**I.** Let

[f_{1}, f_{2}, ldots, f_{m}, dots]

be a sequence of mappings from a common domain (A) into a metric space (left(T, ho^{prime} ight) .) For each (fixed) (x in A,) the function values

[f_{1}(x), f_{2}(x), ldots, f_{m}(x), ldots]

form a sequence of points in the range space (left(T, ho^{prime} ight).) Suppose this sequence converges for each (x) in a set (B subseteq A.) Then we can define a function (f : B ightarrow T) by setting

[f(x)=lim _{m ightarrow infty} f_{m}(x) ext { for all } x in B.]

This means that

[(forall varepsilon>0)(forall x in B)(exists k)(forall m>k) quad
ho^{prime}left(f_{m}(x), f(x)
ight)

Here (k) depends not only on (varepsilon) but also on (x,) since each (x) yields a different sequence (left{f_{m}(x) ight}.) However, in some cases (resembling uniform continuity), (k) depends on (varepsilon) only; i.e., given (varepsilon>0,) one and the same (k) fits all (x) in (B.) In symbols, this is indicated by changing the order of quantifiers, namely,

[(forall varepsilon>0)(exists k)(forall x in B)(forall m>k) quad
ho^{prime}left(f_{m}(x), f(x)
ight)

Of course, **(2) implies (1)**, but the converse fails (see examples below). This suggests the following definitions.

Definition 1

With the above notation, we call (f) the pointwise limit of a sequence of functions (f_{m}) on a set (B(B subseteq A)) iff

[f(x)=lim _{m ightarrow infty} f_{m}(x) ext { for all } x ext { in } B;]

i.e., **formula (1)** holds. We then write

[f_{m} ightarrow f( ext {pointwise}) ext { on } B.]

**In case (2)**, we call the limit uniform (on (B )) and write

[f_{m} ightarrow f( ext {uniformly}) ext { on } B.]

**II.** If the (f_{m}) are real, complex, or vector valued (§3), we can also define (s_{m}=sum_{k=1}^{m} f_{k}) (= sum of the first (m) functions) for each (m), so

[(forall x in A)(forall m) quad s_{m}(x)=sum_{k=1}^{m} f_{k}(x).]

The (s_{m}) form a new sequence of functions on (A.) The pair of sequences

[left(left{f_{m} ight},left{s_{m} ight} ight)]

is called the (infinite) series with general term (f_{m} ; s_{m}) is called its (m) th partial sum. The series is often denoted by symbols like (sum f_{m}, sum f_{m}(x),) etc.

Definition 2

The series (sum f_{m}) on (A) is said to converge (pointwise or uniformly) to a function (f) on a set (B subseteq A) iff the sequence (left{s_{m} ight}) of its partial sums does as well.

We then call (f) the sum of the series and write

[f(x)=sum_{k=1}^{infty} f_{k}(x) ext { or } f=sum_{m=1}^{infty} f_{m}=lim s_{m}]

(pointwise or uniformly) on (B).

Note that series of constants, (sum c_{m},) may be treated as series of constant functions (f_{m},) with (f_{m}(x)=c_{m}) for (x in A.)

If the range space is (E^{1}) or (E^{*},) we also consider infinite limits,

[lim _{m ightarrow infty} f_{m}(x)=pm infty.]

However, a series for which

[sum_{m=1}^{infty} f_{m}=lim s_{m}]

is infinite for some (x) is regarded as divergent (i.e., not convergent) at that (x).

**III.** Since convergence of series reduces to that of sequences (left{s_{m}
ight},) we shall first of all consider sequences. The following is a simple and useful test for uniform convergence of sequences (f_{m} : A
ightarrowleft(T,
ho^{prime}
ight).)

Theorem (PageIndex{1})

Given a sequence of functions (f_{m} : A ightarrowleft(T, ho^{prime} ight),) let (B subseteq A) and

[Q_{m}=sup _{x in B} ho^{prime}left(f_{m}(x), f(x) ight).]

Then (f_{m} ightarrow f( ext {uniformly on } B)) iff (Q_{m} ightarrow 0).

**Proof**If (Q_{m} ightarrow 0,) then by definition

[(forall varepsilon>0)(exists k)(forall m>k) quad Q_{m}

However, (Q_{m}) is an upper bound of all distances ( ho^{prime}left(f_{m}(x), f(x) ight), x in B.)

