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A337510 a(n) = Sum_{k=0..n} T(n,k) where T(n,k) = (T(n-1, k-1) + T(n-1,k))^2.

%I A337510 #28 Jan 16 2025 20:35:58
%S A337510 1,2,6,52,3854,21090612,629815387162156,
%T A337510 561871511512925116799625359336,
%U A337510 446575758106416254441837050759254156476271759098752411181598
%N A337510 a(n) = Sum_{k=0..n} T(n,k) where T(n,k) = (T(n-1, k-1) + T(n-1,k))^2.
%C A337510 Based on Pascal's triangle A007318 by additionally squaring the sum of each term generated.  For example, in Pascal, n=3 gives 1,2,1.  Here n=3 gives, 1^2, (1+1)^2, 1^2 = 1+4+1.
%F A337510 a(n) = Sum_{k=0..n} T(n,k) where T(n,k) = (T(n-1,k-1) + T(n-1,k))^2; T(0,0)=1; T(n,-1):=0; T(n,k):=0, n < k.
%e A337510 1 = 1
%e A337510 1 + 1 = 2
%e A337510 1 + (1 + 1)^2  + 1 = 1 + 4 + 1 = 6
%e A337510 1 + (1 + 4)^2  + (4 + 1)^2 + 1 = 1 + 25 + 25 + 1 = 52
%e A337510 1 + (1 + 25)^2 + (25 + 25)^2 + (25 + 1)^2 + 1 = 1 + 676 + 2500 + 676 + 1 = 3854.
%o A337510 (Python)
%o A337510 def r(i):
%o A337510   t = [[0, 1, 0], [0, 1, 1, 0]]
%o A337510   for n in range(2, i+1):
%o A337510     t.append([0])
%o A337510     for k in range(1, n+2):
%o A337510       t[n].append((t[n-1][k-1] + t[n-1][k])**2)
%o A337510     t[n].append(0)
%o A337510   return(sum(t[i]))
%Y A337510 Cf. A004019, A327563, A007318.
%K A337510 easy,nonn
%O A337510 0,2
%A A337510 _Glen Gilchrist_, Aug 30 2020