OFFSET
0,2
COMMENTS
A110649 is formed from every 2nd term of A084067, which also consists entirely of numbers 1 through 12.
Why are so many a(n) divisible by 3, i.e., 22 of the first 28? - Jonathan Vos Post, Sep 14 2005
FORMULA
G.f. A(x) satisfies: A(x)^4 (mod 8) = g.f. of A084067.
EXAMPLE
A(x) = 1 + 3*x + 2*x^3 + 3*x^5 - 8*x^6 + 30*x^7 - 90*x^8 +..
A(x)^2 = 1 + 6*x + 9*x^2 + 4*x^3 + 12*x^4 + 6*x^5 +...
A(x)^4 = 1 + 12*x + 54*x^2 + 116*x^3 + 153*x^4 + 228*x^5 +..
A(x)^4 (mod 8) = 1 + 4*x + 6*x^2 + 4*x^3 + x^4 + 4*x^5 +...
G(x) = 1 + 12*x + 6*x^2 + 4*x^3 + 9*x^4 + 12*x^5 + 4*x^6 +..
where G(x) is the g.f. of A084067.
PROG
(PARI) {a(n)=local(d=2, m=12, A=1+m*x); for(j=2, d*n, for(k=1, m, t=polcoeff((A+k*x^j+x*O(x^j))^(1/m), j); if(denominator(t)==1, A=A+k*x^j; break))); polcoeff(Ser(vector(n+1, i, polcoeff(A, d*(i-1))))^(1/2), n)}
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Sep 14 2005
STATUS
approved