Uniform Preferential Selection Model for Generating Scale-free Networks

Abstract

It has been observed in real networks that the fraction of nodes P(k) with degree k satisfies the power-law P(k) ∝ k^−γ for k > k_min > 0. However, the degree distribution of nodes in these networks before k_min varies slowly to the extent of being uniform as compared to the degree distribution after k_min. Most of the previous studies focus on the degree distribution after k_min and ignore the initial flatness in the distribution of degrees. In this paper, we propose a model that describes the degree distribution for the whole range of k > 0, i.e., before and after k_min. The network evolution is made up of two steps. In the first step, a new node is connected to the network through a preferential attachment method. In the second step, a certain number of edges between the existing nodes are added such that the end nodes of an edge are selected either uniformly or preferentially. The model has a parameter to control the uniform or preferential selection of nodes for creating edges in the network. We perform a comprehensive mathematical analysis of our proposed model in the discrete domain and prove that the model exhibits an asymptotically power-law degree distribution after k_min and a flat-ish distribution before k_min. We also develop an algorithm that guides us in determining the model parameters in order to fit the model output to the node degree distribution of a given real network. Our simulation results show that the degree distributions of the graphs generated by this model match well with those of the real-world graphs.

Appendix A: Formulation and Derivation of Recurrence Relation for Degree Distribution

In upcoming sections, we will find the exact form of a recurrence relation for the degree distribution and evaluate the relation for the limit $t \to \infty$.

1.1 A.1 Edge-Step for k = 1

A recurrence relation for k = 1 will be formed and its value for $t \to \infty$ will be determined in this section.

Rewriting Eq. 9 in the following form,

$$ N_{1,t+1,m} = \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr) N_{1,t+1,m-1}, $$

and repeatedly using the above relation for $m=1,2,\dots$, we can write N_1,t+ 1 in terms of N_1,t+ 1,0,

$$ N_{1,t+1} = \prod\limits_{m=1}^{{{\varDelta}} e_{t}} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr) N_{1,t+1,0}. $$

Using the relation given by Eq. 4, we get the following expression for N_1,t+ 1,

$$ N_{1,t+1} = \left[\prod\limits_{m=1}^{{{\varDelta}} e_{t}} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr) \right] \biggl(1 + \left( 1- \frac{1}{2e_{t}}\right)N_{1,t} \biggr), $$

which on simplification becomes,

$$ N_{1,t+1} = \prod\limits_{m=1}^{{{\varDelta}} e_{t}} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr) + \biggl(1- \frac{1}{2e_{t}}\biggr)\prod\limits_{m=1}^{{{\varDelta}} e_{t}} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr) N_{1,t}. $$

Using $P_{1,t} = \frac {N_{1,t}}{t}$, above relation reduces to

$$ (t+1)P_{1,t+1} = \prod\limits_{m=1}^{{{\varDelta}} e_{t}} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr) + \biggl(1- \frac{1}{2e_{t}}\biggr)tP_{1,t}\prod\limits_{m=1}^{{{\varDelta}} e_{t}} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{e_{t}+m} \biggr). $$

Replacing Δe_t by α, we get,

$$ \begin{array}{@{}rcl@{}} (t+1)P_{1,t+1} &=& \prod\limits_{m=1}^{\alpha} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{(\alpha+1)(t-1)+m} \biggr)\\ &&+\biggl(1- \frac{1}{2(\alpha+1)(t-1) }\biggr)tP_{1,t} \prod\limits_{m=1}^{\alpha} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{(\alpha+1)(t-1)+ m} \biggr). \end{array} $$

Now consider a term occurring on R.H.S., which can be re-written in the form,

$$ \begin{array}{@{}rcl@{}} &&\prod\limits_{m=1}^{\alpha} \biggl(1 - \frac{2\delta}{t+1}- \frac{1-\delta}{(\alpha+1)(t-1)+m} \biggr)\\ &&= 1 - \sum\limits_{m=1}^{\alpha}\frac{2\delta}{t+1}+ \frac{1-\delta}{(\alpha+1)(t-1)+m}+ \text{ terms of order less than } O\left( \frac{1}{t}\right). \end{array} $$

