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I've got a sequence defined as such:

  • between 0 and 5, $i = 1$
  • between 6 and 10, $i = 1*d$
  • between 11 and 15, $i = 1*d^2$
  • between 16 and 20, $i = 1*d^3$
  • ...

I guess I could write it like this for n=5:

between $k(n)+1$ and $(k+1)n$, $i= 1*d^k$

and I'm looking for the nth term of the induced serie. (ie the sum of the Nth first terms of my sequence)

Actually, I'm trying to solve a real-world problem here. We're selling products with a progressive discount (eg the first 5 items are at full price, then the 5 following items have a 10% discount, then there's an extra 10 percent discount for the next clip of 5, etc).

So far I'm handling this kind of manually (basically, I've created a table with all the possible values up to 1000 - we're not selling Lamborghini's, but still we don't expect a single customer to buy more than 1000 item from us, ever), and I'm quite curious to know how one would solve this using a single math formula.

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Quite a discount! If the regular price of an item is $100$ dollars, then when you get to around three hundred items, the last five are about $2$ cents each, an interesting instance of the magic of compounding. On a more serious note, you could generate a formula for determining total cost if $n$ items are bought instead of unit prices. It is a small variant of summing a (finite) geometric series. – André Nicolas Jul 17 '11 at 15:10

The sequence is $$d^{\lfloor (n-1)/5 \rfloor}, \text{ } n=1,2,3,\dots$$ See Note I didn't include $n=0$ because that doesn't fit the pattern of the rest of your sequence. (There are 6 numbers between 0 and 5 exclusive, unlike 5 numbers in all the other intervals.)

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Thanks for clarifying what the sequence was ; I'm still looking for the sum of the Nth first terms, and I'm a bit stuck because of this flooring thing ... – Brann Jul 17 '11 at 14:45

Answer corrected based on the comments below. The exponent of d is e

 e = 0 if n <= 5 
 e = (n - 1) div 5 if n > 5

In most programming languages the last statement is written as:

e = (n - 1) / 5
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I think that you want e = (n-1) div 5. Or whatever is the integer division (rounded down to an integer) call in your language of choice – Jyrki Lahtonen Jul 17 '11 at 9:51
@Jyrki: IIRC, with the languages these days, p/q for p and q both set to integer types would give integer results anyway. :) – J. M. Jul 17 '11 at 10:05
You both are completely rigth. Thank you for the correction! I have corrected my answer. – Jiri Jul 17 '11 at 10:20

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