what is the difference between sum of first n primes and prime(prime(n?
Let’s learn what is the difference between sum of first n primes and prime(prime(n. The most accurate or helpful solution is served by Mathematics.
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The seq is: -1, 0, -1, 0, -3, 0, -1, 10, 17, 20, 33, 40, 59, 90, 117, 140, 163, 218, 237, ... http://oeis.org/A239731 Is there's a formula looks like $$a(n) =n^2logn/2$$ for this seq?
Answer:
The sum of the first $n$ primes is asymptotic to $\frac{n^2\log n}{2}$. Heuristically, this is because...
mike at Mathematics Mark as irrelevant Undo
Other solutions
nyssy at Yahoo! Answers Mark as irrelevant Undo
Please -- if you can think of a way to reword this please do. I am wondering also if there is a simple name for this phenomenon, or if there are cases where it's not true. Example: Given an array of random numbers with an even number of elements, sorted...
Answer:
Because these are the same as long as you sort the array first. All the differences are guaranteed to...
Shrey Banga at Quora Mark as irrelevant Undo
Given the set [math] \{ a_1, ..., a_n \} [/math] where [math]a_i \in \mathbb{N}[/math], find two subset [math]A[/math] and [math]B[/math] such that: [math]A \cap B = \emptyset[/math] [math] card(A) = card(B) = k > 0[/math] where [math] 2k \le n...
Answer:
Define [math]D[i,j,k] = true[/math] iff there exists some [math]A,B \subset \{a_1, ..., a_i\}[/math...
Mark Gritter at Quora Mark as irrelevant Undo
The sum of the first 50 terms of an arithmetic sequence is 100. If the common difference (d) is increased by 3, what is the new sum?
Answer:
Actually the new sum is not just 100 + 49(3). It is S= 100 + (1+2+3+...+49)*3 = 100 + (50)(49)/2 *3...
Tom at Yahoo! Answers Mark as irrelevant Undo
Hi, I have used the method of differences (also called a telescoping series) to solve various maths problems before but I don't feel I understand the mechanics of it. I began looking for a more general definition of it and found this in the Mathematical...
Answer:
You must have miswritten the second method. If u_n = f(n + 1) - f(n), then \displaystyle \sum_{n = ...
Sridhar Ramesh at Quora Mark as irrelevant Undo
Answer:
The sum of the first 1,000,000 positive even numbers is: 2 + 4 + 6 + 8 + ... + 2,000,000 The sum of...
wiki.answers.com Mark as irrelevant Undo
ChaCha Mark as irrelevant Undo
Answer:
The sum of the first n terms of the series with starting value "a" and common difference ...
Haley at Yahoo! Answers Mark as irrelevant Undo
I'm looking for formulas including (preferably convergent) sums or products over all primes, as in product [p in P] (1/(1-p^(-s))) = zeta(s) = sum [n=1 to infinity] 1/n^s [s>1] I'd like to collect here as many similar facts about the set of all primes...
Answer:
π(all primes) (1 - 1/p^s) = Σ(n = 1 ... ∞) m(n)/n^s, where m is the Mobius Function ...
☮ Vašek at Yahoo! Answers Mark as irrelevant Undo
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