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

%I #9 May 11 2023 10:25:34

%S 1,10,200,7390,260800,10263010,407520920,16758685030,697767370240,

%T 29525605934410,1261570539980200,54419751094210270,

%U 2364396136291654720,103393259758470870770,4545671563318715532280,200804420082143353690390,8907295723280072012247040,396570344897237949249382010

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

%C 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.

%C 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.

%C 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).

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

%H Frits Beukers, <a href="https://doi.org/10.1016/0022-314X(85)90047-2">Some congruences for the Apery numbers</a>, Journal of Number Theory, Vol. 21, Issue 2, Oct. 1985, pp. 141-155. <a href="/A339710/a339710.pdf">local copy</a>

%H Armin Straub, <a href="http://dx.doi.org/10.2140/ant.2014.8.1985">Multivariate Apéry numbers and supercongruences of rational functions</a>, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; <a href="https://arxiv.org/abs/1401.0854">arXiv preprint</a>, arXiv:1401.0854 [math.NT], 2014.

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

%F Conjectures:

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

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

%F 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.

%p A005259 := proc(n) add(binomial(n, k)^2*binomial(n+k,k)^2, k = 0..n) end;

%p E(n,x) := series(exp(n*add(2*A005259(2*k+1)*x^(2*k+1)/(2*k+1), k = 0..10)), x, 21):

%p seq(coeftayl(E(n,x), x = 0, n), n = 0..20);

%Y Cf. A005259, A267220, A362722 - A362733.

%K nonn,easy

%O 0,2

%A _Peter Bala_, May 01 2023