**Hence (2)**follows.Conversely, if

[(forall x in B) quad ho^{prime}left(f_{m}(x), f(x) ight)

then

[varepsilon geq sup _{x in B} ho^{prime}left(f_{m}(x), f(x) ight),]

i.e., (Q_{m} leq varepsilon.)

**Thus (2)**implies[(forall varepsilon>0)(exists k)(forall m>k) quad Q_{m} leq varepsilon]

and (Q_{m} ightarrow 0.) (square)

Examples

(a) We have

[lim _{n ightarrow infty} x^{n}=0 ext { if }|x|<1 ext { and } lim _{n ightarrow infty} x^{n}=1 ext { if } x=1.]

Thus, setting (f_{n}(x)=x^{n},) consider (B=[0,1]) and (C=[0,1)).

We have (f_{n} ightarrow 0) (pointwise) on (C) and (f_{n} ightarrow f( ext { pointwise })) on (B,) with (f(x)=0) for (x in C) and (f(1)=1.) However, the limit is not uniform on (C,) let alone on (B .) Indeed,

[Q_{n}=sup _{x in C}left|f_{n}(x)-f(x) ight|=1 ext { for each } n.]

Thus (Q_{n}) does not tend to (0,) and uniform convergence fails by Theorem 1.

(b) In Example (a), let (D=[0, a], 0

[Q_{n}=sup _{x in D}left|f_{n}(x)-f(x) ight|=sup _{x in D}left|x^{n}-0 ight|=a^{n} ightarrow 0.]

(c) Let

[f_{n}(x)=x^{2}+frac{sin n x}{n}, quad x in E^{1}.]

For a fixed (x),

[lim _{n ightarrow infty} f_{n}(x)=x^{2} quad ext { since }left|frac{sin n x}{n} ight| leq frac{1}{n} ightarrow 0.]

Thus, setting (f(x)=x^{2},) we have (f_{n} ightarrow f) (pointwise) on (E^{1}.) Also,

[left|f_{n}(x)-f(x) ight|=left|frac{sin n x}{n} ight| leq frac{1}{n}.]

Thus ((forall n) Q_{n} leq frac{1}{n} ightarrow 0.) By Theorem 1, the limit is uniform on all of (E^{1}.)

Theorem (PageIndex{2})

Let (f_{m} : A ightarrowleft(T, ho^{prime} ight)) be a sequence of functions on (A subseteq(S, ho).) If (f_{m} ightarrow fleft( ext {uniformly } ext { on a set } B subseteq A, ext { and if the } f_{m} ext { are relatively (or uniformly) } ight.) continuous on (B) , then the limit function (f) has the same property.

**Proof**Fix (varepsilon>0.) As (f_{m} ightarrow f) (uniformly) on (B,) there is a (k) such that

[(forall x in B)(forall m geq k) quad ho^{prime}left(f_{m}(x), f(x) ight)

Take any (f_{m}) with (m>k,) and take any (p in B.) By continuity, there is (delta>0,) with

[left(forall x in B cap G_{p}(delta) ight) quad ho^{prime}left(f_{m}(x), f_{m}(p) ight)

Also, setting (x=p) in (3) gives ( ho^{prime}left(f_{m}(p), f(p) ight)

[egin{aligned} ho^{prime}(f(x), f(p)) & leq ho^{prime}left(f(x), f_{m}(x) ight)+ ho^{prime}left(f_{m}(x), f_{m}(p) ight)+ ho^{prime}left(f_{m}(p), f(p) ight) &

We thus see that for (p in B),

[(forall varepsilon>0)(exists delta>0)left(forall x in B cap G_{p}(delta) ight) quad ho^{prime}(f(x), f(p))

i.e., (f) is relatively continuous at (p( ext { over } B),) as claimed.

Quite similarly, the reader will show that (f) is uniformly continuous if the (f_{n}) are. (square)

**Note 2.** A similar proof also shows that if (f_{m}
ightarrow f) (uniformly) on (B,) and if the (f_{m}) are relatively continuous at a point (p in B,) so also is (f.)

Theorem (PageIndex{3}) (Cauchy criterion for uniform convergence)

Let (left(T, ho^{prime} ight)) be complete. Then a sequence (f_{m} : A ightarrow T, A subseteq(S, ho),) converges uniformly on a set (B subseteq A) iff

[(forall varepsilon>0)(exists k)(forall x in B)(forall m, n>k) quad
ho^{prime}left(f_{m}(x), f_{n}(x)
ight)

**Proof****If (5) holds**then, for any (fixed) (x in B,left{f_{m}(x) ight}) is a Cauchy sequence of points in (T,) so by the assumed completeness of (T,) it has a limit (f(x).) Thus we can define a function (f : B ightarrow T) with[f(x)=lim _{m ightarrow infty} f_{m}(x) ext { on } B.]

