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The following identity can be easily verified using Maple's SumTools:-Summation procedure: for n >= 1, A005809(n) = binomial(3*n,n) = Sum_{k = 0..2*n} n/(n + k)*binomial(n + k,k).

The binomial coefficients A005809(n) are known to satisfy the supercongruences A005809(n*p^r) == A005809(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r (see Meštrović, equation 39). Calculation suggests that the present sequence satisfies the same congruences.

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r.

More generally, for m a positive integer, define a sequence u_m by setting u_m(n) = Sum_{k = 0..m*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

Then we conjecture that each sequence u_m satisfies the above supercongruences. This is the case m = 2. See A333093 (case m = 1) and A352276 case (m = 3).

The following identity can be easily verified using Maple's SumTools:-Summation procedure: for n >= 1, A005810(n) = binomial(4*n,n) = Sum_{k = 0..3*n} n/(n + k)*binomial(n + k,k).

The binomial coefficients A005810(n) are known to satisfy the supercongruences A005810(n*p^r) == A005810(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r (see Meštrović, equation 39).

Calculation suggests that the present sequence satisfies the same congruences.

Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r.

More generally, for m a positive integer, define a sequence u_m by setting u_m(n) = Sum_{k = 0..m*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

Then we conjecture that each sequence u_m satisfies the above supercongruences. This is the case m = 3. See A333093 (case m = 1) and A352275 (m = 2).

Let F(x) = 1 + f(1)*x + f(2)*x^2 + ... be a power series with integer coefficients. The associated sequence u(n) := [x^n] F(x)^n is known to satisfy the Gauss congruences: u(n*p^k) == u(n*p^(k-1)) ( mod p^k ) for any prime p and positive integers n and k. For certain power series F(x), stronger congruences may hold. Examples include F(x) = (1 + x)^2, F(x) = 1/(1 - x) and F(x) = c(x), where c(x) is the o.g.f. of the Catalan numbers A000108. The associated sequences (with some differences of offset) are A000984, A001700 and A025174, respectively.

Here we take F(x) = c(x/(1 + x)) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 9*x^5 + 21*x^6 + ... (cf. A001006 and A086246) and conjecture that the associated sequence a(n) = [x^n] ( c(x/(1 + x)) )^n satisfies the supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(2*k) ) for prime p >= 5 and positive integers n and k. Cf. A333473.

More generally, we conjecture that for any positive integer r and any integer s the sequence a(r,s;n) := [x^(r*n)] ( c(x/(1 + x)) )^(s*n) also satisfies the above congruences.

Note that the sequence b(n) := [x^n] c(x)^n = A025174(n) satisfies the stronger congruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. The sequence d(n) := [x^n] ( (1 + x)*c(x/(1 + x)) )^n = A333093(n) appears to satisfy the same congruences.