###### Asked in Math and Arithmetic

# What number comes next in the sequence 1 2 6 42 1806?

## Answer

###### Wiki User

###### 12/19/2015

Option 1

The sequence is found by multiplying each term by the number one greater than itself. First, 1 * 2 = 2. Then, 2 * 3 = 6. Then, 6 * 7 = 42, and so on.

n*(n+1) = x (where n equals the previous x)

1*(1+1) = 2

2*(2+1) = 6

6*(6+1) = 42

42*(42+1) = 1806

1806*(1806 +1) = **3,263,442**.

Another way to consider it is that each term is being squared then added plus the original amount, so the second term will be equal to the first term squared, then add the first term.

We find the equation (n^2) + n , where n is the previous term.

So the next term is again **3,263,442**.

Option 2 For this option, we take each number and multiply it by the subsequent prime number, beginning in the same values and then diverging from the first choice.

First we take 1 and multiply it by the smallest prime greater than itself - 2, and get 2.

The next prime after 2 is 3 which gives us 2 * 3 = 6.

Six times the next prime, 7, gives us 42.

The number 42 times its nearest greater prime of 43 gives us 1806.

Finally, 1806 times the next prime number (which is 1811) gives
us **3,270,666**.

Option 3 The third approach to this series is a long process but seems logical too.

Lets start. The original series is:

1, 2, 6, 42, 1806, ?

The second digit can be derived by multiplying the first digit by 2. Then the third digit can be derived by multiplying the second digit by 3. Then the fourth digit is third digit times 7. The fifth is fourth digit multiplied by 43.

So we had a second string of series of multipliers which consists: 2,3,7,43,?

To get the fifth multiplier, a pattern can be seen from this series. The second digit is first digit is multiplied by 1 plus 1. The third digit is second digit multiplied by 2 plus 1. The fourth digit is third digit multiplied by 6 plus 1.

This in turn has created the sequence 1, 2, 6 which in the sequential nature of these sequences seems to be created by multiplying by subsequent integers. So the next term would be 6 * 4 = 24.

Going back to the multipliers we find that the fifth digit is 1033 (24 * 43).

2, 3, 7, 43, 1033

Again, the second digit is the first digit multiplied by 1 plus 1. The third digit is the second digit multiplied by 2 plus 1. The fourth digit is the third digit multiplied by 6 plus 1. The fifth digit is now the fourth digit multiplied by 24 plus 1.

Now lets go back to the original series:

1, 2, 6, 42, 1806, ?

Now we see the sixth digit is the fifth digit multiplied by 1033
which gives us **1,865,598**.

* * * * *

Yet another possibility is to fit a polynomial to the given sequence. There are infinitely many polynomials of order 5 or above but consider

T(n) = (1667n^4 - 16554n^3 + 57685n^2 - 82158n + 39384)/24.

For n = 1, 2, 3, 4 and 5 it gives the required 5 numbers and T(6) = 8661.

In reality, short of reading the mind of the person who posed
the question, there is no way of determining which of the
infinitely many solutions is the "correct" one.

3,263,442

10,650,056,950,806 etc

The formula is X2 + X

Simplest way is:

n=n*(n+1).

=== === === === === === other way: 1 2 12*1/2=6 .... 1 2 6
126*2/6=42 ..... 1 2 6 42 12642*6/42=1806 .... 1 2 6 42 1806
12642186*42/1806=**2940042** .... 1 2 6 42 1806
**2940042**

There are at least two possible answers:

1. You could fit the polynomial function:

tn = (1667n4 - 16554n3 + 57685n2 - 82158n + 39384)/24 for n = 1, 2, 3, ...

Following the logic, the 6th term is 8661.

2. One alternative is

t1 = 1 and tn+1 = tn*(tn + 1) for n = 1, 2, 3, ...

That definition gives the next term as 3,263,442.

There are many other solutions.

1806*1807=3263442

*3,263,442* comes after 1806. The series is formed by a number
multiplying itself increased by 1. Here is the equation: X(X+1)
Y

1(1+1) 2

2(2+1) 6

6(6+1) 42

42(42+1) 1806

1806(1806+1) 3,263,442

In simpler numbers,

1 X 2 2

2 X 3 6

6 X 7 42

42 X 43 1806

1806 X 1807 3,263,442

3263442. The progression is n(n+1).

**1 2 6 42 1'806 3'263'442 10'650'056'960'806** 1st is
1

2nd is 1st*(1st+1) -> 1*2

3rd is 2nd*(2nd+1) -> 2*3

4th is 3rd*(3rd+1) -> 6*7

5th is 4th*(4th+1) -> 42*43

...

and so on

3263442

3263442

###### Wiki User

###### 02/10/2010

The next number in the sequence is 3,263,442.

1 x 1 + 1 = 2

2 x 2 + 2 = 6

6 x 6 + 6 = 42

42 x 42 + 42 = 1806

1806 x 1806 + 1806 = 3,263,442