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A062344
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Triangle of binomial(2*n, k) with n >= k.
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7
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1, 1, 2, 1, 4, 6, 1, 6, 15, 20, 1, 8, 28, 56, 70, 1, 10, 45, 120, 210, 252, 1, 12, 66, 220, 495, 792, 924, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 48620
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OFFSET
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0,3
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COMMENTS
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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)
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LINKS
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FORMULA
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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
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EXAMPLE
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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,
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MATHEMATICA
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Flatten[Table[Binomial[2 n, k], {n, 0, 20}, {k, 0, n}]] (* G. C. Greubel, Jun 28 2018 *)
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PROG
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(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
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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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