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A122275
Expansion of q^(-2/3) * eta(q) * eta(q^5)^3 in powers of q.
2
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, -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, 0, 0, 0, -12, 0, 0, -2, 5, 0, 0, 0, 6, 0, 14, 0, 0, -8, 20
OFFSET
0,6
FORMULA
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).
Euler transform of period 5 sequence [ -1, -1, -1, -1, -4, ...].
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
G.f.: Product_{k>0} (1 - x^k) * (1 - x^(5*k))^3.
a(5*n + 3) = a(5*n + 4) = 0. 5*a(n)= A122274(5*n + 3). a(5*n + 1) = -A122274(n).
EXAMPLE
G.f. = 1 - x - x^2 - 2*x^5 + 3*x^6 + 4*x^7 - 3*x^10 - 4*x^12 + 4*x^15 + ...
G.f. = q^2 - q^5 - q^8 - 2*q^17 + 3*q^20 + 4*q^23 - 3*q^32 - 4*q^38 + ...
MATHEMATICA
a[ n_] := SeriesCoefficient[ QPochhammer[ x] QPochhammer[ x^5]^3, {x, 0, n}]; (* Michael Somos, Jun 23 2012 *)
PROG
(PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A) * eta(x^5 + A)^3, n))};
(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)))) };
CROSSREFS
Cf. A122274.
Sequence in context: A350775 A330267 A023868 * A197024 A031235 A090141
KEYWORD
sign
AUTHOR
Michael Somos, Sep 08 2006
STATUS
approved