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7

First we can see that ${\frac{(6k+1)^{k}}{(2k+5)^{k}}}$ behaves asymptotically as $3^k$, while $(z-2i)^k$ is divergent if $\|z-2i\|>1$, however when $\|z-2i\|<1$ it is convergent. For $\|z-2i\|=1$ this criterion is not conclusive. On the other hand we can carefully extend this argument to the full sequence, so we need only \$\|z-2i\| < ...

2

k[] is function defined by you and tri is making UndirectedEdge-s k[j_List] := Block[{t = j[[1]], r = j[[2]], i = j[[3]]},(2 t - 1) 2^(-i + r) 3^(-1 + i)] tri[{t_, r_, i_}] := {{t, r, i} <-> {t, r + 1, i}, {t, r + 1, i} <-> {t, r + 1, i + 1},{t, r + 1, i + 1} <-> {t, r, i}} I remaked your code like this for making Graph. d = 4; data ...

6

I changed tri to List of your code. d = 9; x = {}; For[t = 1, t <= d, t++, If[t < d, AppendTo[x, {t, 1, 1} <-> {t + 1, 1, 1}]]; For[r = 2, r <= d, r++, If[t < d, AppendTo[x, {t, r, 1} <-> {t + 1, r, 1}]]; For[i = 2, i <= r, i++, If[t < d, AppendTo[x, {t, r, i} <-> {t + 1, r, i}]]; AppendTo[x, {t, r, i - 1} ...

5

That is not, properly speaking, a partial fraction decomposition. Those are finite and involve polynomial denominators. What you showed is a form of series representation. That can be done in Mathematica using SeriesCoefficient. In[102]:= SeriesCoefficient[1/(Exp[x] - 1), {x, 0, n}] (* Out[102]= Piecewise[{{BernoulliB[1 + n]/(1 + n)!, n >= -1}}, 0] *)

3

I'll use the function from Factoring polynomials to factors involving complex coefficients by Daniel Lichtblau which can also factor -1 + a x^2 factorCompletely[poly_, x_] := Module[{solns, lcoeff}, solns = Solve[poly == 0, x, Cubics -> False, Quartics -> False]; lcoeff = Coefficient[poly, x^Exponent[poly, x]]; lcoeff*(Times @@ (x - (x /. ...

5

Apart only factorizes the denominator in simple cases. Otherwise it needs some help: Apart[1/Factor[(x^2 + 1), Extension -> I]] (* -(I/(2 (-I + x))) + I/(2 (I + x)) *) Apart[1/Factor[x^2 + 2, Extension -> {I, Sqrt[2]}]] (* 1/(2 Sqrt[2] (Sqrt[2] - I x)) + 1/(2 Sqrt[2] (Sqrt[2] + I x)) *)

1

You can apply your transformation to your sample data and use EmpiricalDistribution on the transformed data without having to use TransformedDistribution: data = RandomVariate[ExponentialDistribution[1], 10^4]; ed = EmpiricalDistribution[data]; edtr = EmpiricalDistribution[Sqrt@data]; Plot[{CDF[ed, x], CDF[edtr, x]}, {x, 0, 4}, PlotLegends -> ...

0

Denominators of first sequence: A003418, which are symmetric sequences of greatest divisors <= ii. The denominators of the differences are these greatest divisors. See here and here. Edit: I missed that ii=9 is 840 instead of 2520 in the first sequence and is a zero instead of 9 in the second. So, the symmetry stops at 9. I'll leave this answer in case ...

8

First I create a set of data to simulate yours. data = RandomVariate[ExponentialDistribution[1], 10^4]; Now you can take advantage of the EmpiricalDistribution function to define a model-free distribution based on your data. edist = EmpiricalDistribution[data]; The core of what you are asking for is to obtain a TransformedDistribution, i.e starting ...

3

my stab at it, unfortunately only a marginal improvement in time: ii=13; Clear[a, b]; b = FoldList[Times, 1, Table[ Exp[MangoldtLambda[n]], {n, 2, ii}]]; a = Prepend[Table[ Limit[ Zeta[s] Total[Exp[Divisors[n]]^(s - 1) MoebiusMu[Divisors[n]]], s -> 1], {n, 2, ii}], 1] ; Monitor[aa = Prepend[Table[ ...

2

I realized now that I included unnecessary many terms of the Dirichlet inverse of the Euler totient. Therefore a better program is: ii = 13 aa = Range[ii]*0; Monitor[Do[ Clear[A, a, b, n, k]; b = Table[Product[Exp[MangoldtLambda[n]], {n, 1, k}], {k, 1, nn}]; a = Table[ If[n == 1, 1, Limit[Zeta[s] Total[ Exp[Divisors[n]]^(s - ...

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