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I'm working on a conjecture and I've come up to a point where I need to express every non-zero positive integers not divisble by 4 with only one parameter (as you would express every even number as 2p). Here the parameter p has to cross the whole natural integers domain $N*$.

Before giving you any further information about the idea I have to express such a set of numbers, I would like to show you something that might come in handy.

Earlier while working on the said conjecture I needed to express (still with only one parameter n) a repeating sequence: (5, 0, 2, 3, 8, 6). I've succeded. I've come up using Fibonacci's sequence to do it. The idea was to notice that -4 the sequence was (1, -4, -2, -1, 4, 2) then write it as $u_n =(-1)^{f(n)}2^{g(n)}$ with $f(n)$ being odd three consecutive time then even the three following, and $g(n)$ alternating between $0,1,2$. Finally I've come up with: $$g(n) =1+\frac {(-1)^{F_n}-(-1)^{F_{n+2}}}{2}$$ $f(n)$ is not interesting here. What's to be noticed is that $h(n)=g(n)-1 =\frac {(-1)^{F_n}-(-1)^{F_{n+2}}}{2}$ alternates between $-1,0,1$.

Now for the idea: non-multiples of 4 are $u_i=1,2,3,5,6,7,9,10,11,13,14,15...$ So basically we can add the $h(n)$ function to numbers $2+6k$ and write $u_i$ as: $$u_i = 2+6{k(i)} + h(i)$$With $i$ in $N*$

Therefore the question is can you find an expression for $k(n)$ that would at least inspire me. Thanks a lot for your answers.

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Let $n = 3k + r;$ where $1\le r \le 3$. In other words $k = \lfloor \frac {n-1}3\rfloor$ and $r = n - 3k$.

Just let $N = 4*k + r$

Or $N = 4 \lfloor \frac {n-1}3\rfloor + n - 3\lfloor \frac {n-1}3\rfloor = n + \lfloor \frac {n-1}3\rfloor$

Thus you get $1,2,3,5,6,7,9,10,......$

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  • $\begingroup$ Oh thanks a lot, that is really clever $\endgroup$ Dec 20 '18 at 2:22
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$$ u(i) = 4\left\lfloor\frac{i}{3}\right\rfloor +\left(i\;\mathrm{mod}\;3\right) + 1 $$ should do the trick for enumerating all natural numbers not divisible by 4.

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