A logical sequence: 1 - 1 - 3 - 6 - 18 - ?

[lfdahl]

What is the next logical number in the sequence:

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

 

[topsquark]

What is the next logical number in the sequence:

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

The next number could literally be anything. We need some kind of reference. What are you working on right now?

-Dan

 

[anemone]

What is the next logical number in the sequence:

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

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?(Wondering)

 

[Opalg]

What is the next logical number in the sequence:

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

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 $a_6 = 48$.

 

[lfdahl]

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?(Wondering)

I don´t know the solution. Yours is quite tricky. Could be right. Thankyou very much!

 

[lfdahl]

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 $a_6 = 48$.

Thankyou very much! I think the simpler the solution the better.

 

[HallsofIvy]

Then what was your purpose in asking this question? What do you mean by "logical number"? There are, as topsquark said, an infinite number of perfectly "logical" formulas that will give those five numbers and then whatever sixth number you want.

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.

 

[topsquark]

What is the next logical number in the sequence:

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

Then 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!

 

[lfdahl]

Then what was your purpose in asking this question? What do you mean by "logical number"? There are, as topsquark said, an infinite number of perfectly "logical" formulas that will give those five numbers and then whatever sixth number you want.

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".

 

[lfdahl]

Then 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!

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 - like

multiply a_n by 2
multiply a_n+1 by 3
multiply a_n+2 by 2
multiply a_n+3 by 3
...etc.