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%I #55 Jun 25 2023 09:39:05
%S 1,1,3,1,8,10,1,15,45,35,1,24,126,224,126,1,35,280,840,1050,462,1,48,
%T 540,2400,4950,4752,1716,1,63,945,5775,17325,27027,21021,6435,1,80,
%U 1540,12320,50050,112112,140140,91520,24310,1,99,2376,24024,126126,378378,672672,700128,393822,92378
%N Triangle T(n,k) = binomial(n,k)*binomial(n+k+1,n+1) read by rows, 0 <= k <= n.
%C Antidiagonal sums are given by A113682. - _Johannes W. Meijer_, Mar 24 2013
%C 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
%C 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
%H David A. Corneth, <a href="/A178301/b178301.txt">Table of n, a(n) for n = 0..10010</a>
%H Author?, <a href="https://dxdy.ru/topic12925.html">Norm of a continuous function</a>, dxdy.ru (in Russian)
%F T(n,k) = A007318(n,k) * A178300(n+1,k+1).
%F From _Peter Bala_, Jun 18 2015: (Start)
%F 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.
%F 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.
%F A182626(n) = -R(n-1,-2) for n >= 1. (End)
%F From _Peter Bala_, Jul 20 2015: (Start)
%F n-th row polynomial R(n,x) = Jacobi_P(n,0,1,2*x + 1).
%F (1 + x)*R(n,x) gives the row polynomials of A123160.
%F (End)
%F 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
%F 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
%F 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
%e n=0: 1;
%e n=1: 1, 3;
%e n=2: 1, 8, 10;
%e n=3: 1, 15, 45, 35;
%e n=4: 1, 24, 126, 224, 126;
%e n=5: 1, 35, 280, 840, 1050, 462;
%e n=6: 1, 48, 540, 2400, 4950, 4752, 1716;
%e n=7: 1, 63, 945, 5775, 17325, 27027, 21021, 6435;
%p A178301 := proc(n,k)
%p binomial(n,k)*binomial(n+k+1,n+1) ;
%p end proc: # _R. J. Mathar_, Mar 24 2013
%p R := proc(n) add((-1)^(n+k)*(2*k+1)*orthopoly:-P(k,2*x+1)/(n+1), k=0..n) end:
%p for n from 0 to 6 do seq(coeff(R(n), x, k), k=0..n) od; # _Peter Luschny_, Aug 25 2021
%t Flatten[Table[Binomial[n,k]Binomial[n+k+1,n+1],{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Aug 23 2014 *)
%o (Maxima) create_list(binomial(n,k)*binomial(n+k+1,n+1),n,0,12,k,0,n); _Emanuele Munarini_, Dec 16 2016
%o (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
%Y Cf. A007318, A047781 (row sums), A178300, A182626, A123160, A132813.
%K easy,nonn,tabl
%O 0,3
%A _Alford Arnold_, May 30 2010
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