To show that (f_{m} ightarrow f) (uniformly) on (B,) we

**use (5) again**. Keeping (varepsilon, k,) (x,) and (m) temporarily fixed, we let (n ightarrow infty) so that (f_{n}(x) ightarrow f(x)). Then by Theorem 4 of Chapter 3, §15, ( ho^{prime}left(f_{m}(x), f_{n}(x) ight) ightarrow p^{prime}left(f(x), f_{m}(x) ight).) Passing to the l**imit in (5), we thus obtain (2)**.The easy proof of the converse is left to the reader (cf. Chapter 3, §17, Theorem 1). (square)

**IV.** If the range space (left(T,
ho^{prime}
ight)) is (E^{1}, C,) or (E^{n}) (*or another normed space), the standard metric applies. In particular, for series we have

[egin{aligned}
ho^{prime}left(s_{m}(x), s_{n}(x)
ight) &=left|s_{n}(x)-s_{m}(x)
ight| &=left|sum_{k=1}^{n} f_{k}(x)-sum_{k=1}^{m} f_{k}(x)
ight| &=left|sum_{k=m+1}^{n} f_{k}(x)
ight| quad ext { for } m

Replacing here (m) by (m-1) and applying Theorem 3 to the sequence (left{s_{m} ight},) we obtain the following result.

Theorem (PageIndex{3'})

Let the range space of (f_{m}, m=1,2, ldots,) be (E^{1}, C,) or (E^{n}) (*or another complete normed space). Then the series (sum f_{m}) converges uniformly on (B) iff

[(forall varepsilon>0)(exists q)(forall n>m>q)(forall x in B) quadleft|sum_{k=m}^{n} f_{k}(x)
ight|

Similarly, via (left{s_{m} ight},) Theorem 2 extends to series of functions. (Observe that the (s_{m}) are continuous if the (f_{m}) are.) Formulate it!

**V.** If (sum_{m=1}^{infty} f_{m}) exists on (B,) one may arbitrarily "group" the terms, i.e., replace every several consecutive terms by their sum. This property is stated more precisely in the following theorem.

Theorem (PageIndex{4})

Let

[f=sum_{m=1}^{infty} f_{m}( ext {pointwise}) ext { on } B.]

Let (m_{1}

[g_{1}=s_{m_{1}}, quad g_{n}=s_{m_{n}}-s_{m_{n-1}}, quad n>1.]

(Thus (g_{n+1}=f_{m_{n}+1}+cdots+f_{m_{n+1}}.)) Then

[f=sum_{n=1}^{infty} g_{n}( ext {pointwise}) ext { on } B ext { as well; }]

similarly for uniform convergence.

**Proof**Let

[s_{n}^{prime}=sum_{k=1}^{n} g_{k}, quad n=1,2, ldots]

Then (s_{n}^{prime}=s_{m_{n}}) (verify!), so (left{s_{n}^{prime} ight}) is a subsequence, (left{s_{m_{n}} ight},) of (left{s_{m} ight} .) Hence (s_{m} ightarrow f( ext { pointwise })) implies (s_{n}^{prime} ightarrow f) (pointwise); i.e.,

[f=sum_{n=1}^{infty} g_{n} ext { (pointwise). }]

For uniform convergence, see Problem 13 (cf. also Problem 19). (square)

## 5 Important Difference between Arithmetic and Geometric Sequence

What is the difference between arithmetic and geometric sequence?

A sequence is a set of numbers arranged in a particular order. These set of numbers are known as terms. The main types of sequence are arithmetic and geometric sequence.

**The main difference between arithmetic and geometric sequence is that arithmetic sequence is a sequence where the difference between two consecutive terms is constant while a geometric sequence is a sequence where the ratio between two consecutive terms is constant.**

## Find sequence formula

The elements of sequence are numbered, starting from 1.

In an **arithmetic progression** the *difference* between one number and the next is always the same. 1 4 7 10 13… is an example of an arithmetic progression that starts with 1 and increases by 3 for each position in the sequence. This sequence can be described using the linear formula *a*_{n}&thinsp= 3*n*&thinsp−&thinsp2 .

In a **geometric progression** the *quotient* between one number and the next is always the same. 2 4 8 16… is an example of a geometric progression that starts with 2 and is doubled for each position in the sequence. This sequence can be described using the exponential formula *a*_{n}&thinsp= 2 *n* .

