# How to replace addition with multiplication to find the next integer value?

Sorry in advance for my lack of mathematical knowledge, I am very new to it.

Yesterday, I posed this question to myself:

"In a world without addition or subtraction, how could we derive the next value in the sequence of natural numbers from $1\to\infty$ with a step size of $1$?"

This lead me to the idea of multiplication to find the next value in a sequence. After analyzing the multipliers between each natural value using: $$\frac{(n+1)}{n}$$

I noticed the pattern of this sequence starts at the high values of $2$ and $1.5$, then converges to a value of $1$.

My two questions:

• Is it right to assume that the sequence of multipliers should have a more predictable sequence?
• Are there more elegant ways of producing the next natural number without addition or subtraction?
• Hi Vita. First of all let me congratulate you with your mathematical curiosity! Keep asking yourself questions like these, and never worry about whether or not they are stupid questions! – Joachim Jul 19 '15 at 19:40
• @john why not ? $p=1\cdot p$. I don't see why we need two "non-trivial numbers". – Dietrich Burde Jul 19 '15 at 19:54
• How do you define "next natural number" in a world without +1? – Siméon Jul 19 '15 at 19:59
• @siméon: hadn't just this been the point/the nerve of the couriosity of the OP? – Gottfried Helms Jul 19 '15 at 20:10
• Something like $n+1=\log_2(2\cdot2^n)$? – g.kov Jul 19 '15 at 20:56

With the function $2^n$ and it's inverse, $\log_2$ available, $n+1=\log_2(2\cdot2^n)$.
If we were allowed to use floor and ceiling functions in this strange world without $+$ and $-$, then perhaps we could use the following function which would generate the next integer after $n$:
$$f(n)=\lceil{(n\times\frac{\lceil n\sqrt2\rceil}{\lfloor n\sqrt2\rfloor})\rceil}$$