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A060922
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Convolution triangle for Lucas numbers A000032(n+1), n >= 0.
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11
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1, 3, 1, 4, 6, 1, 7, 17, 9, 1, 11, 38, 39, 12, 1, 18, 80, 120, 70, 15, 1, 29, 158, 315, 280, 110, 18, 1, 47, 303, 753, 905, 545, 159, 21, 1, 76, 566, 1687, 2568, 2120, 942, 217, 24, 1, 123, 1039, 3612, 6666, 7043, 4311
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OFFSET
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0,2
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COMMENTS
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In the language of Shapiro et al. (see A053121 for the reference) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. G.f. for row polynomials p(n,x) := sum(a(n,m)*x^m,m=0..n) is (1+2*z)/(1-(1+x)*z-(1+2*x)*z^2).
Bisection of this triangle gives triangles A060923 (even part) and A060924 (odd part).
For the m-th column sequence (without leading zeros) one has: a(n+m,m)= (pL1(m,n)*L(n+2)+pL2(m,n)*L(n+1))/(m!*5^m), m >= 0, with the Lucas numbers L(n)=A000032(n), n >= 0 and the row polynomials pL1(n,x) := sum(A061188(n,m)*x^n,m=0..n) and pL2(n,x) := sum(A061189(n,m)*x^m,m=0..n).
Riordan array ((1+2*x)/(1-x-x^2), x*(1+2*x)/(1-x-x^2)). - Philippe Deléham, Jan 21 2014
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LINKS
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FORMULA
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a(n, m)=((n-m+1)*a(n, m-1)+2*(2*n-m)*a(n-1, m-1)+4*(n-1)*a(n-2, m-1))/(5*m), n >= m >= 1, a(n, 0)= A000204(n+1)= A000032(n+1).
G.f. for m-th column: ((1+2*x)/(1-x-x^2))* ((x*(1+2*x))/(1-x-x^2))^m.
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) + T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 21 2014
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EXAMPLE
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p(2,x) = 4+6*x+x^2.
Triangle begins:
1 ;
3, 1;
4, 6, 1;
7, 17, 9, 1;
11, 38, 39, 12, 1;
18, 80, 120, 70, 15, 1;
29, 158, 315, 280, 110, 18, 1;
47, 303, 753, 905, 545, 159, 21, 1;
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MAPLE
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# Uses function PMatrix from A357368. Adds column 1, 0, 0, 0, ... to the left.
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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