A110628 Trisection of A083953 such that the self-convolution cube is congruent modulo 9 to A083953, which consists entirely of 1's, 2's and 3's.

1, 1, 3, 3, 1, 2, 2, 1, 2, 3, 2, 3, 3, 2, 2, 3, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 3, 1, 2, 1, 3, 1, 3, 1, 1, 1, 2, 2, 3, 3, 2, 3, 1, 2, 1, 3, 3, 2, 3, 3, 1, 2, 3, 3, 1, 3, 3, 2, 2, 2, 1, 2, 3, 3, 3, 3, 1, 2, 2, 3, 2, 1, 2, 2, 1, 2, 3, 3, 2, 2, 1, 1, 2, 1, 3, 2, 2, 2, 1, 3, 2, 2, 3, 3, 2, 3, 1, 1, 1, 1, 3, 3

Offset: 0

Author

Robert G. Wilson v and Paul D. Hanna, Aug 08 2005

Keywords

• nonn

Comments

Congruent modulo 3 to A084203 and A104405; the self-convolution cube of A084203 equals A083953.

Crossrefs

Cf. A083953, A111582, A084203, A104405.

Programs

PARI

{a(n)=local(p=3,A,C,X=x+x*O(x^(p*n)));if(n==0,1, A=sum(i=0,n-1,a(i)*x^(p*i))+p*x*((1-x^(p-1))/(1-X))/(1-X^p); for(k=1,p,C=polcoeff((A+k*x^(p*n))^(1/p),p*n); if(denominator(C)==1,return(k);break)))}

Formula

a(n) = A083953(3*n) for n>=0. G.f. satisfies: A(x^3) = G(x) - 3*x*(1+x)/(1-x^3), where G(x) is the g.f. of A083953. G.f. satisfies: A(x)^3 = A(x^3) + 3*x*(1+x)/(1-x^3) + 9*x^2*H(x) where H(x) is the g.f. of A111582.