%I #29 Sep 05 2024 08:09:17
%S 1,1,1,3,4,1,13,19,7,1,67,102,44,10,1,381,593,278,78,13,1,2307,3640,
%T 1795,568,121,16,1,14589,23231,11849,4051,999,173,19,1,95235,152650,
%U 79750,28770,7820,1598,234,22,1,636925,1025965,545680,204760,59650,13642,2392,304,25,1
%N Riordan array (1/(1-xc(2x)),xc(2x)/(1-xc(2x))), c(x) the g.f. of A000108.
%C Deleham triangle Delta(0^n,2-0^n) [see construction in A084938]. The binomial transform of the inverse of this triangle has general element (-2)^(n-k)*C(k,n-k), that is, it is the Riordan array (1,x(1-2x)) [A110509]. Row sums are A052701. Diagonal sums are A110508. Inverse is A110511.
%H G. C. Greubel, <a href="/A110506/b110506.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F T(0,0) = 1, T(n,k) = (Sum_{j=0..n} j*C(2*n-j-1,n-j) * C(j,k) * 2^(n-j))/n.
%F T(n,k) = (-1)^(n-k)*A114189(n,k). - _Philippe Deléham_, Mar 24 2007
%e Rows begin:
%e 1;
%e 1,1;
%e 3,4,1;
%e 13,19,7,1;
%e 67,102,44,10,1;
%e 381,593,278,78,13,1;
%e From _Philippe Deléham_, Dec 01 2015: (Start)
%e Production matrix begins:
%e 1, 1
%e 2, 3, 1
%e 2, 4, 3, 1
%e 2, 4, 4, 3, 1
%e 2, 4, 4, 4, 3, 1
%e 2, 4, 4, 4, 4, 3, 1
%e 2, 4, 4, 4, 4, 4, 3, 1
%e (End)
%t {{1}}~Join~Table[Sum[j Binomial[2 n - j - 1, n - j] Binomial[j, k] 2^(n - j), {j, 0, n}]/n, {n, 9}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Dec 01 2015 *)
%o (PARI) tabl(nn)= {for (n=0, nn, for (k=0, n, if (n==0, x = 0^k, x = sum(j=0, n, j*binomial(2*n-j-1, n-j)*binomial(j, k)*2^(n-j)/n)); print1(x, ", ");); print(););} \\ _Michel Marcus_, Jun 18 2015
%Y Cf. A000108, A052701, A084938, A110508, A110509, A110511, A114189.
%K easy,nonn,tabl
%O 0,4
%A _Paul Barry_, Jul 24 2005