Equation for product of sequence involving 2

Question

Supposing for all even numbers less that or equal to n say 16,

for example the set:

2, 4, 6, 8, 10, 12, 14, 16

2.4.6.8.10.12.14.16 = 10,321,920 = $2^{15}. 3^{2}. 5^{1}. 7^1$

I would like to find an equation that gives me the products of 2's only not the other factors in the case of $n=16$ this would be:

$2\,.\,2\,.\,2^2\,.\,2\,.\,2^3\,.\,2\,.\,2\,.\,2\,.\,2^4\,=\,2^{15}$

Another example is n = 14

2.4.6.8.10.12.14 = 645,120 = $2^{11}. 3^2. 5^1. 7^1$

this would be $2\,.\,2\,.\,2^2\,.\,2\,.\,2^3\,.\,2\,.\,2\,.\,2\,=\,2^{11}$

Can it be done?

Write it as $,2^8 \cdot 8!,$ then see Highest power of a prime $p$ dividing $N!$. — dxiv, Jun 28 '18 at 19:30

score 1 · Accepted Answer · answered Jun 28 '18 at 19:32

1

Hint $$2 \cdot 4 \cdot \cdots \cdot 2n = (2 \cdot 1) \cdot (2 \cdot 2) \cdot \cdots \cdot (2 \cdot n) = (2 \cdot 2 \cdot \cdots \cdot 2) \cdot (1 \cdot 2 \cdot \cdots \cdot n) .$$

Now, see this answer, which gives a formula for the highest power of a prime dividing $n!$.

answered Jun 28 '18 at 19:32

Travis Willse

99,363

thanks I need a bit of time to understand. – onepound Jun 28 '18 at 19:34
You're welcome. What I wrote is just factoring and rearranging---the more substantive part of the problem is contained in the linked answer. – Travis Willse Jun 28 '18 at 19:35

Equation for product of sequence involving 2

1 Answers1