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Consider the sequence:

$0,1,5,12,22,35,51,70,92,117,145,176,\ldots$

Find both a recurrence and a closed form for this sequence.

I've done some research and found out that the majority of these (except 0) are "pentagonal numbers." We haven't learned about these yet, but further research did allow me to find a formula:

$A_n = ((3n^2)-n)/2$

That is the closed form for this sequence, correct? Could someone possibly lead me as to how to get to that equation?

In terms of a recurrence, I'm somewhat lost.

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  • $\begingroup$ Whys is 0 a part of that sequence? $\endgroup$ – Daniel W. Farlow Apr 21 '16 at 3:52
  • $\begingroup$ Not too sure. It was given in the problem. I realize that it is not a pentagonal number. $\endgroup$ – d3m0nxdrums Apr 21 '16 at 3:53
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Let $Sa_n=a_{n+1}$, then we can compute from the table that $(S-1)^3a_n=0$. The solution for this Linear Recurrence Relation is of the form $$ a_n=bn^2+cn+d\tag{1} $$ Using the known values for $n=0,1,2$, we get $$ \begin{align} 0&=d\\ 1&=b+c+d\\ 5&=4b+2c+d \end{align}\tag{2} $$ Solving $(2)$ and substituting into $(1)$ gives $$ \begin{align} a_n &=\frac32n^2-\frac12n\\ &=\frac{(3n-1)n}2\tag{3} \end{align} $$

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When you have a polynomial expressed as a sequence, you can find it by taking differences $$ \begin {array} {l l l l l l} 0&1&5&12&22&35\\1&4&7&10&13\\3&3&3&3 \end {array}$$ where the first line is your sequence and subsequent entries are the difference between the one up and to the right and the one above. The fact that the second differences are constant says you have a second power polynomial. To get the leading term, you take the constant in the second line and divide it by $2!$, so your polynomial starts with $\frac 32n^2$. You can subtract that off and get a linear polynomial, which will be constant in the first line down.

You can use this to get a recurrence. The first line is $A_n-A_{n-1}$ and you can observe it is $1+3(n-1)$, so $A_n=A_{n-1}+3n-2$

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