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%I #29 Jun 03 2023 10:25:05
%S 1,2,24,326,4672,69002,1038984,15856206,244396544,3795731282,
%T 59307908024,931222155030,14680871849152,232236016459098,
%U 3684420837693480,58600075142247326,934064636705476608,14917333936933664674,238641621366613695576,3823510794994321546214,61344017874989324388672
%N a(n) = [x^(3*n)] ( (1 + x)/(1 - x) )^n.
%C a(n) is also equal to [x^n] G(x)^n, where G(x) is the o.g.f. of A027307.
%C Compare with A002003(n) = [x^n] ( (1 + x)/(1 - x) )^n and A103885(n) = [x^(2*n)] ( (1 + x)/(1 - x) )^n = [x^n] S(x)^n, where S(x) is the o.g.f. of the large Schröder numbers A006318.
%C If we define an operator I acting on power series f(x) = 1 + f_1*x + f_2*x^2 + ... by I(f(x)) = 1/x * Revert( x/f(x) ) then S(x) = I( (1 + x)/(1 - x) ) and G(x) = (I o I)( (1 + x)/(1 - x )).
%C It can be shown that a(n) 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 a(n) satisfies the stronger congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Some examples of these congruences are given below.
%C The same congruences may hold more generally for the sequences a(r,s,n) := [x^(r*n)]( (1 + x)/(1 - x) )^(s*n), r a positive integer and s an integer. This is the case a(3,1,n).
%C For fixed m = 1,2,3,..., we conjecture that the sequence b(n) := a(m*n) satisfies a recurrence of the form P(6*m,n)*b(n+1) + P(6*m,-n)*b(n-1) = Q(3*m,n^2)*b(n), where the polynomials P(6*m,n) and Q(3*m,n^2) have degree 6*m. Conjecturally, the 6*m zeros of the polynomial Q(3*m,n^2) belong to the interval [-1, 1] and 6*m - 4 of these zeros appear to be approximated by the rational numbers +- k/(4*m), where 1 <= k <= 4*m - 3, k not a multiple of 4.
%H Seiichi Manyama, <a href="/A333715/b333715.txt">Table of n, a(n) for n = 0..824</a>
%F a(n) = Sum_{k = 0..n} C(n,k)*C(3*n+k-1,n-1).
%F a(n) = (1/3) * Sum_{k = 0..n} C(3*n,n-k)*C(3*n+k-1,k) for n >= 1.
%F a(n) = (1/3) * [x^n] ( (1 + x)/(1 - x) )^(3*n).
%F a(n) = Sum_{k = 1..n} (2^k)*C(n,k)*C(3*n-1,k-1) for n >= 1.
%F P-recursive:
%F P(6,n)*a(n+1) + P(6,-n)*a(n-1) = Q(3,n^2)*a(n), where P(6,n) = (2*n-1)*(3*n+1)*(3*n+2)*(3*n+3)*(35*n^2 - 35*n + 6) and the polynomial Q(3,n) = 4*(7805*n^3 - 7132*n^2 + 1559*n - 72).
%F Congruences: a(p) == 2 ( mod p^3 ) for prime p >= 3.
%F a(n) ~ (223 + 70*sqrt(10))^n / (2^(3/4) * 5^(1/4) * sqrt(Pi*n) * 3^(3*n + 1/2)). - _Vaclav Kotesovec_, Apr 04 2020
%F exp( Sum_{n>=1} a(n)*x^n/n ) = B(x) where B(x) = 1 + x*(B(x)^3 + B(x)^4) is the g.f. of A144097. - _Paul D. Hanna_, May 31 2023
%e Examples of congruences:
%e a(11) - a(1) = 931222155030 - 2 = (2^2)*(11^3)*163*1073069 == ( mod 11^3 )
%e a(3*7) - a(3) = 985413034951400888962602 - 326 = (2^2)*(7^4)*263* 390130947874776863 == 0 ( mod 7^3 )
%e a(5^2) - a(5) = 66292579025690123511768694002 - 69002 = (2^3)*(5^6)*39461* 13439614612035199009 == 0 ( mod 5^6 )
%p seq(add(binomial(n,k)*binomial(3*n+k-1,n-1), k = 0..n), n = 0..20);
%t Table[Binomial[3*n-1, n-1] * Hypergeometric2F1[-n, 3*n, 2*n+1, -1], {n, 0, 20}] (* _Vaclav Kotesovec_, Apr 04 2020 *)
%Y Cf. A002003, A027307, A103885, A144097.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Apr 03 2020
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