ETA: In case someone notices the similarity, this question is not a duplicate of .     Which statement is true about the sequence $\{f(n)\}$:    $$ \left\{  \begin{array}{l} f(1)=1 \\  f(2n)=f(n) \\  f(2n+1)=f(n)+f(n+1) \end{array} \right.  $$   a)All prime numbers appear in $\{f(n)\}$   b)All Mersen primes appear in $\{f(n)\}$   c)A finite set of primes appear in $\{f(n)\}$   d)Just primes of the form $4t-1$ appear in $\{f(n)\}$     It's seen that $f(2n)=1$ but I'm not sure that $f(2n+1)$ generates which primes...
123555
An arithmetic function is defined as follows:$f(1)=1$,  $f(2k)=k$ and $f(2k+1)=f(k)+f(k+1)$. When (for which $k$) is $f(k)$ even?  While it is obvious that $f(4n-1)=f(4n)=2n$, therefore $f(k)$ is even for $k=4n$ and $k=4n-1$, which is possible to prove using induction (isn't it?), I don't know how to prove the converse.  I tried to solve the recurrence or make use of constant difference between certain elements of the sequence $\ldots$ What am I missing?  Thank you.
57235
Let $p_n$ be the $n$th prime number in its natural order. We want to show that  $$p_{n+1} \leq p_1\dotsc p_n + 1 \tag{1}$$  I know $p_1\dotsc p_n + 1 &gt; 1$, so it has at least one prime divisor, but the text says that $p_{n+1}$ cannot exceed the smallest of these divisors, but I cannot understand why, especially when I imagine $p_{n+1}$ being very large. Although some examples confirm that this statement is true, I cannot make a connection as to why the $n+1$ prime number cannot exceed the smallest prime divisor. Once this has been shown to be true, $(1)$ immediately follows.
11111
Let $Y(k)$ be the number consisting of $1$, repeated $k$ times. We know that $Y(2) =11$ is prime. It so happens that $Y(19)$ and $Y(23)$ are also prime.   Are there any more?   Regards,  David
69594
Is this statement true? i.e. If $f(n)$ is injective and $p_n \rightarrow L$, does $p_{f(n)} \rightarrow L$?? (I realize that it's technically only a rearrangement if $f(n)$ is bijective.)   My attempt at a proof:  Since $p_n \rightarrow L$, we have that, for all $n$ except $n \le N$, $d(p_n, L) &lt; \epsilon$.  Let $S = \{n \ | \ f(n)\le N\}$, let $n_0$ be the largest $n \in S$, we know there is such a largest $n$ because $f(n)$ is injective. Now we have that $\forall \ n &gt; n_0 \ f(n) &gt; N$ which implies that $p_{f(n)} \rightarrow L$, as required.  Is this correct? Does my proof work?
42691
Exactly what the name entails. The function $2^n-1$ I see largely tends to generate primes when $n$ is prime. However, a week ago I heard that this was horribly false. Please show me a disproof.
50596
Starting from 2, the sequence of sums of all consecutive primes is:  $$\begin{array}{lcl}2 &amp;=&amp; 2\\     2+3 &amp;=&amp; 5 \\     2+3+5 &amp;=&amp; 10 \\     2+3+5+7 &amp;=&amp; 17 \\     2+3+5+7+11 &amp;=&amp; 28 \\     &amp;\vdots&amp; \end{array} $$ If the $n^\text{th}$ prime is $P_n$, then we can write $S_n=\sum_{i=1}^n P_n$.  I conjecture that the sequence $S_n$ contains an infinite number of primes.  I doubt I'm the first person ever to think this, but I cannot find reference to the idea, nor can I conceive a proof, nor a disproof.  Computationally, it can be verified that     $$S_{13,932}=998,658,581=P_{50,783,012},$$ and the sequence shows no sign of slowing down.  Can you prove the sequence $S_n$ contains an infinite number of primes?
15395
How do I see that $$f_{n+1}f_{n-1} = f_n^2 + (-1)^n$$ for $n \ge 2$, where $f_1 = 1$, $f_2 = 1$, and $f_{n+2} = f_{n+1} + f_n$ for $n \in \mathbb{N}$?
