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A193091
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Augmentation of the triangular array A158405. See Comments.
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26
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1, 1, 3, 1, 6, 14, 1, 9, 37, 79, 1, 12, 69, 242, 494, 1, 15, 110, 516, 1658, 3294, 1, 18, 160, 928, 3870, 11764, 22952, 1, 21, 219, 1505, 7589, 29307, 85741, 165127, 1, 24, 287, 2274, 13355, 61332, 224357, 638250, 1217270, 1, 27, 364, 3262, 21789, 115003
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OFFSET
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0,3
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COMMENTS
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Suppose that P is an infinite triangular array of numbers:
p(0,0)
p(1,0)...p(1,1)
p(2,0)...p(2,1)...p(2,2)
p(3,0)...p(3,1)...p(3,2)...p(3,3)...
...
Let w(0,0)=1, w(1,0)=p(1,0), w(1,1)=p(1,1), and define
W(n)=(w(n,0), w(n,1), w(n,2),...w(n,n-1), w(n,n)) recursively by W(n)=W(n-1)*PP(n), where PP(n) is the n X (n+1) matrix given by
...
row 0 ... p(n,0) ... p(n,1) ...... p(n,n-1) ... p(n,n)
row 1 ... 0 ..... p(n-1,0) ..... p(n-1,n-2) .. p(n-1,n-1)
row 2 ... 0 ..... 0 ............ p(n-2,n-3) .. p(n-2,n-2)
...
row n-1 . 0 ..... 0 ............. p(2,1) ..... p(2,2)
row n ... 0 ..... 0 ............. p(1,0) ..... p(1,1)
...
The augmentation of P is here introduced as the triangular array whose n-th row is W(n), for n>=0. The array P may be represented as a sequence of polynomials; viz., row n is then the vector of coefficients: p(n,0), p(n,1),...,p(n,n), from p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n). For example, (C(n,k)) is represented by ((x+1)^n); using this choice of P (that is, Pascal's triangle), the augmentation of P is calculated one row at a time, either by the above matrix products or by polynomial substitutions in the following manner:
...
row 0 of W: 1, by decree
row 1 of W: 1 augments to 1,1
...polynomial version: 1 -> x+1
row 2 of W: 1,1 augments to 1,3,2
...polynomial version: x+1 -> (x^2+2x+1)+(x+1)=x^2+3x+2
row 3 to W: 1,3,2 augments to 1,6,11,6
...polynomial version:
x^2+3x+2 -> (x+1)^3+3(x+1)^2+2(x+1)=(x+1)(x+2)(x+3)
...
Examples of augmented triangular arrays:
(p(n,k)=1) augments to A009766, Catalan triangle.
Catalan triangle augments to A193560.
Pascal triangle augments to A094638, Stirling triangle.
...
This is the table of g(n,k) in the notation of Carlitz (p. 124). The triangle enumerates two-line arrays of positive integers
............a_1 a_2 ... a_n..........
............b_1 b_2 ... b_n..........
such that
1) max(a_i, b_i) <= min(a_(i+1), b_(i+1)) for 1 <= i <= n-1
2) max(a_i, b_i) <= i for 1 <= i <= n
3) max(a_n, b_n) = k.
(End)
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LINKS
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FORMULA
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T(n,k) = (n-k+1)/n*Sum_{i=0..k} C(n+1,n-k+i+1)*C(2*n+i+1,i) for 0 <= k <= n.
Recurrence equation: T(n,k) = Sum_{i=0..k} (2*k-2*i+1)*T(n-1,i).
(End)
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EXAMPLE
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The triangle P, at A158405, is given by rows
1
1...3
1...3...5
1...3...5...7
1...3...5...7...9...
The augmentation of P is the array W starts with w(0,0)=1, by definition of W. Successive polynomials (rows of W) arise from P as shown here:
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1->x+3, so that W has (row 1)=(1,3);
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x+3->(x^2+3x+5)+3*(x+3), so that W has (row 2)=(1,6,14);
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x^2+6x+14->(x^3+3x^2+5x+7)+6(x^2+3x+5)+14(x+3), so that (row 3)=(1,9,37,79).
...
First 7 rows of W:
1
1 3
1 6 14
1 9 37 79
1 12 69 242 494
1 15 110 516 1658 3294
1 18 160 928 3870 11764 22952
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MATHEMATICA
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p[n_, k_] := 2 k + 1
Table[p[n, k], {n, 0, 5}, {k, 0, n}] (* A158405 *)
m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
TableForm[m[4]]
w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
v[n_] := v[n - 1].m[n]
TableForm[Table[v[n], {n, 0, 6}]] (* A193091 *)
Flatten[Table[v[n], {n, 0, 9}]]
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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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