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This reduces the degree-distribution relationship into the following form,

$$ \begin{array}{@{}rcl@{}} tP_{1,t+1} + P_{1,t+1} &=& 1 + tP_{1,t} - \frac{t}{2(\alpha+1)(t-1)} P_{1,t}\\ &&- tP_{1,t}\sum\limits_{m=1}^{\alpha} \biggl(\frac{2\delta}{t+1}+ \frac{1-\delta}{(\alpha+1)(t-1)+m}\biggr)\\ &&+\text{ terms of order less than } O(1). \end{array} $$

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Applying the limit $t \to \infty$, we get

$$ P_{1} = 1 - \frac{1}{2(\alpha+1)} P_{1} - \sum\limits_{m=1}^{\alpha}\biggl(2\delta+ \frac{1-\delta}{\alpha+1}\biggr)P_{1}, $$

which can be written as,

$$ P_{1} = \frac{2\alpha+2}{4\alpha^{2}\delta + 2\alpha \delta+4\alpha + 3}. $$

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1.2 A.2 Derivation of the Recurrence Relation for Degree Distribution k > 1

An exact form of the recurrence relation will be derived here. The recurrence formula writes N_{k, t+ 1}, number of nodes of degree k at time t + 1, in terms of N_{h, t} for $h=k,k-1, \dots$. The network evolves through both node- and edge- steps. The formulation is valid whether Δe_t is constant or varies over time t.

We introduce the following functions

$$ {{\varPsi}}_{k,m} = \biggl(1 - \frac{2\delta}{t+1}- \frac{k(1-\delta)}{e_{t}+m} \biggr) \text{ and } {{\varPhi}}_{k-1,m} = \biggl(\frac{2\delta}{t+1}+ \frac{(k-1)(1-\delta)}{e_{t}+m} \biggr)$$

and re-write Eq. 8 in the following form forms: for m = 1,

$$ N_{k,t+1,1} = {{\varPsi}}_{k,1} N_{k,t+1,0} + {{\varPhi}}_{k-1,1}N_{k-1,t+1,0}, $$

and in general for adding m − th edge,

$$ N_{k,t+1,m} = {{\varPsi}}_{k,m} N_{k,t+1,m-1} + {{\varPhi}}_{k-1,m} N_{k-1,t+1,m-1}. $$

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Define an operator $\mathcal {D}$ with the following properties,

• P1:

  $$ \mathcal{D}f_{k}= \left\{ \begin{array}{ll} f_{k-1} & \text{if } k > 1 \\ 0 & \text{if } k =1 \end{array} \right. $$

• P2: $\mathcal {D}^{n} f_{k}= \mathcal {D}^{n-1} (\mathcal {D}f_{k}) \text { for } n > 1$

• P3: $\mathcal {D} f_{k} h_{k} = (\mathcal {D}f_{k}) (\mathcal {D}h_{k})$

• P4: $\mathcal {D} (f_{k} + h_{k}) = (\mathcal {D}f_{k})+(\mathcal {D}h_{k})$

Here f_k and h_k can be either N_{k, t, m} ,Ψ_{k, m} or Φ_{k, m}.

$$ \begin{array}{@{}rcl@{}} N_{k,t+1,1} &=& ({{\varPsi}}_{k,1} + {{\varPhi}}_{k-1,1}\mathcal{D})N_{k,t+1,0},\\ N_{k,t+1,2} &=& ({{\varPsi}}_{k,2} + {{\varPhi}}_{k-1,2}\mathcal{D})N_{k,t+1,1},\\ N_{k,t+1,3} &=& ({{\varPsi}}_{k,3} + {{\varPhi}}_{k-1,3}\mathcal{D})N_{k,t+1,2},\\ \end{array} $$

Above three relations can be combined to rewrite N_{k, t+ 1,3} as,

$$ N_{k,t+1,3} = ({{\varPsi}}_{k,3} + {{\varPhi}}_{k-1,3}\mathcal{D})({{\varPsi}}_{k,2} + {{\varPhi}}_{k-1,2}\mathcal{D})({{\varPsi}}_{k,1} + {{\varPhi}}_{k-1,1}\mathcal{D})N_{k,t+1,0}, $$

In the general form, the edge-step gives the following relationship:

$$ N_{k,t+1}= \prod\limits_{i={{\varDelta}} e_{t}}^{1}({{\varPsi}}_{k,i} + {{\varPhi}}_{k-1,i}\mathcal{D})N_{k,t+1,0}. $$

