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Search: id:A133089
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| A133089 |
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Expansion of f(q)^3 in powers of q where f() is a Ramanujan theta function. |
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+0
2
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| 1, 3, 0, -5, 0, 0, -7, 0, 0, 0, 9, 0, 0, 0, 0, 11, 0, 0, 0, 0, 0, -13, 0, 0, 0, 0, 0, 0, -15, 0, 0, 0, 0, 0, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0, 0, 19, 0, 0, 0, 0, 0, 0, 0, 0, 0, -21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -23, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 25, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 27, 0, 0, 0, 0, 0, 0, 0
(list; graph; listen)
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OFFSET
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0,2
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REFERENCES
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S. Ramanujan, Notebooks, Tata Institute of Fundamental Research, Bombay 1957 Vol. 1, see page 266 MR0099904 (20 #6340)
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FORMULA
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Expansion of q^(-1/8) * ( eta(q^2)^3 / ( eta(q) * eta(q^4) ) )^3 in powers of q.
Euler transform of period 4 sequence [ 3, -6, 3, -3, ...].
a(n) = b(8*n+1) where b(n) is multiplicative and b(2^e) = 0^e, b(p^e) = (1+(-1)^e)/2 * p^(e/2) if p == 1, 3 (mod 8), b(p^e) = (1+(-1)^e)/2 * (-p)^(e/2) if p == 5, 7 (mod 8).
G.f. is a period 1 Fourier series which satisfies f(-1 / (256 t)) = 64 (t/i)^(3/2) f(t) where q = exp(2 pi i t).
a(3*n+2) = a(5*n+2) = a(5*n+4) = a(9*n+4) = a(9*n+7) = 0. a(9*n+1) = 3 * a(n). a(25*n+3) = -5 * a(n).
G.f.: Sum_{k>=0} (-1)^[k/2] * (2*k+1) * x^(k*(k+1))/2.
G.f.: ( Product_{k>0} (1 - x^k) * (1 + x^k)^2 / (1 + x^(2*k)) )^3.
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EXAMPLE
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q + 3*q^9 - 5*q^25 - 7*q^49 + 9*q^81 + 11*q^121 - 13*q^169 + ...
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PROGRAM
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(PARI) {a(n) = if(n<0, 0, if( issquare( 8*n+1, &n), (-1)^( (n-1) \ 4) * n))}
(PARI) {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x^2 + A)^3 / eta(x + A) / eta(x^4 + A))^3, n))}
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CROSSREFS
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A010816(n) = -(-1)^n * a(n). A133079(n) = a(3*n).
Sequence in context: A052439 A143073 A010816 this_sequence A136599 A131986 A002656
Adjacent sequences: A133086 A133087 A133088 this_sequence A133090 A133091 A133092
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KEYWORD
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sign
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AUTHOR
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Michael Somos, Sep 09 2007
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