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If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2^*k)) hold for primes p >= 3 and positive integers n and k.

If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1) ) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1) ) (mod p^(2^k)) hold for ..primes p >= 3 and positive integers n and k.

If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1) (mod p^(2^k)) hold for ...

reviewed

editing

proposed

reviewed

editing

proposed

From Peter Bala, May 02 2022: (Start)

Conjecture: a(n) = [x^n] ( (1 + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{j = 0..n-2*k} binomial(n,k)*binomial(2*n,j)*binomial(3*n-2*k-j-1,n-2*k-j).

It appears that a(n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(2*n-k)*binomial(2*k,k)^2* binomial(2*n+k,2*k). Compare with the pair of identities: binomial(2*n,n) = Sum_{k = 0..n} 2^(n-2*k)*binomial(2*k,k)*binomial(n,2*k) and binomial(2*n,n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(4*n-2*k)*binomial(2*k,k)^2*binomial(2*n+k,2*k). - _Peter Bala_, May 02 2022(End)

It appears that a(n)^2 = Sum_{k = 0..2*n} (-1)^k*bimomial2^(2*n-k)*binomial(2*k,k)^2* binomial(2*n + k,2*k)*2^(,2*n-k). Compare with the pair of identities: binomial(2*n,n) = Sum_{k = 0..n} 2^(n-2*k)*binomial(2*k,k)*binomial(n,2*k) and binomial(2*n,n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(4*n-2*k)*binomial(2*k,k)^2*binomial(2*n + k,2*k)*2^(4*n-,2*k). - Peter Bala, May 02 2022