The node-step can be written in a modified form,

$$ N_{k,t+1,0} = \biggl(1 - \frac{k}{2e_{t}} + \frac{k-1}{2e_{t}} \mathcal{D}\biggr)N_{k,t}. $$

implying

$$ N_{k,t+1} = \prod\limits_{i={{\varDelta}} e_{t}}^{1} \biggl({{\varPsi}}_{k,i} + {{\varPhi}}_{k-1,i}\mathcal{D}\biggr)\biggl(1 - \frac{k}{2e_{t}} + \frac{k-1}{2e_{t}} \mathcal{D}\biggr)N_{k,t}. $$

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This is a recurrence relation for k > 1 with $\mathcal {D}^{k} f_{k} = 0$. Note that product components do not commute due to the occurrence of the operator $\mathcal {D}$. If we apply the operator $\mathcal {D}$ to succeeding product components and collect coefficients of N_{k−n, t}’s, the above recurrence relation takes the following form,

$$ N_{k,t+1} = \sum\limits_{n=0}^{{{\varDelta}} e_{t}+1} C_{k-n,{{\varDelta}} e_{t}}N_{k-n,t}, $$

or

$$ (t+1)P_{k,t+1} = \sum\limits_{n=0}^{{{\varDelta}} e_{t}+1} C_{k-n,{{\varDelta}} e_{t}}tP_{k-n,t}. $$

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Next, we find expressions for these coefficients $C_{k-n,{{\varDelta }} e_{t}}$’s,

1.2.1 ${ C_{k, {{\varDelta }} e_{t}}}$

$C_{k, {{\varDelta }} e_{t}}$ is the coefficient of N_{k, t} and can be readily computed,

$$ C_{k,{{\varDelta}} e_{t}} = \biggl(1 - \frac{k}{2e_{t}}\biggr)\prod\limits_{m=1}^{{{\varDelta}} e_{t}} {{\varPsi}}_{k,m}. $$

1.2.2 ${ C_{k-1, {{\varDelta }} e_{t}}}$

Consider $C_{k-1,{{\varDelta }} e_{t}}$, coefficient of the N_k− 1,t. The product given by Eq. 20 is expanded into a sum of products. Products in which $\mathcal {D}$ appears exactly once, form the coefficient of $C_{k-1,{{\varDelta }} e_{t}}$. The coefficient $C_{k-1,{{\varDelta }} e_{t}}$ is the sum of two components, say, $C^{1}_{k-1,{{\varDelta }} e_{t}}$ and $C^{2}_{k-1,{{\varDelta }} e_{t}}$. This is based on the choice, whether $(1 - \frac {k}{2e_{t}})$ or $\frac {k-1}{2e_{t}}\mathcal {D}$ is chosen from the product component $(1 - \frac {k}{2e_{t}} + \frac {k-1}{2e_{t}} \mathcal {D})$ appearing in Eq. 20.

The component $C^{1}_{k-1,{{\varDelta }} e_{t}}$ is given as:

$$ C^{1}_{k-1,{{\varDelta}} e_{t}} = \prod\limits_{i={{\varDelta}} e_{t}}^{1}\biggl({{\varPsi}}_{k,i}\biggr)\biggl(\frac{k-1}{2e_{t}} \biggr) $$

For computing $C^{2}_{k-1,{{\varDelta }} e_{t}}$, we choose $\left (1 - \frac {k}{2e_{t}}\right )$ from the binomial $\left (1 - \frac {k}{2e_{t}} + \frac {k-1}{2e_{t}} \mathcal {D}\right )$. Now, consider the product of binomials, $\prod \limits _{i={{\varDelta }} e_{t}}^{1}({{\varPsi }}_{k,i} + {{\varPhi }}_{k-1,i}\mathcal {D})$. Exactly one term of the form ${{\varPhi }}_{k-1,i}\mathcal {D}$ is selected. If the term is selected at i − th position, we get

${{\varPsi }}_{k,{{\varDelta }} e_{t}} {\dots } {{\varPsi }}_{k,i+1} [{{\varPhi }}_{k-1,i}\mathcal {D}] {{\varPsi }}_{k,i-1}\dots {{\varPsi }}_{k,1}$, which briefly can be written in the form,

$$ \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}} {{\varPsi}}_{k,l} \right] \biggl[{{\varPhi}}_{k-1,i}\mathcal{D}\biggr] \left[\prod\limits_{m=1}^{i-1} {{\varPsi}}_{k,m}\right]. $$

Operator $\mathcal {D}$ will apply on each succeeding term Ψ_{k, m} (Rule P3).

