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A120879 G.f. satisfies: A(x) = A(x^3)*(1 + 3*x + 2*x^2). 1

%I #18 Sep 07 2018 22:00:09

%S 1,3,2,3,9,6,2,6,4,3,9,6,9,27,18,6,18,12,2,6,4,6,18,12,4,12,8,3,9,6,9,

%T 27,18,6,18,12,9,27,18,27,81,54,18,54,36,6,18,12,18,54,36,12,36,24,2,

%U 6,4,6,18,12,4,12,8,6,18,12,18,54,36,12,36,24,4,12,8,12,36,24,8,24,16,3,9

%N G.f. satisfies: A(x) = A(x^3)*(1 + 3*x + 2*x^2).

%C a(n) = 3^A062756(n) * 2^A081603(n), where A062756(n) is the number of 1's and A081603(n) is the number of 2's, in the ternary expansion of n.

%C More generally, if g.f. of {a(n)} satisfies: A(x) = A(x^d)*(1+Sum_{k=1..d-1} c(k)*x^k), then a(n) = Product_{k=1..d-1} c(k)^digits(n,k,d), where digits(n,k,d) is the number of k's in the d-ary expansion of n and d is any integer > 1. This sequence is a simple example for d=3 with c(1)=3 and c(2)=2.

%H Paul D. Hanna, <a href="/A120879/b120879.txt">Table of n, a(n) for n = 0..10000</a>

%F G.f.: A(x) = Product_{n>=0} (1 + x^(3^n))*(1 + 2*x^(3^n)).

%F a(n) = a(floor(n/3)) * 3^((n mod 3) mod 2) * 2^floor((n mod 3)/2) with a(0)=1.

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

%o (PARI) /* Recurrence: */ {a(n)=if(n==0,1,a(n\3)*3^((n%3)%2)*2^((n%3)\2))}

%Y Cf. A120880, A062756, A081603.

%K nonn,look

%O 0,2

%A _Paul D. Hanna_, Jul 11 2006

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