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A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

%I A333096 #17 May 03 2024 15:07:20
%S A333096 1,5,53,647,8373,111880,1525511,21093476,294663349,4148593604,
%T A333096 58770091928,836722722951,11961868391175,171601856667701,
%U A333096 2469036254872996,35615467194043147,514888180699419829,7458193213805231529,108219144962546395364,1572690742149983040857
%N A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4*n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.
%C A333096 The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
%C A333096 We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
%C A333096 More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).
%F A333096 a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1, k) for n >= 1.
%F A333096 a(n) = [x^n] ( (1 + x)*c^4(x/(1 + x)) )^n.
%F A333096 O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + ... = (1/x)*Revert( x/c^4(x) ) is the o.g.f. of A212073.
%F A333096 Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
%F A333096 a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - _Vaclav Kotesovec_, Mar 28 2020
%F A333096 a(n) = Sum_{k = 0..n} 4*n/(4*n+2*k)*binomial(4*n+2*k, k) for n >= 1. - _Peter Bala_, May 03 2024
%e A333096 n-th order Taylor polynomial of c(x)^(4*n):
%e A333096   n = 0: c(x)^0 = 1 + O(x)
%e A333096   n = 1: c(x)^4 = 1 + 4*x + O(x^2)
%e A333096   n = 2: c(x)^8 = 1 + 8*x + 44*x^2 + O(x^3)
%e A333096   n = 3: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + O(x^4)
%e A333096   n = 4: c(x)^16 = 1 + 16*x + 152*x^2 + 1120*x^3 + 7084*x^4 + O(x^5)
%e A333096 Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 44 = 53, a(3) = 1 + 12 + 90 + 544 = 647 and a(4) = 1 + 16 + 152 + 1120 + 7084 = 8373.
%e A333096 The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(4*n), n >= 0, in descending powers of x begins
%e A333096                                          row sums
%e A333096   n = 0 |    1                               1
%e A333096   n = 1 |    4     1                         5
%e A333096   n = 2 |   44     8     1                  53
%e A333096   n = 3 |  544    90    12     1           647
%e A333096   n = 4 | 7084  1120   152    16   1      8373
%e A333096    ...
%e A333096 This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
%e A333096 Examples of congruences:
%e A333096 a(13) - a(1) = 171601856667701 - 5 = (2^4)*3*(7^2)*(13^3)*33208909 == 0 ( mod 13^3 ).
%e A333096 a(3*7) - a(3) = 333475516822140871773101 - 647 = 2*(3^2)*(7^3)* 54012879303877692221 == 0 ( mod 7^3 ).
%e A333096 a(5^2) - a(5) = 15187725485911657497382846255 - 111880 = (3^3)*(5^7)*29* 248279548173268475053 == 0 ( mod 5^6 ).
%p A333096 seq(add(4*n/(4*n+k)*binomial(4*n+2*k-1,k), k = 0..n), n = 1..25);
%p A333096 #alternative program
%p A333096 c:= x → (1/2)*(1-sqrt(1-4*x))/x:
%p A333096 G := (x,n) → series(c(x)^(4*n), x, 126):
%p A333096 seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);
%t A333096 Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%Y A333096 Cf. A000108, A212073, A333090 through A333097.
%K A333096 nonn,easy
%O A333096 0,2
%A A333096 _Peter Bala_, Mar 15 2020