What is the formula for the sequence obeying that pattern? A base-10 analog could be $\{1,2,3,4,5,6,7,8,9, 10,20,30,.., … 10^kn,10^k(n+1),10^k(n+2),..10^{k+1}n, 10^{k+1}(n+1),10^{k+1}(n+2), . . .\}$, where k is an integer starting with 0 stepping by 1.
What if the sequence doesn't increase or decrease locally over the entire domain, such as with a subsequence of {4,5,3} or a rule using sign-flipping or periodicity (e.g. $(-k)^n$ for subsequences with both even and odd or non-integer or $(-k)^{n+1}$ for sequence otherwise without) and it could feature duplicate values; is there a way to apply a rule admitting only the first (or for that matter, only the second, third, or or any rank of choice, preculuding more and more) instances of a value? E.g.: $b_n$={4,5,3, 3,5,1,-3,1} with a rule of |2n|-5 of $n_i$ and simply eliminating duplicate value then continuing to $n_{i+1}$ (instead of continuing to iterate until a valid value arises from operating on $n_i$. The possibility exists also to include omitted(duplicate/sub-duplicate) in the input sequence of (could call it) $n_j$. What is the mathematical or programming way to formulate this? What if you want to take a discrete sequence with rules such as these and regress it to a smooth curve, is that possible?
Best Answer
There are several ways how to achieve what you ask. It depends on what you want from such representation.
If you want your reader to understand how the sequence behaves, then nothing really beats the piecewise bracket: $$ a_k = \begin{cases} 1\cdot 5^{\lfloor k/3\rfloor}&\text{if }k\equiv 0\mod 3,\\ 2\cdot 5^{\lfloor k/3\rfloor}&\text{if }k\equiv 1\mod 3,\\ 3\cdot 5^{\lfloor k/3\rfloor}&\text{if }k\equiv 2\mod 3, \end{cases} $$
If you want to compress this to a single line, you might use unit function $\delta[n]$ or $\mathbb 1_{cond}$: $$ a_k=5^{\lfloor k/3\rfloor}\big(\delta[k\mathop{\mathrm{mod}} 3]+2\delta[(k\mathop{\mathrm{mod}} 3)-1]+3\delta[(k\mathop{\mathrm{mod}} 3)-2]\big)
If you want something that looks continuous, then you can use the following trick. If your sequence has a period of 3, take 3 roots of $x^3=1$ and solve the equation: $$ \begin{cases} a +b+c=1,\\ a x_1^1+bx_2^1+cx_3^1=2,\\ a x_1^2+bx_2^2+cx_3^2=3. \end{cases} $$ with $x_1=1$, $x_2=e^{i2\pi/3}$ and $x_3=e^{-i2\pi/3}$, we get $a=2$, $b=-(i/\sqrt3 + 1)/2$, $c=(i/\sqrt3 - 1)/2$. Thus the sequence: $$ u_k=2-\left(\frac{i}{2\sqrt3}+\frac12\right)e^{\frac{2\pi i}3 k}+\left(\frac{i}{2\sqrt3}-\frac12\right)e^{-\frac{2\pi i}3 k} $$ will be periodic: 1,2,3,1,2,3...
You can do a similar trick with powers of 5. They go 0,0,0,1,1,1,2,2,2... By subtracting the asymptote $k/3$, you will get $0,-\frac13,-\frac23,0,-\frac13,-\frac23,0,-\frac13,-\frac23...$ a periodic sequence again. You can find that this sequence is: $$ v_k = -\frac13+\frac16\left(1-\frac i{\sqrt3}\right)e^{\frac{2\pi i}3 k}+\frac16\left(1+\frac i{\sqrt3}\right)e^{-\frac{2\pi i}3 k}. $$
And finally $a_k=u_k 5^{v_k+k/3}$. If you don't like complex numbers, you can express this with trigonometric functions.