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A341359 Square array T(m,n) read by antidiagonals, satisfying shifted Catalan recurrences: T(m,0) = 1 and T(m,n) = Sum_{k=0..n-1} T(m,k) * T(m,(n-1-k+m) mod n) for all n > 0.

%I A341359 #14 Feb 12 2021 11:38:40
%S A341359 1,1,1,1,1,2,1,1,2,5,1,1,2,5,14,1,1,2,6,15,42,1,1,2,5,26,51,132,1,1,2,
%T A341359 5,30,142,188,429,1,1,2,6,14,305,882,731,1430,1,1,2,5,17,210,3955,
%U A341359 5910,2950,4862,1,1,2,5,22,50,5894,57855,41610,12235,16796,1,1,2,6,30,65,2550,209146,908880,303390,51822,58786
%N A341359 Square array T(m,n) read by antidiagonals, satisfying shifted Catalan recurrences: T(m,0) = 1 and T(m,n) = Sum_{k=0..n-1} T(m,k) * T(m,(n-1-k+m) mod n) for all n > 0.
%C A341359 Each column is periodic, and the period of column n divides A003418(n).
%H A341359 T. Amdeberhan et al. <a href="https://mathoverflow.net/q/380482">Sequences generated by sum & product of terms (with rotating indices): combinatorial?</a>, 2021.
%F A341359 G.f. for row m: p_m(x) + x^(m-1)/2 * ( 1 + sqrt((1 -(4*T(m,m)+1)*x)/(1-x)) ), where p_m(x) = Sum_{n=0..m-1} T(m,n) * x^n.
%e A341359 Rows of the array:
%e A341359 m=0: 1, 1, 2, 5, 14,  42,  132,    429,     1430, ...
%e A341359 m=1: 1, 1, 2, 5, 15,  51,  188,    731,     2950, ...
%e A341359 m=2: 1, 1, 2, 6, 26, 142,  882,   5910,    41610, ...
%e A341359 m=3: 1, 1, 2, 5, 30, 305, 3955,  57855,   908880, ...
%e A341359 m=4: 1, 1, 2, 5, 14, 210, 5894, 209146,  8331582, ...
%e A341359 m=5: 1, 1, 2, 6, 17,  50, 2550, 255050, 32007550, ...
%e A341359 ...
%t A341359 T[m_, 0] := 1; T[m_, n_] := T[m, n] = Sum[T[m, k] * T[m, Mod[n - 1 - k + m, n]], {k, 0, n - 1}]; Table[T[m - n, n], {m, 0, 11}, {n, 0, m}] // Flatten (* _Amiram Eldar_, Feb 09 2021 *)
%Y A341359 Rows: A000108 (m=0), A181768 (m=1).
%Y A341359 Columns: A000012 (n=0 and n=1), A007395 (n=2).
%Y A341359 Cf. A007317, A162326, A082298,A226392, A341360 (diagonal).
%K A341359 nonn,tabl
%O A341359 0,6
%A A341359 _Max Alekseyev_, Feb 09 2021