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+Not too long ago I decided to create a pretty useless project. It shows emoji and color of the day. All of this is calculated by my formulas using my own algorithms on client-side.
+
+Here are some formulas that I used: 
+
+* `t` - timestamp - const - 00:00 of this day in UTC.
+
+### Emojis:
+The formula incorporates certain coefficients that are already substituted. The formula is not fully adapted. Some Unicode characters are missing, so I tried to find the longest consecutive sequences of emojis.
+
+* `tr` - `vector<int>` - const - the start of emoji sequences (in Unicode).
+* `ts` - `vector<int>` - const - the difference between the numbers of the first and last elements in the sequences.
+* `cet(t)` - a function used to calculate the sequence number of an emoji.
+* `cev(t)` - a function used to calculate the decimal value of the emoji in Unicode.
+
+$$cet(t) = \left\lfloor 23 \cdot \left(3 \sqrt[3]{t} + 0.7 \cdot \log(t + 5) \cdot 13 + \frac{t \bmod 86400}{86400} + 11 \cdot \log_2(t) + 17 \cdot \sin\left(\frac{2 \pi t}{86400}\right) + 2 \cdot \cos\left(\frac{2 \pi t}{86400}\right) + \left\lfloor \frac{t}{86400} \right\rfloor^2\right) \right\rfloor \bmod 4$$
+
+$$cev(t) = tr[cet(t)] + \left\lfloor 17 * (3 \cdot sin(2 * pi * t / o.7) + 5 * (3 \sqrt[3]{t} + 13 \cdot \log(t + 11)) \right\rfloor \bmod cet(t)$$
+
+and is output looks like `&#{cev(t)};`.
+
+### Color:
+Here, some cyclic operations are used, making it challenging to represent in a formulaic manner. I'll express it in pseudocode with a mix of mathematical formulas. This pseudocode may seem unconventional, but I am a genius, billionaire, and philanthropist. I have the complete right to use my algorithmic language if I am confident it will be understood by the reader (generally a mix of languages, but I believe it's quite evident).
+
+* `sv(t)` - a function used to fill an array (not implemented as a function).
+* `cf(n)` - a function used to precalculate factorial (not implemented as a function).
+* `num2permutation(k, n)` - a function used to determine the required permutation (from $n!$) of the sequence corresponding to number k in lexicographical order.
+
+```
+func sv(int t) -> vector<int> {
+  vector<int> el(3, 0);
+
+  el[0] = t mod 1000;
+  el[1] = ⌊(t mod 1000000 - el[0]) / 1000⌋;
+  el[2] = ⌊(t - el[1] - el[0]) / 1000000⌋;
+
+  return el;
+}
+```
+
+```
+func cf(int n) -> vector<int> {
+  vector<int> factorials(n + 1, 1);
+  for i from 2 to n + 1 {
+    factorials.push(factorials[i - 1] * i);
+  }
+
+  return factorials;
+}
+```
+
+```
+func num2permutation(int k, int n) -> vector<string> {
+  vector<string> permutation(n, "0");
+  vector<bool> was(n+1, false);
+  int cur_free, already_was;
+
+  for i from 1 to n {
+    already_was = ⌊k / factorials[n - i]⌋;
+    k = k mod factorials[n - i];
+    cur_free = 0;
+
+    for j from 1 to n {
+      if was[j] is false:
+        cur_free += 1;
+
+        if cur_free == already_was + 1:
+          permutation[i - 1] = (el[j - 1] mod 256) -> string;
+          was[j] = true;
+    }
+  }
+
+  return permutation;
+}
+```
+
+The color is output in the RGB format.
+
+---
+
+*p.s. I feel that the task of finding the required permutation can be solved with a lower asymptotic complexity (`color.js`). If you have ideas, please write to me about it.*
+
+*p.s. [v2] I hate MathJax(((*