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A182960 G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ). 1

%I #19 Oct 06 2021 13:24:11

%S 1,5,95,2496,76063,2524161,88534548,3228482908,121171012431,

%T 4649906785719,181614908182551,7196014051078368,288537887780406468,

%U 11686156771344086156,477379538839242423528,19645977861506470428324

%N G.f.: exp( Sum_{n>=1} C(6n-1,2n-1)*x^n/n ).

%C The logarithmic derivative of this sequence is a bisection of the logarithmic derivative of A001764 (ternary trees).

%C To see this, compare the g.f. of this sequence with g.f. of A001764:

%C exp(Sum_{n>=1} C(3n-1,n-1)*x^n/n) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 +...

%F G.f. A(x) satisfies: A(x^2) = F(x)*F(-x) where F(x) = 1 + x*F(x)^3 = Sum_{n>=0} C(3n,n)*x^n/(2n+1) is the g.f. of A001764.

%F Let G(x) = 2*(1+x)^2/(1-2*x+sqrt(1-8*x)), then g.f. A(x) satisfies:

%F * A(x) = G(x*A(x)^3) and A(x/G(x)^3) = G(x);

%F * A(x) = [Series_Reversion( x/G(x)^3 ) / x]^(1/3).

%F The sequence defined by b(n) := [x^n] A(x)^n begins [1, 5, 215, 10463, 537287, 28435880, 1534398353, 83920389642, ...] and conjecturally satisfies the congruence b(p) == b(1) (mod p^3) for prime p >= 5 (checked up to p = 101). - _Peter Bala_, Sep 14 2021

%F From _Vaclav Kotesovec_, Sep 16 2021: (Start)

%F Recurrence: 16*n*(2*n + 1)*(4*n - 1)*(4*n + 1)*(27*n - 31)*a(n) = 18*(69984*n^5 - 220320*n^4 + 267705*n^3 - 156510*n^2 + 42021*n - 3680)*a(n-1) - 59049*(n-1)*(2*n - 3)*(3*n - 5)*(3*n - 4)*(27*n - 4)*a(n-2).

%F a(n) ~ sqrt((sqrt(2) - 1)^(2/3) + (sqrt(2) + 1)^(2/3) - 2) * 3^(6*n + 3/2) / (sqrt(Pi) * n^(3/2) * 2^(4*n + 7/2)). (End)

%e G.f.: A(x) = 1 + 5*x + 95*x^2 + 2496*x^3 + 76063*x^4 + 2524161*x^5 +...

%e log(A(x)) = 5*x + 165*x^2 + 6188*x^3 + 245157*x^4 + 10015005*x^5 +...+ A025174(2n)*x^n/n +...

%e G.f. satisfies: A(x) = G(x*A(x)^3) where G(x) begins:

%e G(x) = 1 + 5*x + 20*x^2 + 96*x^3 + 528*x^4 + 3136*x^5 + 19584*x^6 +...

%t f[n_] := Exp[ Sum[ Binomial[6 i - 1, 2 i - 1]*x^i/i, {i, n}]]; CoefficientList[ Series[f@ 15, {x, 0, 15}], x] (* _Robert G. Wilson v_, Dec 31 2010 *)

%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n,binomial(6*m-1,2*m-1)*x^m/m)+x*O(x^n)),n)}

%o (PARI) {a(n)=polcoeff((serreverse(x*(1-2*x+sqrt(1-8*x+x*O(x^n)))^3/(8*(1+x)^6))/x)^(1/3),n)}

%Y Cf. A182959, A025174, A001764, A060941, A300386 - A300389.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Dec 31 2010

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Last modified March 28 05:39 EDT 2024. Contains 371235 sequences. (Running on oeis4.)