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A161361
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Convolution square root of A000521.
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21
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1, 372, 29250, -134120, 54261375, -6139293372, 854279148734, -128813964933000, 20657907916144515, -3469030105750871000, 603760629237519966018, -108124880417607682194048, 19820541224206810447813500
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OFFSET
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0,2
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COMMENTS
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Triangle A161362 = the corresponding convolution triangle with row sums = A000521.
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LINKS
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FORMULA
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Given A000521: (j = 1/q + 744 + 196884q + 21493760q^2 + 864299970q^3 + ...); multiply by q and take the convolution square root.
G.f. is a period 1 Fourier series which satisfies f(-1 / (4 t)) = f(t) where q = exp(2 Pi i t). - Michael Somos, May 03 2014
a(n) ~ (-1)^n * c * exp(Pi*sqrt(3)*n) / n^(5/2), where c = 0.378271951998085144930610869223050101960774818... = 3^(5/2) * Gamma(1/3)^9 / (2^(7/2) * exp(sqrt(3) * Pi/2) * Pi^(13/2)). - Vaclav Kotesovec, Jul 03 2017, updated Mar 06 2018
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EXAMPLE
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a(2) = 29250 = 1/2 * (A000521(2) - 372^2) = 1/2 * (196884 - 138384) = 29250.
G.f. = 1 + 372*x + 29250*x^2 - 134120*x^3 + 54261375*x^4 - ...
G.f. = 1/q + 372*q + 29250*q^3 - 134120*q^5 + 54261375*q^7 + ...
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MATHEMATICA
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CoefficientList[Series[(65536 + x*QPochhammer[-1, x]^24)^(3/2) / (4096 * QPochhammer[-1, x]^12), {x, 0, 20}], x] (* Vaclav Kotesovec, Sep 23 2017 *)
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PROG
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(PARI) {a(n) = local(A); if( n<0, 0, A = x * O(x^n); A = x * (eta(x^2 + A) / eta(x + A))^24; polcoeff( sqrt(x * (1 + 256*A)^3 / A), n))}; /* Michael Somos, May 03 2014 */
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CROSSREFS
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(q*j(q))^(k/24): A289397 (k=-1), A106205 (k=1), A289297 (k=2), A289298 (k=3), A289299 (k=4), A289300 (k=5), A289301 (k=6), A289302 (k=7), A007245 (k=8), A289303 (k=9), A289304(k=10), A289305 (k=11), this sequence (k=12).
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KEYWORD
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sign
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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