If neither quotient nor difference is constant it might be a good idea to look at the *difference between the differences*. If it turns out that the difference between the differences is constant it means that the sequence can be described using a **second degree polynomial**. 2 5 10 17 26… is an example of such a sequence. If we look at the difference between the five initial numbers we find that they are 3 5 7 9 and, as you can see, the differences between these numbers are 2. This tells us that it is possible to describe the sequence as a second degree polynomial but it does not give us any information about how.

To establish the polynomial we note that the formula will have the following form.

The task now is to find the values of *p*, *q* and *r*. By substituting *n* and *a*_{n} for some elements in the sequence we get a *system of equations*.

To be able to solve a system of equations with three unknown variables there need to be at least three equations. By solving the system of equations above we get *p*&thinsp=&thinsp1 , *q*&thinsp=&thinsp0 and *r*&thinsp=&thinsp1 which gives us the following formula.

Sometimes it can be necessary to use polynomials of higher degree than two but the method is essentially the same. To solve a third degree polynomial the difference between the differences between the differences need to be constant. For fourth degree polynomials we would have to look at yet another level of differences.

Note that as long as you have a finite sequence of numbers it is always possible to find a polynomial that can describe it. If *n* numbers are known it is always possible to find a polynomial of degree *n*&thinsp-&thinsp1 that match all the numbers, but this does not necessarily describe any true pattern of the sequence. For this the polynomial degree would have to be two (preferable three or more) degrees lower than the number of known numbers in the sequence. This is something to think about when using the tool on this page. The reason the tool does not always find a polynomial has to do with technical limitations that makes the numeric precision not good enough for polynomials of higher degrees.

In addition to what has been mentioned already the tool can also recognize the **sequence of prime numbers** and the **Fibonacci sequence**. There are of course many more ways to construct sequences but the ones mentioned here are some of the most common.

## The nth Term

The 'nth' term is a formula with 'n' in it which enables you to find **any** term of a sequence without having to go up from one term to the next.

'* n*' stands for the

*so to find the 50th term we would just substitute 50 in the formula in place of '*

**term number***'.*

**n****There are two types of sequences that you will have to deal with:**

#### Constant Difference Sequences

This is when the difference between terms is always the same.

e.g. 1, 4, 7, 10, . This has a difference which is always 3.

**How do you find the formula for the 'nth' term?**

Well, the three times table has the formula '3n' and the terms in this sequence are two less than the terms in the three times table so the formula is '3n - 2'.

You can always find the 'nth term' by using this formula:

**nth term = dn + (a - d)**

Where **d** is the difference between the terms, **a** is the first term and **n** is the term number.

e.g. 6, 11, 16, 21, . For this sequence **d = 5, a = 6**

So the formula is **nth term = 5n + (6 - 5)**

which becomes **nth term = 5n + 1**

#### Changing Difference Sequences

**What if the difference keeps changing?**

Obviously these are more difficult but once again we can use a formula!

**nth term = a + (n - 1)d + ½(n - 1)(n - 2)c**

This time there is a letter * c* which stands for the

**second difference**(or the difference between the differences and

*is just the difference between the*

**d****first two**numbers.

Putting the right numbers into the formula is reasonably simple (once you've learnt the formula!). Simplifying it requires good Algebra skills so **practice your Algebra!**

**Here's an example:**

Here the difference between the first two numbers is **1** so **d** = **1**

Also the **second differences are 2** so **c** = **2** The first term is **2** so **a** = **2**

Using the formula, **nth term = 2 + (n - 1)x1 + ½(n - 1)(n - 2)x2**

Getting rid of brackets (and noticing that ½ x 2 = 1):

**nth term = 2 + n - 1 + n2 - 3n + 2**

**nth term = n 2 - 2n + 3**

If all that gave you a headache there is an alternative way!

**1.** If the first differences keep changing but the second difference is constant then the formula is something to do with '**n 2** '. Make a table showing the first few terms of '**n 2** '.

**2.** In the next column of your table write the differences between the term of '**n 2** ' and your sequence.

**3.** Find the formula for **this** new sequence using **dn + (a - d)**

**4.** Add it on to **n 2** to give yo your final formula.

**Have a look at this using the sequence above:**

Sequence | 'n 2 ' | Difference |

2 | 1 | 1 |

3 | 4 | -1 |

6 | 9 | -3 |

11 | 16 | -5 |

18 | 25 | -7 |

**For the sequence 1, -1, -3, -5, -7 => a = 1 and d = -2**

So the formula is * -2n + (1 - -2)*which simplifies to

**-2n + 3**Final Formula (Step 4) : * nth term = n 2 -2n + 3*(as we got before!)

## Run the sample

To execute the E1_HelloSequence orchestration, send the following HTTP POST request to the HttpStart function.

