A251573 E.g.f.: exp(3*x*G(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764.

1, 1, 3, 21, 261, 4833, 120303, 3778029, 143531433, 6404711553, 328447585179, 19037277446949, 1230842669484717, 87829738967634849, 6856701559496841159, 581343578623728854397, 53196439113856500195537, 5225543459274294130169601, 548468830470032135590262067, 61258398893626609968686844597

Offset

0,3

Links

Paul D. Hanna, Table of n, a(n) for n = 0..200

Formula

Let G(x) = 1 + x*G(x)^3 be the g.f. of A001764, then the e.g.f. A(x) of this sequence satisfies:

(1) A'(x)/A(x) = G(x)^2.

(2) A'(x) = exp(3*x*G(x)^2).

(3) A(x) = exp( Integral G(x)^2 dx ).

(4) A(x) = exp( Sum_{n>=1} A006013(n-1)*x^n/n ), where A006013(n-1) = binomial(3*n-2,n)/(2*n-1).

(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251583.

(6) A(x) = Sum_{n>=0} A251583(n)*(x/A(x))^n/n! where A251583(n) = 3^(n-1) * (n+1)^(n-3) * (n+3).

(7) [x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3).

a(n) = Sum_{k=0..n} 3^k * n!/k! * binomial(3*n-k-3, n-k) * (k-1)/(n-1) for n>1.

Recurrence (for n>3): 2*(n-3)*(2*n-3)*a(n) = 3*(3*n-8)*(3*n^2 - 13*n + 15)*a(n-1) - 27*(n-2)*a(n-2). - Vaclav Kotesovec, Dec 07 2014

a(n) ~ 3^(3*n-7/2) * n^(n-2) / (2^(2*n-5/2) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014

Example

E.g.f.: A(x) = 1 + x + 3*x^2/2! + 21*x^3/3! + 261*x^4/4! + 4833*x^5/5! +...
such that A(x) = exp(3*x*G(x)^2) / G(x)^2
where G(x) = 1 + x*G(x)^3 is the g.f. of A001764:
G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +...
The e.g.f. satisfies:
A(x) = 1 + x/A(x) + 5*x^2/(2!*A(x)^2) + 54*x^3/(3!*A(x)^3) + 945*x^4/(4!*A(x)^4) + 23328*x^5/(5!*A(x)^5) + 750141*x^6/(6!*A(x)^6) + 29859840*x^7/(7!*A(x)^7) +...+ 3^(n-1)*(n+1)^(n-3)*(n+3) * x^n/(n!*A(x)^n) +...
Note that
A'(x) = exp(3*x*G(x)^2) = 1 + 3*x + 21*x^2/2! + 261*x^3/3! + 4833*x^4/4! +...
LOGARITHMIC DERIVATIVE.
The logarithm of the e.g.f. begins:
log(A(x)) = x + 2*x^2/2 + 7*x^3/3 + 30*x^4/4 + 143*x^5/5 +...
and so A'(x)/A(x) = G(x)^2.
TABLE OF POWERS OF E.G.F.
Form a table of coefficients of x^k/k! in A(x)^n as follows.
n=1: [1, 1,  3,   21,   261,   4833,  120303,   3778029, ...];
n=2: [1, 2,  8,   60,   744,  13536,  330912,  10232928, ...];
n=3: [1, 3, 15,  123,  1557,  28179,  680427,  20771235, ...];
n=4: [1, 4, 24,  216,  2832,  51552, 1237248,  37404288, ...];
n=5: [1, 5, 35,  345,  4725,  87285, 2094975,  62949825, ...];
n=6: [1, 6, 48,  516,  7416, 139968, 3378528, 101278944, ...];
n=7: [1, 7, 63,  735, 11109, 215271, 5250987, 157613463, ...];
n=8: [1, 8, 80, 1008, 16032, 320064, 7921152, 238878720, ...]; ...
in which the main diagonal begins (see A251583):
[1, 2, 15, 216, 4725, 139968, 5250987, 238878720, ...]
and is given by the formula:
[x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3) for n>=0.

Mathematica

Flatten[{1, 1, Table[Sum[3^k * n!/k! * Binomial[3*n-k-3, n-k] * (k-1)/(n-1), {k, 0, n}], {n, 2, 20}]}] (* Vaclav Kotesovec, Dec 07 2014 *)
Flatten[{1, 1, RecurrenceTable[{27*(n-2)*a[n-2]-3*(3*n-8)*(15-13*n+3*n^2)*a[n-1]+2*(n-3)*(2*n-3)*a[n]==0, a[2]==3, a[3]==21}, a, {n, 20}]}] (* Vaclav Kotesovec, Dec 07 2014 *)

Prog

(PARI) {a(n) = local(G=1); for(i=1, n, G=1+x*G^3 +x*O(x^n)); n!*polcoeff(exp(3*x*G^2)/G^2, n)}
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n) = if(n==0|n==1, 1, sum(k=0, n, 3^k * n!/k! * binomial(3*n-k-3, n-k) * (k-1)/(n-1) ))}
for(n=0, 20, print1(a(n), ", "))

Crossrefs

Cf. A251583, A251663, A001764, A006013.

Cf. Variants: A243953, A251574, A251575, A251576, A251577, A251578, A251579, A251580.

Sequence in context: A262939 A232470 A349961 * A265002 A012131 A322224

Adjacent sequences: A251570 A251571 A251572 * A251574 A251575 A251576

Keyword

nonn

Author

Paul D. Hanna, Dec 06 2014

Status

approved