How do I find the next number if the given pattern is $$1,2,3,2,3,4,1,2,6,23,14,19,64,69,12,78,152,93,108,?$$ (Find the question mark)
2026-03-25 17:31:55.1774459915
Patterns of no formula.
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Let $p(n)$ be the $n$th term in the sequence. Clearly, this sequence follows the formula:
$$\begin{align} p(x) &= \frac{600631 x^{19}}{121645100408832000}-\frac{791723 x^{18}}{800296713216000}+\frac{196988587 x^{17}}{2134124568576000}\\ &-\frac{41785811 x^{16}}{7846046208000}+\frac{8219611 x^{15}}{38626689024}-\frac{49026370303 x^{14}}{7846046208000}+\frac{26296057821373 x^{13}}{188305108992000}\\ &-\frac{1098593289863 x^{12}}{452656512000}+\frac{320897391017407 x^{11}}{9656672256000}-\frac{39606777445183 x^{10}}{109734912000}\\ &+\frac{30088961291838131 x^9}{9656672256000}-\frac{(12871880314235441 x^8)}{603542016000}+\frac{(5410873671286319827 x^7)}{47076277248000}-\frac{(708875674839982733 x^6)}{1471133664000}+\frac{4033947669590964373 x^5)}{2615348736000}-\frac{(599274486262658993 x^4)}{163459296000} \\&\frac{+(49104110859304547 x^3)}{7916832000}-\frac{(2153634755170519 x^2)}{308756448}+\frac{(3188726258687 x)}{692835}-1320490 \end{align}$$
Thus, $p(20)=42$.
Ok, so that was a joke. However, this illustrates an important point--you can find some formula for $p(n)$ such that $p(20)$ is any value you wish. Without other context or information, coming up with the $20$th term in this sequence is not a well-defined question.