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A178301
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Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.
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4
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1, 1, 3, 1, 8, 10, 1, 15, 45, 35, 1, 24, 126, 224, 126, 1, 35, 280, 840, 1050, 462, 1, 48, 540, 2400, 4950, 4752, 1716, 1, 63, 945, 5775, 17325, 27027, 21021, 6435, 1, 80, 1540, 12320, 50050, 112112, 140140, 91520, 24310, 1, 99, 2376, 24024, 126126, 378378, 672672, 700128, 393822, 92378
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OFFSET
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0,3
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COMMENTS
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The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial binomial(x+n,n)*binomial(x+n,n-1) in the basis made of the binomial(x+i,i). - F. Chapoton, Nov 01 2022
Chapoton's observation above is correct: the precise expansion is binomial(x+n,n)*binomial(x+n,n-1) = Sum_{k = 0..n-1} (-1)^k*T(n-1,n-1-k)*binomial(x+2*n-1-k,2*n-1-k), as can be verified using the WZ algorithm. For example, n = 4 gives binomial(x+4,4)*binomial(x+4,3) = 35*binomial(x+7,7) - 45*binomial(x+6,6) + 15*binomial(x+5,5) - binomial(x+4,4). - Peter Bala, Jun 24 2023
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LINKS
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FORMULA
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n-th row polynomial R(n,x) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k+1,n+1)*x^k = Sum_{k = 0..n} (-1)^(n+k)*binomial(n+1,k+1)*binomial(n+k+1,n+1)*(1 + x)^k.
Recurrence: (2*n - 1)*(n + 1)*R(n,x) = 2*(4*n^2*x + 2*n^2 - x - 1)*R(n-1,x) - (2*n + 1)(n - 1)*R(n-2,x) with R(0,x) = 1, R(1,x) = 1 + 3*x.
A182626(n) = -R(n-1,-2) for n >= 1. (End)
n-th row polynomial R(n,x) = Jacobi_P(n,0,1,2*x + 1).
(1 + x)*R(n,x) gives the row polynomials of A123160.
(End)
G.f.: (1+x-sqrt(1-2*x+x^2-4*x*y))/(2*(1+y)*x*sqrt(1-2*x+x^2-4*x*y)). - Emanuele Munarini, Dec 16 2016
R(n,x) = Sum_{k=0..n} (-1)^(n+k)*(2*k+1)*P(k,2*x+1)/(n+1), where P(k,x) is the k-th Legendre polynomial (cf. A100258) and P(k,2*x+1) is the k-th shifted Legendre polynomial (cf. A063007). - Max Alekseyev, Jun 28 2018; corrected by Peter Bala, Aug 08 2021
Polynomial g(n,x) = R(n,-x)/(n+1) delivers the maximum of f(1)^2/(Integral_{x=0..1} f(x)^2 dx) over all polynomials f(x) with real coefficients and deg(f(x)) <= n. This maximum equals (n+1)^2. See dxdy.ru link. - Max Alekseyev, Jun 28 2018
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EXAMPLE
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n=0: 1;
n=1: 1, 3;
n=2: 1, 8, 10;
n=3: 1, 15, 45, 35;
n=4: 1, 24, 126, 224, 126;
n=5: 1, 35, 280, 840, 1050, 462;
n=6: 1, 48, 540, 2400, 4950, 4752, 1716;
n=7: 1, 63, 945, 5775, 17325, 27027, 21021, 6435;
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MAPLE
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binomial(n, k)*binomial(n+k+1, n+1) ;
R := proc(n) add((-1)^(n+k)*(2*k+1)*orthopoly:-P(k, 2*x+1)/(n+1), k=0..n) end:
for n from 0 to 6 do seq(coeff(R(n), x, k), k=0..n) od; # Peter Luschny, Aug 25 2021
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MATHEMATICA
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Flatten[Table[Binomial[n, k]Binomial[n+k+1, n+1], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Aug 23 2014 *)
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PROG
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(Maxima) create_list(binomial(n, k)*binomial(n+k+1, n+1), n, 0, 12, k, 0, n); Emanuele Munarini, Dec 16 2016
(PARI) R(n, x) = sum(k=0, n, (-1)^(n+k) * (2*k+1) * pollegendre(k, 2*x+1)) / (n+1); \\ Max Alekseyev, Aug 25 2021
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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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