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%I #30 May 01 2023 15:03:10
%S 1,7,109,1951,36993,724007,14457421,292732671,5987886081,123440423047,
%T 2560421160109,53373725431583,1117198199782785,23465732683090471,
%U 494330214846965389,10440064992542621951,220978578227187097601,4686426367646858888711,99559270036968523118317
%N a(n) is the n-th order Taylor polynomial (centered at 0) of S(x)^(3*n) evaluated at x = 1, where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the Schröder numbers A006318.
%C 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 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 More generally, for each integer m, we conjecture that the sequence a_m(n) defined as the n-th order Taylor polynomial of S(x)^(m*n) evaluated at x = 1 satisfies the same congruences. For cases, see A333090 (m = 1) and A333091 (m = 2). For similarly defined sequences see A333093 through A333097.
%F a(n) = [x^n] ( (1 + x)*S^3(x/(1 + x)) )^n.
%F O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 6*x + 66*x^2 + 902*x^3 + 13794*x^4 + ... = (1/x) * series reversion of ( x/S^3(x) ).
%F Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
%F a(n) ~ 3*sqrt(85 + 21*sqrt(17)) * (349 + 85*sqrt(17))^n / (68 * sqrt(Pi*n) * 2^(5*n)). - _Vaclav Kotesovec_, Mar 28 2020
%e n-th order Taylor polynomial of S(x)^(3*n):
%e n = 0: S(x)^0 = 1 + O(x)
%e n = 1: S(x)^3 = 1 + 6*x + O(x^2)
%e n = 2: S(x)^6 = 1 + 12*x + 96*x^2 + O(x^3)
%e n = 3: S(x)^9 = 1 + 18*x + 198*x^2 + 1734*x^3 + O(x^4)
%e n = 4: S(x)^12 = 1 + 24*x + 336*x^2 + 3608*x^3 + 33024*x^4 + O(x^5)
%e Setting x = 1 gives a(0) = 1, a(1) = 1 + 6 = 7, a(2) = 1 + 12 + 96 = 109, a(3) = 1 + 18 + 198 + 1734 = 1951 and a(4) = 1 + 24 + 336 + 3608 + 33024 = 36993.
%e The triangle of coefficients of the n-th order Taylor polynomial of S(x)^(2*n), n >= 0, in descending powers of x begins
%e row sums
%e n = 0 | 1 1
%e n = 1 | 6 1 7
%e n = 2 | 96 12 1 109
%e n = 3 | 1734 198 18 1 1951
%e n = 4 | 33024 3608 336 24 1 36993
%e ...
%e This is a Riordan array belonging to the Hitting time subgroup of the Riordan group. The first column sequence [1, 6, 96, 1734, 33024, 648006, ...] = [x^n] S(x)^(3*n), and may also satisfy the above congruences.
%e Examples of congruences:
%e a(13) - a(1) = 23465732683090471 - 7 = (2^5)*(3^4)*(13^3)*83*911*54497 == 0 ( mod 13^3 ).
%e a(3*7) - a(3) = 962815680123979633351467303 - 1951 = (2^3)*(7^3)*29*41* 1832861*161008076794727 == 0 ( mod 7^3 ).
%e a(5^2) - a(5) = 201479167004032422703424646224007 - 724007 = (2^5)*(5^6)* 402958334008064845406849291 == 0 ( mod 5^6 ).
%p S:= x -> (1/2)*(1-x-sqrt(1-6*x+x^2))/x:
%p G := (x,n) -> series(S(x)^(3*n), x, 101):
%p seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);
%t Table[SeriesCoefficient[((1+x)*(1 - 3*x*(1+x) + (x^2 + x - 1)*Sqrt[1 - 4*x*(1+x)]) / (2*x^3))^n, {x, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 28 2020 *)
%Y Cf. A006318, A333090 through A333097.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Mar 22 2020
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