This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A376467 #21 Oct 07 2024 03:22:53
%S A376467 1,3,2,13,18,6,63,132,90,20,321,900,930,420,70,1683,5910,8190,5600,
%T A376467 1890,252,8989,37926,65940,60480,30870,8316,924,48639,239624,501228,
%U A376467 577080,395010,160776,36036,3432,265729,1497096,3660300,5072760,4358970,2378376,804804,154440,12870,1462563,9274410,25951860,42060480,43513470,29801772,13513500,3912480,656370,48620
%N A376467 Triangular array read by rows: A063007 * A007318.
%C A376467 Note that the n-th row generating polynomial of A063007 is equal to P(n,2*x + 1), where P(n,x) denotes the n-th Legendre polynomial.
%C A376467 The matrix product A063007 * A007318^(-1) is equal to a signed version of A063007 and A007318^(-1) * A063007 = A115951.
%F A376467 T(n, k) = Sum_{j = k..n} binomial(n, j)*binomial(n+j, j)*binomial(j, k).
%F A376467 (n - k)*T(n, k) = 3*(2*n - 1)*T(n-1, k) - (n + k - 1)*T(n-2, k).
%F A376467 T(n, k) = (1/k!) * (d/dx)^k (P(n, 2*x+1)) evaluated at x = 1, where P(n,x) denotes the n-th Legendre polynomial.
%F A376467 G.f. for triangle: 1/sqrt(1 - 6*t + t^2 - 4*t*x) = 1 + (3 + 2*x)*t + (13 + 18*x + 6*x^2)*t^2 + ....
%F A376467 G.f. for column k: binomial(2*k, k) * x^k/(1 - 6*x + x^2)^(k+1/2).
%F A376467 T(n, k) is divisible by binomial(2*k, k) and the array ( T(n, k)/binomial(2*k, k) )n,k >= 0 is the Riordan array (1/sqrt(1 - 6*x + x^2), x/(1 - 6*x + x^2)).
%F A376467 T(n, k) is divisible by binomial(n+k, k) and the array ( T(n, k)/binomial(n+k, k) )n,k >= 0 is the Riordan array A118384.
%F A376467 T(n, n) = binomial(2*n, n); T(n, n-1) = 3*(2*n-1)!/(n-1)!^2 = 3 * A002457(n-1) for n >= 1.
%F A376467 The n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*(1 + x)^k = P(n, 2*x+3) = hypergeom([-n, n+1], [1], -1-x).
%F A376467 Recurrence: n*R(n, x) = (2*x + 3)*(2*n - 1)*R(n-1, x) - (n - 1)*R(n-2, x) with R(0, x) = 1.
%F A376467 If we set R(-1,x) = 1, we can run the recurrence backwards to give R(-n, x) = Sum_{k = 0..n} binomial(-n, k)*binomial(-n+k, k)*(1 + x)^k = R(n-1, x).
%F A376467 R(n, x) = (-1)^n * R(n, -x-3).
%F A376467 R(n, x) = 1/n! * (d/dx)^n( ((1 + x)*(2 + x))^n ).
%F A376467 R(1, x) = 3 + 2*x divides R(2*n+1, x) in the polynomial ring Z[x].
%e A376467 Triangle begins
%e A376467 n\k| 0 1 2 3 4 5 6 7
%e A376467 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e A376467 0 | 1
%e A376467 1 | 3 2
%e A376467 2 | 13 18 6
%e A376467 3 | 63 132 90 20
%e A376467 4 | 321 900 930 420 70
%e A376467 5 | 1683 5910 8190 5600 1890 252
%e A376467 6 | 8989 37926 65940 60480 30870 8316 924
%e A376467 7 | 48639 239624 501228 577080 395010 160776 36036 3432
%e A376467 ...
%p A376467 A376467 := proc(n, k); add(binomial(n, j)*binomial(n+j, j)*binomial(j, k), j = k..n) end:
%p A376467 seq(print(seq(A376467(n, k) , k = 0..n)), n = 0..10);
%Y A376467 A000984 (main diagonal), A002457( (1/3)*first subdiagonal ), A001850 (Column 0), A002695 ( (1/2)*Column 1 ), A277660 ( (1/3)*Column 2 ), A006442 (row sums).
%Y A376467 Cf. A063007, A007318, A115951, A118384.
%K A376467 nonn,tabl,easy
%O A376467 0,2
%A A376467 _Peter Bala_, Sep 30 2024