$$ \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}} {{\varPsi}}_{k,l} \right] \biggl[{{\varPhi}}_{k-1,i}\biggr] \left[\prod\limits_{m=1}^{i-1} {{\varPsi}}_{k-1,m}\right]\mathcal{D}. $$

Now, the chosen position for Φ can vary from 1 to Δe_t. Resultantly, we get the sum of all such products and find $C^{2}_{k-1,{{\varDelta }} e_{t}}$,

$$ C^{2}_{k-1,{{\varDelta}} e_{t}} = \sum\limits_{i=1}^{{{\varDelta}} e_{t}}{{\varPhi}}_{k-1,i} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}} {{\varPsi}}_{k,l} \prod\limits_{m=1}^{i-1} {{\varPsi}}_{k-1,m} \right] \mathcal{D}\left( 1 - \frac{k}{2e_{t}}\right). $$

Eventually, coefficients of N_k− 1,t is

$$ C_{k-1,{{\varDelta}} e_{t}} = \left[\prod\limits_{i=1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,i}\right]\biggl(\frac{k-1}{2e_{t}} \biggr)+\sum\limits_{i=1}^{{{\varDelta}} e_{t}}{{\varPhi}}_{k-1,i} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}} {{\varPsi}}_{k,l} \prod\limits_{m=1}^{i-1} {{\varPsi}}_{k-1,m} \right] \biggl(1 - \frac{k-1}{2e_{t}}\biggr). $$

1.2.3 ${ C_{k-2, {{\varDelta }} e_{t}}}$

Now consider $C_{k-2,{{\varDelta }} e_{t}}$, the coefficient of N_k− 2,t. As was assumed previously, $C_{k-2,{{\varDelta }} e_{t}}$ is the sum of two components. For one component, we choose $(1 - \frac {k}{2e_{t}})$ from the the binomial $(1 - \frac {k}{2e_{t}} + \frac {k-1}{2e_{t}} \mathcal {D})$. To get an coefficient in N_k− 2,t, we have to select exactly two terms of the form ${{\varPhi }}_{k-1,i}\mathcal {D}$ from the product of binomials $\prod \limits _{i={{\varDelta }} e_{t}}^{1} \biggl ({{\varPsi }}_{k,i} + {{\varPhi }}_{k-1,i}\mathcal {D}\biggr )$. If we select these terms at i − th and j − th positions with i > j, we get,

${{\varPsi }}_{k,{{\varDelta }} e_{t}} {\dots } {{\varPsi }}_{k,i+1} ({{\varPhi }}_{k-1,i}\mathcal {D}) {{\varPsi }}_{k,i-1}{\dots } {\dots } {{\varPsi }}_{k,j+1} ({{\varPhi }}_{k-1,j}\mathcal {D}) {{\varPsi }}_{k,j-1}\dots {{\varPsi }}_{k,1}$

Operator $\mathcal {D}$ will apply on each succeeding term Ψ_{k, m} and Φ_{k, m} (Rule P3). In a brief form,

$$ \begin{array}{@{}rcl@{}} &&\left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,l}\right] \biggl[{{\varPhi}}_{k-1,i}\mathcal{D}\biggr] \left[\prod\limits_{m=j+1}^{i-1} {{\varPsi}}_{k,m}\right] \biggl[{{\varPhi}}_{k-1,j}\mathcal{D}\biggr] \left[\prod\limits_{n=1}^{j-1} {{\varPsi}}_{k,n}\right]\\ &&=\left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,l}\right] {{\varPhi}}_{k-1,i} \left[\prod\limits_{m=j+1}^{i-1}{{\varPsi}}_{k-1,m}\right] {{\varPhi}}_{k-2,j} \left[\prod\limits_{n=1}^{j-1} {{\varPsi}}_{k-2,n}\right] \mathcal{D}^{2}. \end{array} $$

Summing all such products,

$$ \sum\limits_{i={{\varDelta}} e_{t}}^{2}\sum\limits_{j=i-1}^{1} {{\varPhi}}_{k-1,i} {{\varPhi}}_{k-2,j} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,l}\right] \left[\prod\limits_{m=j+1}^{i-1} {{\varPsi}}_{k-1,m}\right] \left[\prod\limits_{n=1}^{j-1} {{\varPsi}}_{k-2,n}\right]. $$

