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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.
12
1, 1, 3, 21, 261, 4833, 120303, 3778029, 143531433, 6404711553, 328447585179, 19037277446949, 1230842669484717, 87829738967634849, 6856701559496841159, 581343578623728854397, 53196439113856500195537, 5225543459274294130169601, 548468830470032135590262067, 61258398893626609968686844597
OFFSET
0,3
LINKS
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), ", "))
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 06 2014
STATUS
approved