4

The following pattern yields a finite sequence of natural numbers, whose sum equal its product:

  • $A_1=k$
  • $A_2=2$
  • $A_3,\dots,A_k=1$

A few examples:

  • $A_n=2,2$
  • $A_n=3,2,1$
  • $A_n=4,2,1,1$
  • $A_n=5,2,1,1,1$
  • $A_n=6,2,1,1,1,1$

Is there any other such sequence of natural numbers, which does not follow the pattern above?

Thanks

barak manos
  • 43,109
  • 2
    Couldn't you take any finite sequence of positive integers followed by enough 1's to bring the sum up to the product? – Ned Jun 27 '14 at 21:35
  • @Ned I think that's what he meant by the pattern above. Was my initial thought too :D – DanZimm Jun 27 '14 at 21:38
  • @Ned: Yeah, sorry about that... dumb question really... It should have been "except for the pattern that ", but was already answered (in the same way), so I can no longer change it. – barak manos Jun 27 '14 at 21:42

1 Answers1

4

Take any finite sequence of natural numbers $x_1,x_2,\ldots,x_n$ such that $x_i\ne1$ for all $i$. Then $\prod\limits_{i=1}^nx_i\ge\sum\limits_{i=1}^nx_i$ and we can add to our sequence $\prod\limits_{i=1}^nx_i-\sum\limits_{i=1}^nx_i$ ones in order to satisfy the sum being equal to the product.

Note that this accounts for all possible such (unordered) sequences, apart from the simplest possible one: $x_1=1$.

For example take $x_1=3,x_2=3$. Then we can make the sequence $3,3,1,1,1$.

Peter Woolfitt
  • 21,114
  • Yep, sorry about that... good point, and in fact, a really dumb question, which I should erase in order to prevent myself the embarrassment... Nevertheless, since you answered it correctly, I don't quite have a choice now do I? – barak manos Jun 27 '14 at 21:40
  • 3
    @barakmanos I don't think it is a stupid question at all. – Listing Jun 27 '14 at 21:41
  • @Listing: Thank you. It should probably be rephrased as "except for the pattern ". – barak manos Jun 27 '14 at 21:43