OFFSET
0,3
FORMULA
a(n) = A105558(n)/(n+1) = A105556(2n,n)/(n+1) = [N^(n+1)](n+1,1)/(n+1) for n>=0, where N^(n+1) is the (n+1)-th matrix power of the Narayana triangle N=A001263.
G.f.: A(x) = Series_Reversion[x/B(x)]/x where B(x) = A(x/B(x)) = Sum_{n>=0} x^n/[n!*(n+1)!/2^n].
G.f. satisfies: A(x) = C(x*A(x)^2) and C(x) = A(x/C(x)^2) where C(x) is the g.f. of A155927.
EXAMPLE
G.f.: A(x) = 1 + x + 4*x^2/3 + 37*x^3/18 + 621*x^4/180 + 16526*x^5/2700 +...+ a(n)*x^n/[n!*(n+1)!/2^n] +...
B(x) = 1 + x + 1/3*x^2 + 1/18*x^3 + 1/180*x^4 +...+ x^n/[n!*(n+1)!/2^n] +... where
A(x) = B(x*A(x)) and B(x) = A(x/B(x)) ;
1/B(x) = 1 - x + 2*x^2/3 - 7*x^3/18 + 39*x^4/180 - 321*x^5/2700 +...+ (-1)^n*A103365(n)*x^n/[n!*(n+1)!/2^n] +...
Also, A(x) = C(x*A(x)^2) where:
C(x) = 1 + x - 2*x^2/3 + 19*x^3/18 - 379*x^4/180 + 12726*x^5/2700 +...+ A155927(n)*x^(n+1)/[n!*(n+1)!/2^n] +...
A(x)^2 = 1 + 2*x + 11*x^2/3 + 122*x^3/18 + 2302*x^4/180 + 66482*x^5/2700 +...
PROG
(PARI) {a(n)=local(F=sum(k=0, n, x^k/(k!*(k+1)!/2^k))+x*O(x^n)); polcoeff(serreverse(x/F)/x, n)*n!*(n+1)!/2^n}
(PARI) {a(n)=local(N=matrix(n+1, n+1, m, j, if(m>=j, binomial(m-1, j-1)*binomial(m, j-1)/j))); sum(j=0, n, (N^n)[n+1, j+1])/(n+1)}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 30 2009
STATUS
approved