A122274 - OEIS

%I #9 Aug 26 2015 15:21:28

%S 1,-3,0,5,0,-1,-4,0,-5,0,8,10,0,-5,0,-20,7,0,0,0,2,13,0,0,0,12,-16,0,

%T -10,0,0,-20,0,15,0,10,14,0,20,0,-11,-24,0,0,0,-10,-4,0,0,0,8,0,0,-15,

%U 0,40,13,0,0,0,-22,20,0,-20,0,-21,-16,0,0,0,-28,40,0,0,0,10,26,0,20,0,2,-6,0,-25,0,-9,0,0,-10

%N Expansion of q^(-1/3) * eta(q)^3 * eta(q^5) in powers of q.

%F Euler transform of period 5 sequence [ -3, -3, -3, -3, -4, ...].

%F Define c(3*k + 1) = A122274(k), c(3*k + 2) = sqrt(-5) * A122275(k), c(3*k) = 0. Then c(n) is multiplicative with c(3^e) = 0^e, c(p^e) = p^(e/2)*(1 + (-1)^e)/2 if p == 7, 13 (mod 15), c(p^e) = (-p)^(e/2)*(1 + (-1)^e)/2 if p == 11, 14 (mod 15), c(p^e) = c(p) * c(p^(e-1)) - p * c(p^(e-2)) if p == 1, 4 (mod 15), c(p^e) = c(p) * c(p^(e-1)) + p * c(p^(e-2)) if p == 2, 8 (mod 15), where c(p) is determined by c(2) = 1, or c(p) = 2*x when p = x^2 + 15*y^2, x == 1 (mod 3) for p == 1, 4 (mod 15) and c(p) = 2*x * sqrt(-5) when p = 5*x^2 + 3*y^2, x == 2 (mod 3) for p == 2, 8 (mod 15).

%F G.f. is a period 1 Fourier series which satisfies f(-1 / (45 t)) = 10125^(1/2) (t/i)^2 g(t) where q = exp(2 Pi i t) and g() is the g.f. for A122275. - _Michael Somos_, Jun 23 2012

%F G.f.: Product_{k>0} (1 - x^k)^3 * (1 - x^(5*k)).

%F a(5*n + 2) = a(5*n + 4) = 0. a(5*n + 3) = 5 * A122275(n). -a(n) = A122275(5*n + 1).

%e G.f. = 1 - 3*x + 5*x^3 - x^5 - 4*x^6 - 5*x^8 + 8*x^10 + 10*x^11 - 5*x^13 + ...

%e G.f. = q - 3*q^4 + 5*q^10 - q^16 - 4*q^19 - 5*q^25 + 8*q^31 + 10*q^34 + ...

%t a[ n_] := SeriesCoefficient[ QPochhammer[ x]^3 QPochhammer[ x^5], {x, 0, n}]; (* _Michael Somos_, Jun 23 2012 *)

%o (PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^3 * eta(x^5 + A), n))}

%o (PARI) {a(n) = my(A, p, e, w, x, y, z, a0, a1); if( n<0, 0, n = 3*n + 1; A = factor(n); w = quadgen(-20); prod( k=1, matsize(A)[1], [p, e] = A[k, ]; if( p==3, 0, z = p * kronecker( 5, p); if( p==2, x=1, p==5, x=-1, (p%15) != 2^valuation( p%15, 2), x=0, if( valuation( p%15, 2)%2, for( y=1, sqrtint(p\3), if( issquare( (p-3*y^2)/5, &x), break)), for( y=1, sqrtint(p\15), if( issquare( p - 15*y^2, &x), break))); x *= 2*(-1)^(p%3 != x%3)); a0=1; a1 = y = x * if( p%3==1, 1, w); for( i=2, e, x = y*a 1 - z*a0; a0=a1; a1=x); a1))) };

%Y Cf. A122275.

%K sign

%O 0,2

%A _Michael Somos_, Sep 08 2006