Equivalently,

$$ \left( \sum\limits_{i=2}^{{{\varDelta}} e_{t}}\sum\limits_{j=1}^{i-1} {{\varPhi}}_{k-1,i} {{\varPhi}}_{k-2,j} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,l}\right] \left[\prod\limits_{m=j+1}^{i-1} {{\varPsi}}_{k-1,m}\right] \left[\prod\limits_{n=1}^{j-1} {{\varPsi}}_{k-2,n}\right]\right)\mathcal{D}^{2} \biggl(1 - \frac{k}{2e_{t}}\biggr), $$

which becomes

$$ \left( \sum\limits_{i=2}^{{{\varDelta}} e_{t}}\sum\limits_{j=1}^{i-1} {{\varPhi}}_{k-1,i} {{\varPhi}}_{k-2,j} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,l}\right] \left[\prod\limits_{m=j+1}^{i-1} {{\varPsi}}_{k-1,m}\right] \left[\prod\limits_{n=1}^{j-1} {{\varPsi}}_{k-2,n}\right] \right) \biggl(1 - \frac{k-2}{2e_{t}}\biggr)\mathcal{D}^{2}. $$

Adding contribution from the second component,

$$ \begin{array}{@{}rcl@{}} &&C_{k-2,{{\varDelta}} e_{t}} = \sum\limits_{i=1}^{{{\varDelta}} e_{t}}{{\varPhi}}_{k-1,i} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}} {{\varPsi}}_{k,l} \prod\limits_{m=1}^{i-1} {{\varPsi}}_{k-1,m} \right] \biggl(\frac{k-2}{2e_{t}}\biggr) \mathcal{D}^{2}\\ &&+ \left[\sum\limits_{i=2}^{{{\varDelta}} e_{t}}\sum\limits_{j=1}^{i-1} {{\varPhi}}_{k-1,i} {{\varPhi}}_{k-2,j} \left[\prod\limits_{l=i+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,l}\right] \left[\prod\limits_{m=j+1}^{i-1} {{\varPsi}}_{k-1,m}\right] \left[\prod\limits_{n=1}^{j-1} {{\varPsi}}_{k-2,n}\right]\right]\\&& \left( 1 - \frac{k-2}{2e_{t}}\right)\mathcal{D}^{2}. \end{array} $$

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1.2.4 ${C_{k-n,{{\varDelta }} e_{t}}}$ for $n = 1,2,3,\dots ,{{\varDelta }} e_{t}$

The general form for $C_{k-n,{{\varDelta }} e_{t}}$ can be obtained in the similar way. If we select n indices for Φ terms with $i_{n} > i_{n-1}>{\dots } >i_{1}$, then we get the coefficient,

$$ \begin{array}{@{}rcl@{}} &&C_{k-n,{{\varDelta}} e_{t}}\\ &&=\sum\limits_{i_{n}=n}^{{{\varDelta}} e_{t}}\sum\limits_{i_{n-1}=n-1}^{i_{n}-1}{\dots} \sum\limits_{i_{1}=1}^{i_{1}-1} \left[\prod\limits_{j=1}^{n}{{\varPhi}}_{k-j,i_{j}}\right] \left[\prod\limits_{m_{1}=i_{n}+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,m_{1}}\right]\left[\prod\limits_{m_{2}=i_{n-1}+1}^{i_{n}-1}{{\varPsi}}_{k-1,m_{1}} \right] \\ &&\quad\dots\left[\prod\limits_{m_{n+1}=1}^{i_{1} - 1}{{\varPsi}}_{k-n,m_{1}}\right]\left( 1 - \frac{k-n}{2e_{t}}\right)\\ &&\quad+\sum\limits_{i_{n-1}=n-1}^{{{\varDelta}} e_{t}}\sum\limits_{i_{n-2}=n-2}^{i_{n-1}-1}{\dots} \sum\limits_{i_{1}=1}^{i_{1}-1} \left[\prod\limits_{j=1}^{n-1}{{\varPhi}}_{k-j,i_{j}}\right] \left[\prod\limits_{m_{1}=i_{n-1}+1}^{{{\varDelta}} e_{t}}{{\varPsi}}_{k,m_{1}}\right]\left[\prod\limits_{m_{2}=i_{n-2}+1}^{i_{n-1}-1}{{\varPsi}}_{k-1,m_{1}} \right] \\ &&\quad\dots\left[\prod\limits_{m_{n}=1}^{i_{1} - 1}{{\varPsi}}_{k-n+1,m_{1}}\right]\biggl(\frac{k-n}{2e_{t}}\biggr), \end{array} $$

