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a(n) = Sum_{k=0..n} T(n,k) where T(n,k) = (T(n-1, k-1) + T(n-1,k))^2.
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%I #28 Jan 16 2025 20:35:58

%S 1,2,6,52,3854,21090612,629815387162156,

%T 561871511512925116799625359336,

%U 446575758106416254441837050759254156476271759098752411181598

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

%C 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 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 1 = 1

%e 1 + 1 = 2

%e 1 + (1 + 1)^2 + 1 = 1 + 4 + 1 = 6

%e 1 + (1 + 4)^2 + (4 + 1)^2 + 1 = 1 + 25 + 25 + 1 = 52

%e 1 + (1 + 25)^2 + (25 + 25)^2 + (25 + 1)^2 + 1 = 1 + 676 + 2500 + 676 + 1 = 3854.

%o (Python)

%o def r(i):

%o t = [[0, 1, 0], [0, 1, 1, 0]]

%o for n in range(2, i+1):

%o t.append([0])

%o for k in range(1, n+2):

%o t[n].append((t[n-1][k-1] + t[n-1][k])**2)

%o t[n].append(0)

%o return(sum(t[i]))

%Y Cf. A004019, A327563, A007318.

%K easy,nonn

%O 0,2

%A _Glen Gilchrist_, Aug 30 2020