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The convolution triangle of the nonnegative integers. - Peter Luschny, Oct 07 2022

# Uses function PMatrix from A357368. Adds a row and column above and to the left.

PMatrix(10, n -> n); # Peter Luschny, Oct 07 2022

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The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598. (End)

For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)

G.f.: x*y / (1 - (2 + y)*x + x^2). To get row n, expand this in powers of x then expand the coefficient of x^n in increasing powers of y.

From Philippe Deléham, Feb 16 2004: (Start)

If indexing begins at 0 we have

T(n,k) = (n+k+1)!/((n-k)!*(2k+1))!.

T(n,k) = Sum_{j>=0} T(n-1-j, k-1)*(j+1) with T(n, 0) = n+1, T(n, k) = 0 if n < k.

If indexing begins at 0 we have T(n, k) = (n+k+1)!/((n-k)!*(2k+1))!. T(n, k) = Sum_{j>=0} T(n-1-j, k-1)*(j+1) with T(n, 0) = n+1, T(n, k) = 0 if n < k. T(n, k) = T(n-1, k-1) + T(n-1, k) + Sum_{j>=0} (-1)^j*T(n-1, k+j)*A000108(j) with T(n, k) = 0 if k < 0, T(0, 0)=1 and T(0, k) = 0 for k>0. G.f. for the column k: Sum_{n>=0} T(n, k)*x^n = (x^k)/(1-x)^(2k+2). Row sums: Sum_{k >= 0} T(n, k) = A001906(n+1). - _Philippe Deléham_, Feb 16 2004

G.f. for the column k: Sum_{n>=0} T(n, k)*x^n = (x^k)/(1-x)^(2k+2).

Row sums: Sum_{k>=0} T(n, k) = A001906(n+1). (End)

For the transition matrix R (T with offset [0,0]) defined in a comment above, row n=2: F(12*m)/L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. - Wolfdieter Lang, Aug 30 2012

T(n,m) = Sum_{k=0..n-m} (binomial(2*k,n-m)*binomial(m+k,k)*(-1)^(n-m+k)* binomial(n+1,m+k+1)). - Vladimir Kruchinin, Apr 13 2016

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Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.

For the transition matrix R (T with offset [0,0]) defined in a comment above, row n=2: F(12*m)/L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. - Wolfdieter Lang, Aug 30 2012

For the transition matrix R (T with offset [0,0]) defined in a comment above, row n=2: F(12*m)/L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. - Wolfdieter Lang, Aug 30 2012

For another version see A128908. - Philippe Deléham, Mar 27 2012

From Philippe Deléham, Mar 27 2012: (Start)

(0, 2, -1/2, 1/2, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, ...) begins:

1

0, 1

0, 2, 1

0, 3, 4, 1

0, 4, 10, 6, 1

0, 5, 20, 21, 8, 1

0, 6, 35, 56, 36, 10, 1

0, 7, 56, 126, 120, 55, 12, 1

0, 8, 84, 252, 330, 220, 78, 14, 1. (End)

Row sums give A001906. With signs: A053122.

Cf. A128908, A053123, A049310. Row sums give A001906, A119900, A085478, A029653, A106195.

With signs: A053122.

Cf. A119900. - Philippe Deléham, Dec 02 2008

Cf. A085478.

Cf. A029653, A106195.

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