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#### lfdahl

##### Well-known member

- Nov 26, 2013

- 740

What is the next logical number in the sequence:

1, 1, 3, 6, 18, ??

1, 1, 3, 6, 18, ??

- Thread starter lfdahl
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- Thread starter
- #1

- Nov 26, 2013

- 740

What is the next logical number in the sequence:

1, 1, 3, 6, 18, ??

1, 1, 3, 6, 18, ??

- Aug 30, 2012

- 1,250

The next number could literally be anything. We need some kind of reference. What are you working on right now?What is the next logical number in the sequence:

1, 1, 3, 6, 18, ??

-Dan

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- Feb 14, 2012

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I would say, given the first two terms as $1, 1$, the sequenceWhat is the next logical number in the sequence:

1, 1, 3, 6, 18, ??

n=1 | n=2 | n=3 | n=4 | n=5 | n=6 |

1 | 1 | 3 | 6 | 18 | - |

can be viewed as

n=1 | n=2 | n=3 | n=4 | n=5 | n=6 |

1 | 1 | $1+1+1^0=3$ | $1+3+2^1=6$ | $3+6+3^2=18$ | - |

i.e., for $n>2$, $a_n=a_{n-2}+a_{n-1}+ (n-2)^{n-3}$.

Hence, the next number $(a_6)$ would be $a_6=a_4+a_5+(6-2)^{6-3}=6+18+4^3=88$.

Does this look right,

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- #4

- Feb 7, 2012

- 2,792

The OEIS lists several possibilities, the simplest of which is the sequence given by $a_n = a_{n-1} + (n-1)a_{n-2}$, with $a_1=a_2=1$. The next term is then $a_6 = 48$.What is the next logical number in the sequence:

1, 1, 3, 6, 18, ??

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- #5

- Nov 26, 2013

- 740

I don´t know the solution. Yours is quite tricky. Could be right. Thankyou very much!I would say, given the first two terms as $1, 1$, the sequence

n=1 n=2 n=3 n=4 n=5 n=6 1 1 3 6 18 -

can be viewed as

n=1 n=2 n=3 n=4 n=5 n=6 1 1 $1+1+1^0=3$ $1+3+2^1=6$ $3+6+3^2=18$ -

i.e., for $n>2$, $a_n=a_{n-2}+a_{n-1}+ (n-2)^{n-3}$.

Hence, the next number $(a_6)$ would be $a_6=a_4+a_5+(6-2)^{6-3}=6+18+4^3=88$.

Does this look right,Dan?

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- #6

- Jan 29, 2012

- 1,151

In fact, you can assign any number you want as the sixth number and then find the unique fifth degree polynomial that will give those six numbers.

- Aug 30, 2012

- 1,250

Then there's the "classical" fit scheme: The given numbers specify a 4th degree polynomial (or higher):What is the next logical number in the sequence:

1, 1, 3, 6, 18, ??

\(\displaystyle \frac{3}{8}x^4 - \frac{47}{12} x^3 + \frac{121}{8} x^2 - \frac{283}{12} x + 13\)

This gives the next number as 56.

We can "force" the next number to be anything we want. For example:

\(\displaystyle - \frac{37}{120} x^5 + 5x^4 - \frac{241}{8} x^3 + \frac{169}{2} x^2 - \frac{1621}{15} x + 50\)

gives the next number as 19.

Or you can add logarithmic terms, trig terms, etc. (The coefficients will be really messy, but it can be done.)

-Dan

PS Fond gratitude to Mathematica for slugging through the simultaneous equations!

- Thread starter
- #9

- Nov 26, 2013

- 740

In fact, you can assign any number you want as the sixth number and then find the unique fifth degree polynomial that will give those six numbers.

Yes, you´re absolutely right. I expected a quite simple explanation far from fifth degree polynomials etc. This was my purpose of asking using the term "logical".

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- #10

- Nov 26, 2013

- 740

Again: Thankyou for a thorough answer allthough this solution is far from what I assumed to be the answer. I´d expected a very simple obvious logical pattern - likeThen there's the "classical" fit scheme: The given numbers specify a 4th degree polynomial (or higher):

\(\displaystyle \frac{3}{8}x^4 - \frac{47}{12} x^3 + \frac{121}{8} x^2 - \frac{283}{12} x + 13\)

This gives the next number as 56.

We can "force" the next number to be anything we want. For example:

\(\displaystyle - \frac{37}{120} x^5 + 5x^4 - \frac{241}{8} x^3 + \frac{169}{2} x^2 - \frac{1621}{15} x + 50\)

gives the next number as 19.

Or you can add logarithmic terms, trig terms, etc. (The coefficients will be really messy, but it can be done.)

-Dan

PS Fond gratitude to Mathematica for slugging through the simultaneous equations!

multiply a

multiply a

multiply a

multiply a

...etc.