|
%I #18 Jan 04 2013 05:06:10
%S 1,1,2,3,9,15,75,133,676,1712,7295,16797,100889,208013,1083061,
%T 3186268,14634359,35357671,209715154,477638701,2679305455,7638371497,
%U 35224117709,91482563641,568102780247,1311836293464,6979729709485,21256008509080,103440124113965
%N G.f.: 1 + Sum_{n>=1} x^n * exp( Sum_{k>=1} binomial(2*n*k-1, n*k) * x^(n*k)/k ).
%H Paul D. Hanna, <a href="/A206777/b206777.txt">Table of n, a(n) for n = 0..500</a>
%F G.f.: Sum_{n>=0} x^n * Product_{k=0..n-1} C(u(n)^k*x), where u(n) = exp(2*Pi*I/n) is an n-th root of unity, and C(x) = (1-sqrt(1-4*x))/(2*x) is the g.f. of the Catalan numbers (A000108).
%e G.f.: A(x) = 1 + x + 2*x^2 + 3*x^3 + 9*x^4 + 15*x^5 + 75*x^6 + 133*x^7 +...
%e Let C(x) = (1-sqrt(1-4*x))/(2*x), then the g.f. A(x) equals the series:
%e A(x) = 1 + x*C(x) + x^2*C(x)*C(-x) + x^3*C(x)*C(u(3)*x)*C(u(3)^2*x) + x^4*C(x)*C(I*x)*C(-x)*C(-I*x) + x^5*C(x)*C(u(5)*x)*C(u(5)^2*x)*C(u(5)^3*x)*C(u(5)^4*x) +...
%e where u(n) = exp(2*Pi*I/n).
%e The expansions of P(n) = Product_{k=0..n-1} C(u(n)^k*x) begin:
%e P(1) = C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 +...
%e P(2) = C(x)*C(-x) = 1 + 3*x^2 + 22*x^4 + 211*x^6 + 2306*x^8 +...
%e P(3) = 1 + 10*x^3 + 281*x^6 + 10580*x^9 + 457700*x^12 + 21475122*x^15 +...
%e P(4) = 1 + 35*x^4 + 3830*x^8 + 570451*x^12 + 98118690*x^16 +...
%e P(5) = 1 + 126*x^5 + 54127*x^10 + 32006130*x^15 + 21932146139*x^20 +...
%e P(6) = 1 + 462*x^6 + 782761*x^12 + 1841287756*x^18 +...
%e P(7) = 1 + 1716*x^7 + 11501478*x^14 + 107761838756*x^21 +...
%e P(8) = 1 + 6435*x^8 + 170974710*x^16 + 6386000085075*x^24 +...
%e such that A(x) = 1 + x*C(x) + x^2*P(2) + x^3*P(3) + x^4*P(4) +...
%o (PARI) {a(n)=polcoeff(1+sum(m=1,n,x^m*exp(sum(k=1,n\m,binomial(2*m*k-1,m*k)*x^(m*k)/k)+x*O(x^n))),n)}
%o for(n=0,40,print1(a(n),", "))
%o (PARI) {a(n)=local(CATALAN=(1-sqrt(1-4*x+x^2*O(x^n)))/(2*x));A=1+sum(m=1,n+1,x^m*CATALAN*round(prod(k=1,m-1,subst(CATALAN,x,exp(2*Pi*I*k/m)*x+x*O(x^n)))));polcoeff(A,n)}
%o for(n=0,40,print1(a(n),", "))
%Y Cf. A205504, A000108.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Feb 12 2012
| 