) of the triangle equals (1+x)^n + n*(1+x)^(n-1) for n >= 0 and satisfies d/dx(R(n,x)) = n*R(n-1,x), as well as R(n,x+y) = Sum_{k = 0..n} binomial(n,k)*R(k,x)*y^(n-k). The row polynomials are a Sheffer sequence of Appell type.
Matrix inverse of the triangle is a signed version of A073107. (End)
With offset 0 and D = d/dx, the raising operator for the signed row polynomials P(n,x) is RP = x - d{log[e^D/(1-D)]}/dD = x - 1 - 1/(1-D) = x - 2 - D - D^2 + ..., i.e., RP P(n,x) = P(n+1,x).
The e.g.f. for the signed array is (1-t) * e^(-t) * e^(x*t).
From the Appell formalism, the row polynomials PI(n,x) of A073107 are the umbral inverse of this entry's row polynomials; that is, P(n,PI(.,x)) = x^n = PI(n,P(.,x)) under umbral composition. (End)
As a triangle, we let S(n,k) = T(n-k+1, k+1) = (n-k+1)*binomial(n, k) for n >= 0 and 0 <= k <= n). See the example below.
As stated above by Peter Bala, Sum_{n,k >= 0} S(n,k)*z^n*x^k = (1 - x*z)/(1 - z -x*z)^2.
Also, Sum_{n, k >= 0} S(n,k)*z^n*x^k/n! = (1+z)*exp((1+x)*z).
As he also states, the n-th row polynomial is R(n,x) = Sum_{k = 0..n} S(n, k)*x^k = (1 + x)^n + n*(1 + x)^(n-1).
If we define the signed triangle S*(n,k) = (-1)^(n+k) * S(n,k) = (-1)^(n+k) * T(n-k+1, k+1), as Tom Copeland states, Sum_{n,k >= 0} S^*(n,k)*t^n*x^k/n! = (1-t)*exp((1-x)*(-t)) = (1-t) * e^(-t) * e^(x*t).
Apparently, S*(n,k) = A103283(n,k).
As he says above, the signed n-th row polynomial is P(n,x) = (-1)^n*R(n,-x) = (x - 1)^n - n*(x - 1)^(n-1).
According to Gary W. Adamson, P(n,x) is "the monic characteristic polynomial of the n X n matrix with 2's on the diagonal and 1's elsewhere." (End)
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EXAMPLE
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Array T(m,n) (with rows m >= 1 and columns n >= 1) begins as follows:
1 1 1 1 1 1 ...
2 4 6 8 10 12 ...
3 9 18 30 45 63 ...
4 16 40 80 140 224 ...
5 25 75 175 350 630 ...
...
Triangle S(n,k) = T(n-k+1, k+1) begins
.n\k.|....0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....2....1
..2..|....3....4....1
..3..|....4....9....6....1
..4..|....5...16...18....8....1
..5..|....6...25...40...30...10....1
..6..|....7...36...75...80...45...12....1
...
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MATHEMATICA
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nmax = 10;
T = Transpose[CoefficientList[# + O[z]^(nmax+1), z]& /@ CoefficientList[(1 - x z)/(1 - z - x z)^2 + O[x]^(nmax+1), x]];
row[n_] := T[[n+1, 1 ;; n+1]];
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PROG
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(GAP) nmax:=14;; T:=List([1..nmax], n->List([1..nmax], k->k*Binomial(n+k-2, n-1)));;
b:=List([2..nmax], n->OrderedPartitions(n, 2));;
a:=Flat(List([1..Length(b)], i->List([1..Length(b[i])], j->T[b[i][j][1]][b[i][j][2]]))); # Muniru A Asiru, Aug 07 2018
(Sage) # uses[riordan_array from A256893]
riordan_array((1+x)*exp(x), x, 8, exp=true) # Peter Luschny, Nov 02 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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