cp's OEIS Frontend

A370258 Triangle read by rows: T(n, k) = binomial(n, k)*binomial(2*n+k, k), 0 <= k <= n.

%I A370258 #23 May 04 2024 09:23:33
%S A370258 1,1,3,1,10,15,1,21,84,84,1,36,270,660,495,1,55,660,2860,5005,3003,1,
%T A370258 78,1365,9100,27300,37128,18564,1,105,2520,23800,107100,244188,271320,
%U A370258 116280,1,136,4284,54264,339150,1139544,2089164,1961256,735471,1,171,6840,111720,921690,4239774,11306064
%N A370258 Triangle read by rows: T(n, k) = binomial(n, k)*binomial(2*n+k, k), 0 <= k <= n.
%C A370258 Compare with A063007(n, k) = binomial(n, k)*binomial(n+k, k), the table of coefficients of the shifted Legendre polynomials P(n, 2*x + 1).
%F A370258 n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*x^k = (1 + x)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*(x/(1 + x))^k = Sum_{k = 0..n} A110608(n, n-k)*x^k*(1 + x)^(n-k).
%F A370258 (x - 1)^n * R(n, 1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n, n-k)*x^k = the n-th row polynomial of A110608.
%F A370258 R(n, x) = hypergeom([-n, 2*n + 1], [1], -x).
%F A370258 Second-order differential equation: ( (1 + x)^n * (x + x^2)*R(n, x)' )' = n*(2*n + 1)*(1 + x)^n * R(n, x), where the prime indicates differentiation w.r.t. x.
%F A370258 Equivalently, x*(1 + x)*R(n, x)'' + ((n + 2)*x + 1)*R(n, x)' - n*(2*n + 1)*R(n, x)' = 0.
%F A370258 Analog of Rodrigues' formula for the shifted Legendre polynomials:
%F A370258 R(n, x) = 1/(1 + x)^n * 1/n! * (d/dx)^n (x*(1 + x)^2)^n.
%F A370258 Analog of Rodrigues' formula for the Legendre polynomials:
%F A370258 R(n, (x-1)/2) = 1/(n!*2^n) * 1/(1 + x)^n *(d/dx)^n ((x - 1)*(x + 1)^2)^n.
%F A370258 Orthogonality properties:
%F A370258 Integral_{x = -1..0} (1 + x)^n * R(n, x) * R(m, x) dx = 0 for n > m.
%F A370258 Integral_{x = -1..0} (1 + x)^n * R(n, x)^2 dx = 1/(3*n + 1).
%F A370258 Integral_{x = -1..0} (1 + x)^(n+m) * R(n, x) * R(m, x) dx = 0 for m >= 2*n + 1 or m <= (n - 1)/2.
%F A370258 Integral_{x = -1..0} (1 + x)^k * R(n, x) dx = 0 for n <= k <= 2*n - 1;
%F A370258 Integral_{x = -1..0} (1 + x)^(2*n) * R(n, x) dx = (2*n)!*n!/(3*n+1)! = 1/A090816(n).
%F A370258 Recurrence for row polynomials:
%F A370258 2*n*(2*n - 1)*((9*n - 12)*x + 8*n - 11)*(1 + x)*R(n, x) = (9*(3*n - 1)*(3*n - 2)*(3*n - 4)*x^3 + 3*(3*n - 1)*(3*n - 2)*(20*n - 27)*x^2 + 6*(3*n - 2)*(20*n^2 - 34*n + 9)*x + 2*(32*n^3 - 76*n^2 + 50*n - 9))*R(n-1, x) - 2*(n - 1)*(2*n - 3)*((9*n - 3)*x + 8*n - 3)*R(n-2, x), with R(0, x) = 1, R(1, x) = 1 + 3*x.
%F A370258 Conjecture: exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + 3*t)*x + (1 + 8*t + 12*t^2)*x^2 + ... is the o.g.f. for A102537. If true, then it would follows that, for each integer t, the sequence u = {R(n,t) : n >= 0} satisfies the Gauss congruences u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all primes p and positive integers m and r.
%F A370258 R(n, 1) = A114496(n); R(n, -1) = (-1)^n * A000984(n).
%F A370258 R(n, 2) = A339710(n); R(n, -2) = (-1)^n * A026000(n).
%F A370258 (2^n)*R(n, -1/2) = A234839(n).
%e A370258 Triangle begins
%e A370258 n\k| 0    1     2      3       4       5       6       7
%e A370258 - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e A370258  0 | 1
%e A370258  1 | 1    3
%e A370258  2 | 1   10    15
%e A370258  3 | 1   21    84     84
%e A370258  4 | 1   36   270    660     495
%e A370258  5 | 1   55   660   2860    5005    3003
%e A370258  6 | 1   78  1365   9100   27300   37128   18564
%e A370258  7 | 1  105  2520  23800  107100  244188  271320  116280
%e A370258  ...
%p A370258 seq(print(seq(binomial(n, k)*binomial(2*n+k, k), k = 0..n)), n = 0..10);
%Y A370258 A114496 (row sums), A000984 (alt. row sums unsigned), A005809 (main diagonal), A090763 (first subdiagonal), A014105 (column 1).
%Y A370258 Cf. A026000, A063007, A102537, A110608, A339710.
%K A370258 nonn,tabl,easy
%O A370258 0,3
%A A370258 _Peter Bala_, Feb 13 2024