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From Peter Bala, Jan 09 2022: (Start)
Conjectures: for k >= 2, the number of k-ary words of length n such that the number of 1's <= the number of 0's is equal to the coefficient of x^n in the expansion of ( k*x + 1/(1 + x) )^n, and satisfies the recurrence u(0) = 1, u(1) = k-1 and n*u(n) = (k-2)*(2*n-1)*u(n-1) - k*(k-4)*(n-1)* u(n-2) + k^(n-1) for n >= 2.
For cases see A027306 (k = 2), A027914 (k = 3) and A032443 (k = 4). (End)
a(n) = Sum_{k = 0..n} Sum_{j = 0..n} C(n, j)*C(2*j, j+k).
From Peter Bala, Jan 07 08 2022: (Start)
a(n) = (1/2)*(5^n + A026375(n)) = (1/2)* (5^n + Sum_{k = 0..n} binomial(n,k) *binomial(2*k-1,,k)*(-1/5)^k.
a(n) = (1/2)*(5^n)*(1 + Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(-1/5)^k).
a(0) = 1, a(1) = 4 and n*a(n) = 3*(2*n-1)*a(n-1) - 5*(n-1)*a(n-2) + 5^(n-1) for n >= 2.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. (End)
Binomial transform of A032443. (End)
seq( (1/2)*(5^n + add(binomial(n, k)*binomial(2*k, k), k = 0..n)), n = 0..30); # Peter Bala, Jan 08 2022
a(n) = sumSum_{k=0..n, sum} Sum_{j=0..n, } C(n, j)*C(2*j, j+k)}}.
From Peter Bala, Jan 07 2022: (Start)
a(n) = 5^n * Sum_{k = 0..n} binomial(n,k)*binomial(2*k-1,k)*(-1/5)^k.
a(n) = [x^n] ( 5*x + 1/(1 + x) )^n.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. (End)
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Table[Sum[Sum[Binomial[n, j]Binomial[2j, j+k], {j, 0, n}], {k, 0, n}], {n, 0, 25}] (* From _Harvey P. Dale, _, Dec 16 2011 *)
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