# Linear recurrence calculator

## World's simplest number tool

Quickly generate a linear recurrence sequence in your browser. To get your sequence, just specify the initial values, coefficients and the length of the sequence in the options below, and this utility will generate that many linear recurrence series numbers. Created by developers from team Browserling.

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Linear recurrence calculator tool
What is a linear recurrence calculator?

This is an online browser-based utility for generating linear recurrence series. A linear recurrence is a recursive relation of the form xₙ = Axₙ₋₁ + Bxₙ₋₂ + Cxₙ₋₃ + Dxₙ₋₄ + Exₙ₋₅ + …. Constants A, B, C, D, E are real numbers, and xₙ is expressed in terms of the previous n elements of the series. That is, each term of the sequence is a linear function of earlier terms in the sequence. For example, the Fibonacci sequence is a linear recurrence series. It is given by the linear equation xₙ = Axₙ₋₁ + Bxₙ₋₂, where A = B = 1 (all other coefficients C = D = E = 0), with initial values x₁ = 0 and x₂ = 1. Substituting the initial values into the recurrent formula, you can find the series that forms the Fibonacci numbers. Other examples of linear recurrence equations are the Lucas numbers, Pell numbers, and Padovan numbers. In this tool, you can generate a linear recurrence with up to five terms in the sum. To specify the recurrence relation, you only need to adjust the coefficients A, B, C, D, E. If any coefficient is not specified, then this term is not used in the recurrence formula. To specify the starting values, you need to fill in the xₙ fields. You can also specify how many terms of recurrence equation you need and what symbol you want to separate them with. That's numberwang!

Linear recurrence calculator examples
Click to use

Fibonacci Relation

In this example, we generate a second-order linear recurrence relation. We set A = 1, B = 1, and specify initial values equal to 0 and 1. From these conditions, we can write the following relation xₙ = xₙ₋₁ + xₙ₋₂. This relation is a well-known formula for finding the numbers of the Fibonacci series.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Lucas Numbers

In this example, we're using the same linear recurrence as in the previous example because A = B = 1. However, here the first initial value is 2 and the second is 1. So, we get the recursion xₙ = xₙ₋₁ + xₙ₋₂, with x₁ = 2, x₂ = 1. These conditions correspond to the recurrence formula that calculates the Lucas number sequence.

2
1
3
4
7
11
18
29
47
76

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Padovan Sequence

In this example, we calculate a third-order linear recurrence equation. It has the following coefficients: A = 0, B = 1, C = 1, and initial values: x₁ = 1, x₂ = 1, x₃ = 1. So, we get the linear equation xₙ = xₙ₋₂ + xₙ₋₃, which forms the Padovan sequence. We calculate the following numbers up to x₂₀ term and separate them with a semicolon character.

1; 1; 1; 2; 2; 3; 4; 5; 7; 9; 12; 16; 21; 28; 37; 49; 65; 86; 114; 151

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Pell Numbers

In this example, we generate Pell numbers. Pell numbers are calculated by the following recurrence: xₙ = 2xₙ₋₁ + xₙ₋₂, where x₁ = 0, x₂ = 1. That is, a doubled previous term plus another previous term forms the next term. For example, 2*1 + 0 = 2, 2*2 + 1 = 5, 2*5 + 2 = 12, and so on. We generate twelve Pell numbers, and place "=>" symbol between them.

0 => 1 => 2 => 5 => 12 => 29 => 70 => 169 => 408 => 985 => 2378 => 5741

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Pell-Lucas Series

In this example, we calculate the Pell-Lucas series (also known as companion Pell series). Pell-Lucas numbers are calculated according to the same recurrence formula as in the previous example, but here both initial values are equal to 2. As the initial values are even, and the equation is linear and has no odd terms, the Pell-Lucas numbers are always even.

2 2 6 14 34 82 198 478 1154 2786 6726 16238 39202 94642 228486

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Perrin Numbers

In this example, we generate Perrin numbers. The linear relation for the Perrin series is the same as for the Padovan series (see example above) but with different initial conditions – x₁ = 3, x₂ = 0, x₃ = 2. We calculate forty Perrin numbers and separate them by the "~" character.

3 ~ 0 ~ 2 ~ 3 ~ 2 ~ 5 ~ 5 ~ 7 ~ 10 ~ 12 ~ 17 ~ 22 ~ 29 ~ 39 ~ 51 ~ 68 ~ 90 ~ 119 ~ 158 ~ 209 ~ 277 ~ 367 ~ 486 ~ 644 ~ 853 ~ 1130 ~ 1497 ~ 1983 ~ 2627 ~ 3480 ~ 4610 ~ 6107 ~ 8090 ~ 10717 ~ 14197 ~ 18807 ~ 24914 ~ 33004 ~ 43721 ~ 57918

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Negafibonacci Numbers

In this example, we set one coefficient to a negative value and get the following recurrence formula xₙ = xₙ₋₂ - xₙ₋₁. This means that the next series member is calculated as the difference between the two previous ones. We use the initial values 0 and 1 for the first two terms, and here's how we calculate some of following terms: 0 - 1 = -1, 1 - (-1) = 2, -1 - 2 = -3, and so on. As you can see, these numbers are members of the Negafibonacci sequence. We find 15 elements and separate them with a comma character.

0, 1, -1, 2, -3, 5, -8, 13, -21, 34, -55, 89, -144, 233, -377

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Periodic Series

In this example, the recurrence relation is very similar to the previous example, but here we set the coefficient B = -1, and the first two values both to 1. Let's try to calculate a few subsequent values. F₃ = 1 - 1 = 0, F₄ = 0 - 1 = -1, F₅ = -1 - 0 = -1, F₆ = -1 - (-1) = 0, F₇ = 0 - (-1) = 1, F₈ = 1 - 0 = 1. As you can see in terms 7 and 8, we have returned to our initial values. What that means is that the sequence will periodically repeat and we get a cycle.

1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1; 0; -1; -1; 0; 1; 1

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Fifth-order Linear Relation

In this example, we use a linear equation with five members. The number of members in the equation determines the order of the series. In this case, it's the fifth order series. From the given coefficients we get the following relation xₙ = xₙ₋₁ - xₙ₋₂ + xₙ₋₃ - xₙ₋₄ + xₙ₋₅, with initial values x₁ = x₃ = x₅ = 0.5, and x₂ = x₄ = 1. We've selected the initial conditions so that the series is periodic. That is, after every 6 terms the result repeats again. We generate 100 numbers here and separate them with a space.

0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1 0.5 -0.5 0.5 1 0.5 1

**Required options**

Coefficient A.

Number x

_{n-1}.
Coefficient B.

Number x

_{n-2}.
Coefficient C.

Number x

_{n-3}.
Coefficient D.

Number x

_{n-4}.
Coefficient E.

Number x

_{n-5}.
How many x

_{n}numbers to calculate?
Which symbol to delimit
the output terms with?
(By default the new-line \n.)

Pro tips
Master online number tools

You can pass options to this tool using their codes as query arguments and it will automatically compute output. To get the code of an option, just hover over its icon. Here's how to type it in your browser's address bar. Click to try!

https://onlinenumbertools.com/linear-recurrence-calculator?&A=1&B=1&C=&D=&E=&xn-1=0&xn-2=1&xn-3=&xn-4=&xn-5=&count=25&separator=%2C%20

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_{n}= Ax_{n-1}+ Bx_{n-2}+ Cx_{n-3}+ Dx_{n-4}+ Ex_{n-5}