login
G.f. A(x) = Product_{n>=0} F(n), where F(0) = x, F(1) = 1+x, F(2) = 1 + x*(1+x), and F(n+1) = 1 + (F(n-1) - 1)*(F(n) - 1)*F(n) for n > 1.
0

%I #6 Dec 31 2021 12:00:42

%S 1,2,3,6,13,25,47,97,202,405,804,1620,3317,6892,14473,30380,63169,

%T 129986,265696,541597,1104411,2258693,4640651,9583818,19888842,

%U 41440324,86582460,181122452,378791172,791010101,1648083517,3425009503,7100406046,14690106611,30348419552

%N G.f. A(x) = Product_{n>=0} F(n), where F(0) = x, F(1) = 1+x, F(2) = 1 + x*(1+x), and F(n+1) = 1 + (F(n-1) - 1)*(F(n) - 1)*F(n) for n > 1.

%e G.f.: A(x) = x + 2*x^2 + 3*x^3 + 6*x^4 + 13*x^5 + 25*x^6 + 47*x^7 + 97*x^8 + 202*x^9 + 405*x^10 + 804*x^11 + 1620*x^12 + ...

%e GENERATING METHOD.

%e Define F(n), a polynomial in x of order Pell(n), by the following recurrence involving Fibonacci numbers:

%e F(0) = x,

%e F(1) = (1 + x),

%e F(2) = (1 + x * (1+x)),

%e F(3) = (1 + x^2 * (1+x) * F(2)),

%e F(4) = (1 + x^3 * (1+x)^2 * F(2) * F(3)),

%e F(5) = (1 + x^5 * (1+x)^3 * F(2)^2 * F(3) * F(4)),

%e F(6) = (1 + x^8 * (1+x)^5 * F(2)^3 * F(3)^2 * F(4) * F(5)),

%e F(7) = (1 + x^13 * (1+x)^8 * F(2)^5 * F(3)^3 * F(4)^2 * F(5) * F(6)),

%e ...

%e F(n+1) = 1 + (F(n-1) - 1) * (F(n) - 1) * F(n)

%e ...

%e Then the g.f. A(x) equals the infinite product:

%e A(x) = x * F(1) * F(2) * F(3) * ... * F(n) * ...

%e that is,

%e A(x) = x * (1+x) * (1 + x*(1+x)) * (1 + x^2*(1+x)*(1 + x*(1+x))) * (1 + x^3*(1+x)^2*(1 + x*(1+x))*(1 + x^2*(1+x)*(1 + x*(1+x)))) * ...

%o (PARI) {F(n) = my(G=x); if(n==0,G=x,if(n==1,G=1+x,if(n==2,G=1+x*(1+x), G = 1 + (F(n-2) - 1)*(F(n-1) - 1)*F(n-1))));G}

%o {a(n) = my(A = prod(k=0,round(log(n)/log(3/2))+1, F(k) +x*O(x^n))); polcoeff(A,n)}

%o for(n=1,30,print1(a(n),", "))

%Y Cf. A350432.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Dec 30 2021