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G.f.: exp( Sum_{n>=1} 3^b(n) * x^n/n ) where b(n) = highest exponent of 3 in 2^n+1.
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%I #12 Apr 19 2012 10:20:14

%S 1,3,5,9,15,21,29,39,49,63,81,99,123,153,183,219,261,303,353,411,469,

%T 537,615,693,781,879,977,1089,1215,1341,1485,1647,1809,1989,2187,2385,

%U 2607,2853,3099,3375,3681,3987,4323,4689,5055,5457,5895,6333,6813,7335,7857,8421

%N G.f.: exp( Sum_{n>=1} 3^b(n) * x^n/n ) where b(n) = highest exponent of 3 in 2^n+1.

%F G.f. satisfies: A(x) = (1-x^2)*(1-x^3)/(1-x)^3 * A(x^3).

%F Define TRISECTIONS: A(x) = T_0(x^3) + x*T_1(x^3) + x^2*T_2(x^3), then:

%F (1) T_1(x)/T_0(x) = 3*(1+x)/(1+5*x),

%F (2) T_2(x)/T_0(x) = (5+x)/(1+5*x),

%F (3) T_0(x)/T_0(x^3) = (1+x)*(1+5*x)*(1-x^3)^2 / ((1-x)^3*(1+5*x^3)),

%F (4) T_1(x)/T_1(x^3) = (1+x)^2*(1-x^3)^2 / ((1-x)^3*(1+x^3)),

%F (5) T_2(x)/T_2(x^3) = (1+x)*(5+x)*(1-x^3)^2 / ((1-x)^3*(5+x^3)),

%F (6) A(x) = (1-x)/(1+5*x)*T_0(x) = (1-x)/(1+x)*T_1(x)/3 = (1-x)/(5+x)*T_2(x).

%e G.f.: A(x) = 1 + 3*x + 5*x^2 + 9*x^3 + 15*x^4 + 21*x^5 + 29*x^6 + 39*x^7 +...

%e The g.f. satisfies:

%e A(x)/A(x^3) = 1 + 3*x + 5*x^2 + 6*x^3 + 6*x^4 + 6*x^5 +...+ 6*x^n +...

%e The logarithm of the g.f. begins:

%e log(A(x)) = 3*x + x^2/2 + 9*x^3/3 + x^4/4 + 3*x^5/5 + x^6/6 + 3*x^7/7 + x^8/8 + 27*x^9/9 + x^10/10 + 3*x^11/11 + x^12/12 +...+ 3^b(n)*x^n/n +...

%e where b(n) = highest exponent of 3 in 2^n+1, for n>=1, and begins:

%e b = [1,0,2,0,1,0,1,0,3,0,1,0,1,0,2,0,1,0,1,0,2,0,1,0,1,0,4,...].

%e The g.f.s of the TRISECTIONS begin:

%e T_0(x) = 1 + 9*x + 29*x^2 + 63*x^3 + 123*x^4 + 219*x^5 + 353*x^6 +...

%e T_1(x) = 3 + 15*x + 39*x^2 + 81*x^3 + 153*x^4 + 261*x^5 + 411*x^6 +...

%e T_2(x) = 5 + 21*x + 49*x^2 + 99*x^3 + 183*x^4 + 303*x^5 + 469*x^6 +...

%e where T_1(x)/T_0(x) = 3*(1+x)/(1+5*x), T_2(x)/T_0(x) = (5+x)/(1+5*x).

%o (PARI) {a(n)=polcoeff(exp(sum(m=1, n+1, 3^valuation(2^m+1, 3)*x^m/m)+x*O(x^n)), n)}

%o for(n=0, 65, print1(a(n), ", "))

%o (PARI) {a(n)=local(A=1+x);for(i=1,ceil(log(n+1)/log(3)),A=(1-x^2)*(1-x^3)/(1-x)^3*subst(A,x,x^3+x*O(x^n)));polcoeff(A,n)}

%Y Cf. A182000, A161809.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Apr 18 2012