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A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3*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 A333095 #19 May 04 2024 14:49:57
%S A333095 1,4,34,337,3554,38754,431521,4874377,55639010,640177033,7412165034,
%T A333095 86256322816,1007980394849,11820510331777,139032549536551,
%U A333095 1639506780365337,19376785465043938,229458302589724067,2721958273545613513,32339465512495259708,384758834631081248554
%N A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3*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 A333095 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 A333095 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 A333095 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), A333096 (m = 4), A333097 (m = 5).
%F A333095 a(n) = Sum_{k = 0..n} 3*n/(3*n+k)*binomial(3*n+2*k-1, k) for n >= 1.
%F A333095 a(n) = [x^n] ( (1 + x)*c^3(x/(1 + x)) )^n.
%F A333095 O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + ... = (1/x)*Revert( x/c^3(x) ) is the o.g.f. of A118970.
%F A333095 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 A333095 a(n) ~ 5^(5*n + 3/2) / (7 * 2^(8*n + 3/2) * sqrt(Pi*n)). - _Vaclav Kotesovec_, Mar 28 2020
%F A333095 a(n) = Sum_{k = 0..n} 3*n/(3*n+2*k)*binomial(3*n+2*k, k) for n >= 1. - _Peter Bala_, May 03 2024
%e A333095 n-th order Taylor polynomial of c(x)^(3*n):
%e A333095   n = 0: c(x)^0 = 1 + O(x)
%e A333095   n = 1: c(x)^3 = 1 + 3*x + O(x^2)
%e A333095   n = 2: c(x)^6 = 1 + 6*x + 27*x^2 + O(x^3)
%e A333095   n = 3: c(x)^9 = 1 + 9*x + 54*x^2 + 273*x^3 + O(x^4)
%e A333095   n = 4: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + 2907*x^4 + O(x^5)
%e A333095 Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 27 = 34, a(3) = 1 + 9 + 54 + 273 = 337 and a(4) = 1 + 12 + 90 + 544 + 2907 = 3554.
%e A333095 The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
%e A333095                                               row sums
%e A333095   n = 0 |    1                                    1
%e A333095   n = 1 |    3       1                            4
%e A333095   n = 2 |   27       6    1                      34
%e A333095   n = 3 |  273      54    9   1                 337
%e A333095   n = 4 | 2907     544   90  12   1            3554
%e A333095    ...
%e A333095 This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
%e A333095 Examples of supercongruences:
%e A333095 a(13) - a(1) = 11820510331777 - 4 = 3*11*(13^3)*(43^2)*88177 == 0 ( mod 13^3 ).
%e A333095 a(3*7) - a(3) = 4583419703934987639046 - 337 = (3^2)*(7^4)*2441* 86893477573061 == 0 ( mod 7^3 ).
%e A333095 a(5^2) - a(5) = 93266278848727959965820004 - 38754 = 2*(5^7)*19* 31416009717466260199 == 0 ( mod 5^6 ).
%p A333095 seq(add(3*n/(3*n+k)*binomial(3*n+2*k-1,k), k = 0..n), n = 1..25);
%p A333095 #alternative program
%p A333095 c:= x → (1/2)*(1-sqrt(1-4*x))/x:
%p A333095 G := (x,n) → series(c(x)^(3*n), x, 101):
%p A333095 seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);
%t A333095 Join[{1}, Table[3*Binomial[5*n-1, n] * HypergeometricPFQ[{1, -4*n, -n}, {1/2 - 5*n/2, 1 - 5*n/2}, 1/4]/4, {n, 1, 20}]] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%Y A333095 Cf. A000108, A118970, A333090 through A333097.
%K A333095 nonn,easy
%O A333095 0,2
%A A333095 _Peter Bala_, Mar 15 2020