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%I #11 Aug 26 2015 15:31:33
%S 1,-1,-1,0,0,-2,3,4,0,0,-3,0,-4,0,0,4,-5,-2,0,0,8,0,6,0,0,0,1,-8,0,0,
%T -12,4,7,0,0,-8,0,-2,0,0,2,5,7,0,0,10,0,0,0,0,4,-8,-16,0,0,4,-10,10,0,
%U 0,0,0,-12,0,0,-2,5,0,0,0,6,0,14,0,0,-8,20
%N Expansion of q^(-2/3) * eta(q) * eta(q^5)^3 in powers of q.
%H M. D. Rogers, <a href="http://arxiv.org/abs/0806.3590">Hypergeometric formulas for lattice sums and Mahler measures</a>. see p. 4 eq. (1.8)
%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 Euler transform of period 5 sequence [ -1, -1, -1, -1, -4, ...].
%F G.f. is a period 1 Fourier series which satisfies f(-1 / (45 t)) = 405^(1/2) (t/i)^2 g(t) where q = exp(2 Pi i t) and g() is the g.f. for A122274. - _Michael Somos_, Jun 23 2012
%F G.f.: Product_{k>0} (1 - x^k) * (1 - x^(5*k))^3.
%F a(5*n + 3) = a(5*n + 4) = 0. 5*a(n)= A122274(5*n + 3). a(5*n + 1) = -A122274(n).
%e G.f. = 1 - x - x^2 - 2*x^5 + 3*x^6 + 4*x^7 - 3*x^10 - 4*x^12 + 4*x^15 + ...
%e G.f. = q^2 - q^5 - q^8 - 2*q^17 + 3*q^20 + 4*q^23 - 3*q^32 - 4*q^38 + ...
%t a[ n_] := SeriesCoefficient[ QPochhammer[ x] QPochhammer[ x^5]^3, {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) * eta(x^5 + A)^3, n))};
%o (PARI) {a(n) = my(A, p, e, w, x, y, z, a0, a1); if( n<0, 0, n = 3*n + 2; A = factor(n); w = quadgen(-20); simplify( 1/w * 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. A122274.
%K sign
%O 0,6
%A _Michael Somos_, Sep 08 2006
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