You can divide by 2 or multiply by 1 / 2. Welcome to the forum.

]]>Welcome to the forum.

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Thanks for the link. I understand how you want to count them say for ()()(). But what is the question you are asking?

I do see the formula you are conjecturing on page 5 of that pdf.

]]>also i didnt understand whats between ε0 and ω1]]>

The solutions found in post #10 are correct but I am afraid their might be more that ht method I used there is missing. I hate using reasoning in math, that is why I dislike and mistrust the whole concept of proof. Trouble is, computation although more reliable can leave you hanging.

]]>A=2x3x5x7x11x13x17........ up to

m=Any multiple

A-pm= A number factorable by a factor of p

if p has one.To get A-pm, go through the primes, subtracting p as many times as you like.

Example:

p=129

A=2x3x5x7x11

2x3x5x7=210 210-129=81

81x11=891 891-(129x6)=117

A-129m=117

117 will have a factor the same as p if it has any.

117/129=39/43 129 is NOT prime (Common denominator =3)

In this way we can find out if p is composite without ever having to use a number

. Might be useful for computers.]]>The number they want next is

.Welcome to the forum.

]]>For part 2:

As you want dA/dt I would work with dV/dt = dV/dA . dA /dt

As D = 2 when H = 2. => r = 0.5h.

Express both V and A in terms of h and eliminate h to get V in terms of A.

Differentiate that.

Bob

]]>Since the limit of -x^2 as x approaches 0 is 0 and the limit of x^2 as x approaches 0 is 0 and

stays between them, the limit of it as x approaches 0 is also 0.Intuitively we can understand by the following drawing that Mathegocart provided:

If your function the green line always stays between the red and the blue line, then as the red and the blue line get closer together the green line gets sandwiched in between.

]]>Example:

p=130

Rd. Up to nearest prime= 13

Next prime after that = 17

17-1=16

Largest prime gap <130 = 16 (Correct)

This works because the greatest number of composites between two primes occurs when factors are not combined. So what could have been two composites is actually just one, like 15=3x5. To create the greatest possible number of composites I start at 2 not 0. 0 has an infinite number of prime factors, and so the greatest gap between the next repeat will occur after 0. Starting with the smallest composite which is NOT combined factors, I move up. Deleting all numbers factorable by primes less than the square root. The first time I attain TWO primes is when I reach the second prime after the square root. So this -1 is the gap required to create two non-composites with greatest possible occurrence of composites.

Largest prime gap <p =

Rd. Up to second nearest prime -1.]]>