The previous HTTP snippet assumes there is an entry in the host.json file which removes the default api/ prefix from all HTTP trigger functions URLs. You can find the markup for this configuration in the host.json file in the samples.

For example, if you're running the sample in a function app named "myfunctionapp", replace "

The result is an HTTP 202 response, like this (trimmed for brevity):

At this point, the orchestration is queued up and begins to run immediately. The URL in the Location header can be used to check the status of the execution.

The result is the status of the orchestration. It runs and completes quickly, so you see it in the *Completed* state with a response that looks like this (trimmed for brevity):

As you can see, the runtimeStatus of the instance is *Completed* and the output contains the JSON-serialized result of the orchestrator function execution.

You can implement similar starter logic for other trigger types, like queueTrigger , eventHubTrigger , or timerTrigger .

Look at the function execution logs. The E1_HelloSequence function started and completed multiple times due to the replay behavior described in the orchestration reliability topic. On the other hand, there were only three executions of E1_SayHello since those function executions do not get replayed.

## Recommendations for interpreting the loss of function PVS1 ACMG/AMP variant criterion

The 2015 ACMG/AMP sequence variant interpretation guideline provided a framework for classifying variants based on several benign and pathogenic evidence criteria, including a pathogenic criterion (PVS1) for predicted loss of function variants. However, the guideline did not elaborate on specific considerations for the different types of loss of function variants, nor did it provide decision-making pathways assimilating information about variant type, its location, or any additional evidence for the likelihood of a true null effect. Furthermore, this guideline did not take into account the relative strengths for each evidence type and the final outcome of their combinations with respect to PVS1 strength. Finally, criteria specifying the genes for which PVS1 can be applied are still missing. Here, as part of the ClinGen Sequence Variant Interpretation (SVI) Workgroup's goal of refining ACMG/AMP criteria, we provide recommendations for applying the PVS1 criterion using detailed guidance addressing the above-mentioned gaps. Evaluation of the refined criterion by seven disease-specific groups using heterogeneous types of loss of function variants (n = 56) showed 89% agreement with the new recommendation, while discrepancies in six variants (11%) were appropriately due to disease-specific refinements. Our recommendations will facilitate consistent and accurate interpretation of predicted loss of function variants.

**Keywords:** ACMG/AMP ClinGen PVS1 loss of function variant interpretation.

© 2018 Wiley Periodicals, Inc.

### Figures

Figure 1.. PVS1 decision tree.

Figure 1.. PVS1 decision tree.

Refer to text for detailed description. NMD, nonsense-mediated decay LoF,…

## Forward references#

A forward reference is when you compose tasks, using string references, that haven't been registered yet. This was a common practice in older versions, but this feature was removed to achieve faster task runtime and promote the use of named functions.

In newer versions, you'll get an error, with the message "Task never defined", if you try to use forward references. You may experience this when trying to use exports for your task registration *and* composing tasks by string. In this situation, use named functions instead of string references.

During migration, you may need to use the forward reference registry. This will add an extra closure to every task reference and dramatically slow down your build. **Don't rely on this fix for very long**.

## Special considerations

### Annotation of multiple assemblies

When multiple assemblies of good quality are available for a given organism, annotation of all is done in coordination. To ensure that matching regions across assemblies are annotated the same way, assemblies are aligned to each other before the annotation.

- Assembly-assembly alignment results are used to rank the transcript and the curated genomic alignments: for a given query sequence, alignments to corresponding regions of two assemblies receive the same rank.
- Corresponding loci of multiple assemblies are assigned the same GeneID and locus type.

Assembly-assembly alignments are available through the NCBI Genome Remapping Service.

### Re-annotation

Organisms are periodically re-annotated when new evidence is available (e.g. RNA-Seq) or when a new assembly is released. Special attention is given to tracking of models and genes from one release of the annotation to the next. Previous and current models annotated at overlapping genomic locations are identified and the locus type and GeneID of the previous models are taken into consideration when assigning GeneIDs to the new models. If the assembly was updated between the two rounds of annotation, the assemblies are aligned to each other and the alignments used to match previous and current models in mapped regions.

