A362723 a(n) = [x^n] ( E(x)/E(-x) )^n where E(x)= exp( Sum_{k >= 1} A005259(k)*x^k/k ).

1, 10, 200, 7390, 260800, 10263010, 407520920, 16758685030, 697767370240, 29525605934410, 1261570539980200, 54419751094210270, 2364396136291654720, 103393259758470870770, 4545671563318715532280, 200804420082143353690390, 8907295723280072012247040, 396570344897237949249382010

Offset: 0

Author

Peter Bala, May 01 2023

Keywords

• nonn
• easy

Comments

It is known that the sequence of Apéry numbers A005259 satisfies the Gauss congruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.

One consequence is that the power series expansion of E(x) = exp( Sum_{k >= 1} A005259(k)*x^k/k ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + ... has integer coefficients. See A267220. For a proof see, for example, Beukers, Proposition, p 143. Therefore, the power series expansion of E(x)/E(-x) also has integer coefficients and so a(n) = [x^n] ( E(x)/E(-x) )^n is an integer.

In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).

We conjecture below that {a(n)} satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.

Links

• Frits Beukers, Some congruences for the Apery numbers, Journal of Number Theory, Vol. 21, Issue 2, Oct. 1985, pp. 141-155. local copy
• Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.

Crossrefs

Cf. A005259, A267220, A362722 - A362733.

Programs

Maple

A005259 := proc(n) add(binomial(n, k)^2*binomial(n+k,k)^2, k = 0..n) end;
E(n,x) := series(exp(n*add(2*A005259(2*k+1)*x^(2*k+1)/(2*k+1), k = 0..10)), x, 21):
seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

Formula

a(n) = [x^n] exp( Sum_{k >= 1} n*( 2*A005259(2*k+1)*x^(2*k+1) )/(2*k+1) ).

Conjectures:

1) the supercongruence a(p) == a(1) (mod p^3) holds for all primes p >= 5 (checked up to p = 101).

2) for n >= 2, a(n*p) == a(n) (mod p^2) holds for all primes p >= 5.

3) for n >= 1, r >= 2, the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 5.