OFFSET
0,3
COMMENTS
From Wolfdieter Lang, Sep 19 2012: (Start)
The triangle a(n,k) appears in the formula F(2*l+1)^(2*n) = (sum(a(n,k)*L(2*(n-k)*(2*l+1)),k=0..n-1) + a(n,n))/5^n, n>=0, l>=0, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
The signed triangle as(n,k):=a(n,k)*(-1)^k appears in the formula F(2*l)^(2*n) = (sum(as(n,k)*L(4*(n-k)*l),k=0..n-1) + as(n,n))/5^n, n>=0, l>=0. Proof with the Binet-de Moivre formula for F and L and the binomial formula. (End)
LINKS
G. C. Greubel, Rows n = 0..100 of triangle, flattened
E. H. M. Brietzke, An identity of Andrews and a new method for the Riordan array proof of combinatorial identities, Discrete Math., 308 (2008), 4246-4262.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
FORMULA
a(n,k) = a(n,k-1)*((2n+1)/k-1) with a(n,0)=1.
G.f.: 1/((1-sqrt(1-4*x*y))^4/(16*x*y^2) + sqrt(1-4*x*y) - x). - Vladimir Kruchinin, Jan 26 2021
EXAMPLE
Rows start
(1),
(1,2),
(1,4,6),
(1,6,15,20)
etc.
Row n=2, (1,4,6):
F(2*l+1)^4 = (1*L(4*(2*l+1)) + 4*L(2*(2*l+1)) + 6)/25,
F(2*l)^4 = (1*L(8*l) - 4*L(4*l) + 6)/25, l>=0, F=A000045, L=A000032. See a comment above. - Wolfdieter Lang, Sep 19 2012
MATHEMATICA
Flatten[Table[Binomial[2 n, k], {n, 0, 20}, {k, 0, n}]] (* G. C. Greubel, Jun 28 2018 *)
PROG
(Maxima) create_list(binomial(2*n, k), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */
(PARI) for(n=0, 20, for(k=0, n, print1(binomial(2*n, k), ", "))) \\ G. C. Greubel, Jun 28 2018
(Magma) [[Binomial(2*n, k): k in [0..n]]: n in [0..20]]; // G. C. Greubel, Jun 28 2018
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Henry Bottomley, Jul 06 2001
STATUS
approved