## 4.12: Sequences and Series of Functions

**Terms of Use Contact Person:** Donna Roberts

We saw in Sequences - Basic Information, that sequences can be expressed in various forms.

This page will look at one of those forms, the recursive form.

If you are on the *n* th rung, you must have stepped on the *n*-1 st rung. *a*_{n} = *a*_{n-1} + "step up"

A recursive formula is written with **two parts**: a statement of the first term along with a statement of the formula relating successive terms.

Sequence: <10, 15, 20, 25, 30, 35, . >. Find a recursive formula.

This example is an arithmetic sequence (the same number, 5, is added to each term to get to the next term).

Sequence: <3, 6, 12, 24, 48, 96, . >. Find a recursive formula.

This example is a geometric sequence (the same number, 2, is multiplied times each term to get to the next term).

Sequence: <0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . >

This example is neither an arithmetic sequence nor a geometric sequence.

While we have seen recursive formulas for arithmetic sequences and geometric sequences, there are also recursive forms for sequences that do not fall into either of these categories.

The sequence shown in this example is a famous sequence called the Fibonacci sequence.

Is there a pattern for the Fibonacci sequence?

Yes. After the first two terms, each term is the sum of the previous two terms.

Let's say you have a long list of sequence numbers to mark items, such as check numbers in bank statements, normally we scrolling through and locate the missing sequence numbers manually. Sometimes this is quite arduous and time-consuming. You may think of tricky ways to deal with it. Yes, there are several easy ways to identify and locate missing numbers sequence in Excel 2007, Excel 2010, and Excel 2013 quickly and conveniently.

**Kutools for Excel**’s **Find Missing Sequence Number** feature can helper you quickly and easily find the missing sequence, and insert the missing numbers or blank rows into the existing data sequence, or fill background color when encountering the missing sequence.

**Kutools for Excel**: with more than 200 handy Excel add-ins, free to try with no limitation in 60 days. ** Download and free trial Now! **

#### Identify missing numbers sequence with IF formula

As we all known, most of sequence numbers are with fixed increment of 1, such as 1, 2, 3, …, N. Therefore, if you can identify the number is not less 1 than its following number, there is a missing number.

We will show you the tutorials with an example as following screenshot shows:

1. In a blank cell, enter the formula of **=IF(A3-A2=1,"","Missing")** , and press the **Enter** key. In this case, we enter the formula in Cell B2.

If there is no missing numbers, this formula will return nothing if missing numbers exist, it will return the text of "Missing" in active cell.

2. Select the cell B2 and drag the fill handle over the range of cells that you want to contain this formula. Now it identifies the missing numbers with the text of "Missing" in corresponding cells of Column B. See the following screenshot:

#### Identify missing numbers sequence with an array formula

Sometimes it requires not only identifying missing numbers sequence, but also listing missing numbers too. You can deal it with following steps:

1. in the adjacent cell, please enter the formula **= SMALL(IF(ISNA(MATCH(ROW(A$1:A$30),A$1:A$30,0)),ROW(A$1:A$30)),ROW(A1))**

* A1:A30* = range of numbers, the sequence to check against is from 1 to 30

2. Press the **Ctrl + Shift + Enter** Keys together to finish the formula. Copy down the formula until you get #NUM! errors meaning all missing numbers have been listed. See screenshot:

#### Identify missing numbers sequence with Kutools for Excel quickly

The above methods only can identify the missing pure number sequence, if you have the sequence such as AA-1001-BB, AA-1002-BB, they may not work successfully. But, don't worry, **Kutools for Excel**’s powerful feature – **Find Missing Sequence Number** can help you quickly identify the missing sequence.

**Note:** To apply this **Find Missing Sequence Number**, firstly, you should download the **Kutools for Excel**, and then apply the feature quickly and easily.

After installing **Kutools for Excel**, please do as this:

1. Select the data sequence that you want to find the missing sequence.

2. Click **Kutools** > **Insert** > **Find Missing Sequence Number**, see screenshot:

3. In the **Find Missing Sequence Number** dialog box:

(1.) If you choose **Inserting new column with following missing marker** option, all the missing sequence numbers have been marked with the text **Missing** in a new column next to your data. See screenshot:

(2.) If you choose **Inserting missing sequence number** option, all the missing numbers have been inserted into the sequence list. See screenshot:

(3.) If you choose **Inserting blank rows when encounting missing sequence numbers** option, all blank rows are inserted when there are missing numbers . See screenshot:

(4.) If you choose **Fill background color** option, the location of the missing numbers will be highlighted at once. See screenshot: