Is there a way to find the relation between numbers? I have 2 groups of numbers.

Group A

5 --> 20
6 --> 32
7 --> 112
8 --> 192
9 --> 576
10 --> 1024

Group B

7 --> 56
8 --> 80
9 --> 432
10 --> 672

Is there a way to find relation between numbers and if there's no way to do that except mentally, what is the relation in the two groups?


Unfortunately, there is often no definitive way to determine a pattern within some given set of numbers, since patterns may be as arbitrary as you like. However, there are some ways of seeing patterns a little more clearly.

By way of example, I'm going to rewrite these relations in a slightly different way. Using that as a hint, try and find the relations yourself. I'll place the answer in a spoiler below.

Group A:

$5\to 2^2\cdot5$

$6\to 2^5$

$7\to 2^4\cdot7$

$8\to 2^6\cdot3$

$9\to 2^6\cdot9$

$10\to 2^{10}$

Group B:

$7\to 2^3\cdot7$

$8\to 2^4\cdot5$

$9\to 2^4\cdot3\cdot9$

$10\to 2^5\cdot3\cdot7$

Group A:

When $n$ is odd, $n\to n2^{n-2}$. When $n$ is even, $n\to2^{n-3}(n-2)$

Group B:

When $n$ is odd, $n\to 2^{n-6}\cdot n(n-3)$. When $n$ is even, $n\to 2^{n/2}(n-3)(n-7)$

  • $\begingroup$ thank you, jazzy. i have found a relation for all numbers in group a it's 2^n-3 (n-2+2(n mod 2)) but isn't there such one for group b without separating odd and even . $\endgroup$ Mar 14 '17 at 7:39
  • $\begingroup$ oh i found one for group b too : 2^((n-6(n mod 2)) / (-(n mod 2)+2)) x (n-7 + 7(n mod 2)) x (n-3) it's a bit long but it works . thanks $\endgroup$ Mar 14 '17 at 8:02
  • $\begingroup$ If the equations are going to get that complicated, you might as well lagrange group B to get f(x)=−(220x^3)/3+1924x^2−49328x/3+46032 which is actually shorter than what you had... i think. $\endgroup$
    – Wen1now
    Mar 14 '17 at 8:43
  • $\begingroup$ I don't like the answer for B. You are splitting it into 2 sequences, which means you only have 2 samples for each, which isn't enough to determine a relation. You might as well just do a linear equation, so odd= 188n-1260 and even= 296n-2288 $\endgroup$
    – Kruga
    Mar 14 '17 at 8:55
  • $\begingroup$ I agree that it's a little fishy since there are so few samples. I feel that the answer to Group B suffices due to the context given by group A, though certainly I'm sure that there are more than a few distinct solutions because the given sequence is so short. $\endgroup$ Mar 14 '17 at 9:00

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