%I #18 Oct 31 2024 01:34:41
%S 1,5,49,685,11807,232771,5031415,116222699,2822056474,71230971002,
%T 1854334597966,49503739725470,1349449053997654,37438177610268014,
%U 1054371288632733022,30081609844254013942,867990837575171240525,25295504900322451251793
%N Expansion of exp( Sum_{n >= 1} A005259(n)*x^n/n ).
%C The expansion of exp( Sum_{n >= 1} A005259(n-1)*x^n/n ) may also have only integer coefficients. See the Example section.
%C This is proved in Beukers, p. 143. - _Peter Bala_, Mar 06 2020
%H F. 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.
%H P. Peart and W.-J. Woan, <a href="http://dx.doi.org/10.1016/S0166-218X(99)00166-3">A divisibility property for a subgroup of Riordan matrices</a>, Discrete Applied Mathematics, Vol. 98, Issue 3, Jan 2000, 255-263.
%F n*a(n) = Sum_{k = 0..n-1} A005259(n-k)*a(k).
%F O.g.f.: A(x) = exp( Sum_{n >= 1} A005259(n)*x^n/n ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + 232771*x^5 + 5031415*x^6 + ....
%F The o.g.f. A(x) satisfies 1 + x* d/dx(log(A(x)) = Sum_{n >= 0} A005259(n)*x^n.
%F Let F(x) = x/( series reversion of x*A(x) ) = 1 + 5*x + 24*x^2 + 200*x^3 + 2430*x^4 + 36096*x^5 + 605620*x^6 + 11024496*x^7 + 212758245*x^8 + .... Then A005259(n) = [x^n]( F(x)^n ). See the example section.
%F From _Peter Bala_, Oct 17 2024: (Start)
%F For integer m, define a sequence {u_m(n) : n >= 0} by u_m(n) = [x^n] A(x)^(m*n). Conjecture: the supercongruences u_m(n*p^r) == u_m(n*p^(r-1)) (mod p^(2*r)) hold for all primes p >= 5 and positive integers n and r.
%F For integer m, define a sequence {v_m(n) : n >= 0} by v_m(n) = [x^n] F(x)^(m*n). Conjecture: the supercongruences v_m(n*p^r) == v_m(n*p^(r-1)) (mod p^(2*r)) hold for all primes p >= 5 and positive integers n and r. (End)
%e exp( Sum_{n >= 1} A005259(n-1)*x^n/n ) = 1 + 3*x + 27*x^2 + 390*x^3 + .... The coefficient list begins [1, 3, 27, 390, 7038, 144550, 3232294, 76768575, 1907248655, 49067872253, 1298071849821, 35131589549434, 969031280176162, 27161049933788970, 771837331889465586, 22196147941000214583, 644991902540171273463, ...].
%e The Apéry number sequence A005259 begins [1, 5, 73, 1445, 33001, 819005, ...]. Let F(x) = x/( series reversion of x*A(x) ) = 1 + 5*x + 24*x^2 + 200*x^3 + 2430*x^4 + ....
%e Truncation of F(x)^n:
%e F(x)^0: 1
%e F(x)^1: 1 + 5*x
%e F(x)^2: 1 + 10*x + 73*x^2
%e F(x)^3: 1 + 15*x + 147*x^2 + 1445*x^3
%e F(x)^4: 1 + 20*x + 246*x^2 + 2740*x^3 + 33001*x^4
%e The coefficient array
%e 1
%e 5 1
%e 73 10 1
%e 1445 147 15 1
%e 33001 2740 246 20 1
%e ...
%e has the Apéry numbers as the first column.
%e It is a Riordan array belonging to the hitting-time subgroup of the Riordan group - see Peart and Woan.
%p #A267220
%p #define the Apéry numbers
%p A005259 := proc (n) option remember; if n = 0 then 1 elif n = 1 then 5 else ((34*(n-1)^3+51*(n-1)^2+27*n-22)*A005259(n-1)-(n-1)^3*A005259(n-2))/n^3 end if; end proc:
%p exp(add(A005259(n)*x^n/n, n = 1 .. 17)):
%p seq(coeftayl(%, x = 0, n), n = 0 .. 17);
%Y Cf. A005259, A267219.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Jan 12 2016