A351858 a(n) = [x^n] (1 + x + x^2)^(3*n)/(1 + x)^(2*n).

1, 1, 7, 19, 103, 376, 1825, 7547, 35175, 153838, 708132, 3181091, 14616481, 66582283, 306501377, 1407473269, 6497464679, 29991098982, 138844558150, 643215119214, 2985368996228, 13868212710623, 64508509024241, 300324344452479, 1399598738196897, 6527698842078501

Offset: 0

Author

Peter Bala, Feb 27 2022

Keywords

• nonn
• easy

Comments

Given an integer sequence (g(n))n>=1, there exists a formal power series G(x), with rational coefficients, such that g(n) = [x^n] G(x)^n. The power series G(x) has integer coefficients iff the Gauss congruences g(n*p^r) == g(n*p^(r-1)) (mod p^r) hold for all primes p and positive integers n and r.

The central binomial coefficient binomial(2*n,n) = A000984(n) may be defined using the coefficient extraction operator as binomial(2*n,n) = [x^n] ((1 + x)^2)^n and hence the Gauss congruences hold for A000984. Moreover, it is known that the stronger supercongruences A000984(n*p^r) == A000984(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. See Meštrović, equation 39.

We define an infinite family of sequences as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. In particular, G_1(x) = (1 + x)^2 and the sequence u_1 is the sequence of central binomial coefficients. The present sequence is the case k = 2. See A351859 for the case k = 3.

Conjecture: for k >= 2, each sequence u_k satisfies the same supercongruences as the central binomial coefficients.

More generally, if r is a positive integer and s an integer then the sequence defined by u_k(r,s;n) = [x^(r*n)] G_k(x)^(s*n) may satisfy the same supercongruences.

Examples

			Examples of supercongruences:
a(5) - a(1) = 376 - 1 = 3*(5^3) == 0 (mod 5^3)
a(2*7)- a(2) = 306501377 - 7 = 2*5*(7^3)*193*463 == 0 (mod 7^3)
A(5^2) - a(5) = 6527698842078501 - 376 = (5^6)*17*107*229671647 == 0 (mod 5^6)
		

References

• R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Links

• R. Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.

Crossrefs

Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351859, A372211, A372212, A372213.

Programs

Maple

seq(add(add((-1)^(n-k-j)*binomial(n,k)*binomial(3*n,j)* binomial(4*n-2*j-k-1,n-k-j), j = 0..n-k), k = 0..n), n = 0..25);

Mathematica

A351858[n_]:=Sum[(-1)^(n-k-j)Binomial[n,k]Binomial[3n,j]Binomial[4n-2j-k-1,n-k-j],{k,0,n},{j,0,n-k}];Array[A351858,25,0] (* Paolo Xausa, Oct 04 2023 *)
a[n_]:=SeriesCoefficient[(1 + x + x^2)^(3*n)/(1 + x)^(2*n),{x,0,n}]; Array[a,26,0] (* Stefano Spezia, Apr 30 2024 *)

Formula

a(n) = Sum_{k = 0..n} Sum_{j = 0..n-k} (-1)^(n-k-j)*C(n,k)*C(3*n,j)*C(4*n-2*j-k-1,n-k-j).

Conjecture: a(n) = Sum_{k = 0..floor(n/2)} C(3*n,k)*C(n-k,k).

The o.g.f. A(x) = 1 + x + 7*x^2 + 19*x^3 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2)^3/(1 + x)^2) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.

Let F(x) = (1/x)*Series_Reversion( x*(1 + x)^2/(1 + x + x^2)^3 ) = 1 + x + 4*x^2 + 10*x^3 + 40*x^4 + 133*x^5 + 536*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).

a(n) ~ sqrt(2/9 + 2*sqrt(53/47)*cos(arccos(1259*sqrt(47/53)/1696)/3)/9) * (2*sqrt(164581)*cos(arccos(-90631279/(1316648*sqrt(164581)))/3)/81 - 293/81)^n / sqrt(Pi*n). - Vaclav Kotesovec, Jun 05 2022