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 ).
(5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251583.
(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
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 06 2014
STATUS
approved