A112574 G.f. A(x) satisfies: A(x)^2 equals the g.f. of A110649, which consists entirely of numbers 1 through 12.

1, 3, 0, 2, 0, 3, -8, 30, -90, 290, -930, 3000, -9696, 31461, -102420, 334467, -1095510, 3598464, -11852026, 39136629, -129548493, 429817733, -1429178703, 4761992751, -15898024868, 53174651133, -178168302693, 597971203902, -2010093276240, 6767100270918

Offset: 0

Author

Paul D. Hanna, Sep 14 2005

Keywords

• sign

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

Examples

			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.
		

Crossrefs

Cf. A110649, A084067.

Programs

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)}

Formula

G.f. A(x) satisfies: A(x)^4 (mod 8) = g.f. of A084067.