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Re: different results

• To: mathgroup at smc.vnet.net
• Subject: [mg125328] Re: different results
• From: Roger Bagula <roger.bagula at gmail.com>
• Date: Wed, 7 Mar 2012 05:35:00 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201203050742.CAA29927@smc.vnet.net> <jj4qvt$ij8$1@smc.vnet.net>

Bob Hanlon got the definitive result.
Here is a 7th root version that gets Bob Hanlon's inverse root
and removes most of the zeros:
f[x_] = Product[1 + x^(8*i - 1), {i, 1, Infinity}]
g[x_] = Product[(1 - x^(8*i - 1)), {i, 1, Infinity}]
h[x_] = (1 + 2*f[x])/(-1 + g[x])
a = Table[SeriesCoefficient[
   Series[x*h[x^(1/7)], {x, 0, 50}], n], {n, 0, 50}]
ListPlot[a]
{-3, -2, 0, 0, 0, 0, 0, 0, 0, -3, 5, -21, 16, -20, -6, 13, -4, 3,
-21, 53, -205, 294, -481, 583, -633, 519, -164, -334, 765, -1901,
3707, -7263, 12241, -18113, 23208, -25644, 23554, -15343, -4280,
38143, -94739, 184290, -319135, 503453, -724127, 940484, -1084478,
1050440, -714961, -97011, 1629268}
On Mar 6, 3:04 am, Bob Hanlon <hanlonr... at gmail.com> wrote:
> f[x_] = Product[1 + x^(8*i - 1), {i, 1, Infinity}];
> g[x_] = Product[(1 - x^(8*i - 1)), {i, 1, Infinity}];
> h[x_] = (1 + 2*f[x])/(-1 + g[x]);
>
> Series[h[x], {x, 0, 10}] // Normal
>
> -2 - 3/x^7 + 3*x - 3*x^8
>
> The "first" coefficient for h[x] is for the x^-7 term. If you want all
> of the coefficients you need to look at x^7*h[x]
>
> CoefficientList[Series[x^7*h[x], {x, 0, 107}], x]
>
> {-3, 0, 0, 0, 0, 0, 0, -2, 3, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 0, -4, 3, \
> 0, 0, 0, 0, 0, 0, -3, -3, 0, 0, 0, 0, 0, -4, 2, 3, 0, 0, 0, 0, 0, -3, -9, -3, \
> 0, 0, 0, 0, -4, 1, 5, 3, 0, 0, 0, 0, 1, -15, -9, -3, 0, 0, 0, -4, 0, 6, 5, 3, \
> 0, 0, 0, 5, -20, -21, -12, -3, 0, 0, -4, -1, 16, 11, 11, 3, 0, 0, 9, -20, \
> -40, -36, -18, -3, 0, -4, -6, 31, 27, 25, 14, 3, 0, 13, -15, -61}
>
> Bob Hanlon
>
>
>
>
>
>
>
> On Mon, Mar 5, 2012 at 2:42 AM, Roger Bagula <roger.bag... at gmail.com> wrote:
> > Two versions of the code:
> > Table[SeriesCoefficient[
> > Series[(1 + 2*Product[1 + x^(8*i - 1), {i, 1, Infinity}])/(-1 +
> >  Product[(1 - x^(8*i - 1)), {i, 1, Infinity}]), {x, 0, 100}],
> > n], {n, 0, 100}]
>
> > f[x_] = Product[1 + x^(8*i - 1), {i, 1, Infinity}]
> > g[x_] = Product[(1 - x^(8*i - 1)), {i, 1, Infinity}]
> > h[x_] = (1 + 2*f[x])/(-1 + g[x])
A 7th root result that uses Bob /Hanlon's solution:
f[x_] = Product[1 + x^(8*i - 1), {i, 1, Infinity}]
g[x_] = Product[(1 - x^(8*i - 1)), {i, 1, Infinity}]
h[x_] = (1 + 2*f[x])/(-1 + g[x])
a = Table[SeriesCoefficient[
   Series[x*h[x^(1/7)], {x, 0, 50}], n], {n, 0, 50}]
ListPlot[a]

> > a = Table[SeriesCoefficient[
> >   Series[h[x], {x, 0, 100}], n], {n, 0, 100}]
>
> > Discussion at:https://oeis.org/draft/A208150
> > I get as I submitted:
> > DATA    -2, 3, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 0, -4, 3, 0, 0, 0,
> > 0, 0, 0, -3, -3, 0, 0, 0, 0, 0, -4, 2, 3, 0, 0, 0, 0, 0, -3, -9, -3,
> > 0, 0, 0, 0, -4, 1, 5, 3, 0, 0, 0, 0, 1, -15, -9, -3, 0, 0, 0, -4, 0,
> > 6, 5, 3, 0, 0
>
> > An OEIS editor gets:
> > [-3, 0, 0, 0, 0, 0, 0, -2, 3, 0, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 0,
> > -4, 3, 0, 0, 0, ...]
> > Roger Bagula

• References:
  • different results
    • From: Roger Bagula <roger.bagula@gmail.com>