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or in the simplest form,

$$ \begin{array}{@{}rcl@{}} C_{k-n,{{\varDelta}} e_{t}} &=& \sum\limits_{i=1}^{{\binom{{{\varDelta}} e_{t}}{n}}}\left[\prod\limits_{j=1}^{n} {{\varPhi}} \prod\limits_{j=1}^{{{\varDelta}} e_{t}- n} {{\varPsi}} \right]\biggl(1 - \frac{k-n}{2e_{t}}\biggr) \\&&+\sum\limits_{i=1}^{{\binom{{{\varDelta}} e_{t}}{n-1}}}\left[\prod\limits_{j=1}^{n-1} {{\varPhi}} \prod\limits_{j=1}^{{{\varDelta}} e_{t}- n+1} {{\varPsi}} \right]\biggl(\frac{k-n}{2e_{t}}\biggr). \end{array} $$

1.3 A.3 Evaluation of the Recurrence Relation

This section determines the stationary degree distribution. It finds the limiting value ($\lim \limits t \to \infty$) of the recurrence relation (21). To accomplish this, we replace Δe_t by α and evaluate the limit $\lim \limits _{t \to \infty } t~C_{k-n, {{\varDelta }} e_{t}}$ for $n=0,1,2,\dots$.

$$ {{\varPsi}}_{k,m} = \biggl(1 - \frac{2\delta}{t+1}- \frac{k(1-\delta)}{e_{t}+m} \biggr) \text{ and } {{\varPhi}}_{k-1,m} = \biggl(\frac{2\delta}{t+1}+ \frac{(k-1)(1-\delta)}{e_{t}+m} \biggr) $$

1.3.1 C _{k, α}

$$ \begin{array}{@{}rcl@{}} tC_{k,\alpha} &=& t\biggl(1 - \frac{k}{2e_{t}}\biggr)\prod\limits_{m=1}^{\alpha} \biggl(1 - \frac{2\delta}{t+1}- \frac{k(1-\delta)}{(\alpha+1)(t-1)+m} \biggr)\\ &=& t\biggl(1 - \frac{k}{2(\alpha+1)(t-1)}\biggr) \biggl(1-\sum\limits_{m=1}^{\alpha} \bigl(\frac{2\delta}{t+1}+ \frac{k(1-\delta)}{(\alpha+1)(t-1)+ m} \bigr) + O\left( \frac{1}{t^{2}}\right) \biggr). \end{array} $$

Now expanding further and ignoring terms of order $\frac {1}{t}$ and less,

$$ \begin{array}{@{}rcl@{}} \lim\limits_{t \to \infty} tC_{k,\alpha}P_{k,t}& =& \lim\limits_{t \to \infty} t \biggl(1-\sum\limits_{m=1}^{\alpha} \bigl(\frac{2\delta}{t+1}+ \frac{k(1-\delta)}{(\alpha+1)(t-1)+ m} \bigr) - \frac{kt}{2(\alpha+1)(t-1)}\biggr)P_{k,t}\\ &=& \lim\limits_{t \to \infty}tP_{k,t} - \sum\limits_{m=1}^{\alpha} \biggl(2\delta+ \frac{k(1-\delta)}{\alpha+1} \biggr)P_{k} - \frac{k}{2(\alpha+1)} P_{k}\\ &=& \lim\limits_{t \to \infty}tP_{k,t} - \biggl(2\delta \alpha+ \frac{k(1-\delta)\alpha}{\alpha+1} + \frac{k}{2(\alpha+1)} \biggr)P_{k}\\ &=& \lim\limits_{t \to \infty}tP_{k,t} - \frac{(2\alpha -2\alpha \delta +1)k +4\alpha^{2} \delta +4\alpha \delta}{2(\alpha+1)}P_{k}. \end{array} $$

${{\varPsi }}_{k,m} = \bigl (1 - \frac {2\delta }{t+1}- \frac {k(1-\delta )}{e_{t}+m} \bigr )$ and ${{\varPhi }}_{k-1,m} = \bigl (\frac {2\delta }{t+1}+ \frac {(k-1)(1-\delta )}{e_{t}+m} \bigr )$

1.3.2 C _k− 1,α

C_k− 1,α is the coefficient of the term N_k− 1,t. It is the sum of two components.

$$ \begin{array}{@{}rcl@{}} t C_{k-1,\alpha} &=& t\left[\prod\limits_{i=1}^{\alpha}{{\varPsi}}_{k,i}\right]\biggl(\frac{k-1}{2(\alpha+1)(t-1)} \biggr)\\ && +t\sum\limits_{i=1}^{\alpha}{{\varPhi}}_{k-1,i} \left[\prod\limits_{l=i+1}^{\alpha} {{\varPsi}}_{k,l} \prod\limits_{m=1}^{i-1} {{\varPsi}}_{k-1,m} \right] \biggl(1 - \frac{k-1}{2(\alpha+1)(t-1)}\biggr). \end{array} $$

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Now expanding further and ignoring terms of order $\frac {1}{t}$ and less than it,

$$ tC_{k-1,\alpha} = t\biggl(\frac{k-1}{2(\alpha+1)(t-1)} \biggr)+t\sum\limits_{i=1}^{\alpha}\biggl(\frac{2\delta}{t+1}+ \frac{(k-1)(1-\delta)}{(\alpha+1)(t-1)+i} \biggr) (1)(1) (1). $$

$$ \begin{array}{@{}rcl@{}} \lim\limits_{t \to \infty} tC_{k-1,\alpha}P_{k-1,t} &=& \lim\limits_{t \to \infty} \biggl(\frac{(k-1)t}{2(\alpha+1)(t-1)} +\sum\limits_{i=1}^{\alpha}\bigl(\frac{2\delta t}{t+1}+ \frac{t(k-1)(1-\delta)}{(\alpha+1)(t-1)+i} \bigr) \biggr)P_{k-1,t}\\ &=& \biggl (\frac{k-1}{2(\alpha+1)} + \sum\limits_{i=1}^{\alpha}\left( 2\delta + \frac{(k-1)(1-\delta)}{\alpha+1} \right) \biggr)P_{k-1}\\ &=& \biggl(\frac{k-1}{2(\alpha+1)} + 2\delta \alpha + \frac{(k-1)(1-\delta)\alpha}{\alpha+1} \biggr) P_{k-1}\\ &=& \frac{ k-1 + 2 \delta \alpha(\alpha+1) + 2(k-1)(1-\delta)\alpha} {2(\alpha+1)} P_{k-1}\\ &=& \frac{(2\alpha -2\alpha \delta +1)k +4\alpha^{2} \delta +6\alpha \delta- 2\alpha-1 }{2(\alpha+1)}P_{k-1}. \end{array} $$

1.3.3 C _{k−n, α} for n = 2, 3,...α

$$ \begin{array}{@{}rcl@{}} C_{k-n,\alpha} &=& \sum\limits_{i=1}^{{\binom{\alpha}{n}}}\left[\prod\limits_{j=1}^{n} {{\varPhi}} \prod\limits_{j=1}^{{{\varDelta}} e_{t}- n} {{\varPsi}} \right]\biggl(1 - \frac{k-n}{2e_{t}}\biggr) \\&&+\sum\limits_{i=1}^{{\binom{\alpha}{n-1}}}\left[\prod\limits_{j=1}^{n-1} {{\varPhi}} \prod\limits_{j=1}^{\alpha- n+1} {{\varPsi}} \right] \biggl(\frac{k-n}{2e_{t}}\biggr). \end{array} $$

C_{k−n, α} has order $O(\frac {1}{t^{n}})$. Thus tC_{k−n, α} approaches to 0 on applying the limit for n = 2, 3,...α.

Now we have all the limits required to determine the recurrence relation for the degree distribution:

$$ \begin{array}{@{}rcl@{}} \lim\limits_{t \to \infty} [ t P_{k,t+1} + P_{k,t+1}] &=& \lim\limits_{t \to \infty} t P_{k,t} - \frac{(2\alpha -2\alpha \delta +1)k +4\alpha^{2} \delta +4\alpha \delta}{2(\alpha+1)}P_{k} \\&&+ \frac{(2\alpha -2\alpha \delta +1)k +4\alpha^{2} \delta +6\alpha \delta- 2\alpha-1 }{2(\alpha+1)}P_{k-1}. \end{array} $$

$$ P_{k} = \frac{(2\alpha -2\alpha \delta +1)k +4\alpha^{2} \delta +6\alpha \delta- 2\alpha-1 }{(2\alpha -2\alpha \delta +1)k +4\alpha^{2} \delta +4\alpha \delta + 2(\alpha+1)}P_{k-1}. $$

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