119261
Recall that the Fibonacci sequence is defined by $F_0=0, F_1=1$ and $F_n =F_n−1 +Fn−2$, for $n\geq 2$.  Prove that for all $n\geq 1$ we have $F_n&lt;{\left(\frac 74\right)}^n$.
29148
Let $f_0=2$ and $f_n=f_{n-1}+\ln(f_{n-1})$. What is a good asymptotic to the sequence $f_n$? With good I mean much better than $f_n \sim \dfrac{3n \ln(2)\ln(n)}{2}$.
65396
I need something explained or corrected:  In my number theory class we proved that there are an infinite number of primes using Euler's Phi Totient. It went something like this:  Let $M = p_1 p_2 \dots p_n$ be the product of all primes. Consider $1 &lt; A \le M$:  Some prime must divide $A$, call it $q$. Since $q$ must be one of the primes, $q$ must divide $M.$  So the $gcd(A,M) &gt; 1.$ Thus $\phi(M) = 1$ ...?   Which is not even and contradicts the theorem that $\phi(N)$ is even for $N&gt;2.$ Therefore there exists an infinite number of primes.  I get confused by the statement "Thus $\phi(M) = 1$"....  Did i possibly copy this proof down wrong? or am I missing something? Thank you in advance.   Edit: by consider a such that...i meant consider an integer '$a$' I replaced it with $A$ to hopefully make it more clear.   Edit2: I'm sorry I am not familiar with using the equation editor. This is not a homework assignment, just studying for my exam. I just want to be able to understand this or clearify it.
18316
Prove that any two consecutive terms of the Fibonacci sequence are relatively prime.   My attempt:    We have $f_1 = 1, f_2 = 1, f_3 = 2, \dots$, so obviously $\gcd(f_1, f_2) = 1$. Suppose that $\gcd(f_n, f_{n+1}) = 1$; we will show that $\gcd(f_{n+1}, f_{n+2}) = 1$. Consider $\gcd(f_{n+1}, f_{n+2}) = \gcd(f_{n+1}, f_{n+1} + f_n)$ because $f_{n+2} = f_{n+1} + f_n.$ Then $\gcd(f_{n+1}, f_{n+1} + f_n) = \gcd(f_{n+1}, f_{n}) = 1$ (gcd property).  Hence, $\gcd(f_n, f_{n+1}) = 1$ for all $n &gt; 0$.  Am I on the right track? Any feedback would be greatly appreciated.  Thanks,
45738
I have seen the past threads but I think I have another proof, though am not entirely convinced.  Suppose there are only finitely many primes $p_1, ..., p_n$ of the form $6k-1$ and then consider the number  $N = (p_1.p_2. ... .p_n)^2 - 1$  If all of the primes dividing N are of the form $6k+1$ then $N$ leaves remainder $1$ upon division by $6$. However, the RHS has remainder $1-1 = 0$ or $-1-1 = -2$ upon division by $6$, depending on whether $n$ is even or odd.  Therefore there must be at least one prime of the form $6k-1$ dividing $N$, and thus some $p_j$ divides N, and so divides $(p_1.p_2. ... .p_n)^2 - N$ which is $1$, a contradiction.  Apologies for this question if it appears unnecessary I'm just beginning number theory and trying to get to grips with it.
73842
Let $T(n,x)$ be the nth Chebyshev polynomial of the first kind and let $U(n-1,x)$ the $(n-1)$th Chebyshev polynomial of the second kind. Would any one kindly help show that   1) $n$ is prime iff $T(n,x)$ is irreducible in $\mathbb{Z}[x]$.  2) $n$ is prime iff $U(n-1,x)$, expressed in powers of $(x^2-1)$, is irreducible in $\mathbb{Z}[x]$.   Many Thanks!!    No "ordering" is implied here. The wording of the question was done in similar fashion  as any question in any math/research question. I do not see how a person who is asking for help would be ordering people to help.   Any way back to the topic,  For the first part of the question, I noticed that if n is prime, then T(n,x) satisfies  Eisenstein's Irreducibility Criterion. But I am not sure how to show if T(n,x) is irreducible then n is prime.
97373
Consider numbers of the type $n=2^m+1$. Prove that such an $n$ is prime only if $n=F_K$ for some $ k \in N$, where $F_k$ is a Fermat Prime. Consider numbers of the type $n=a^m-1$ where $a&gt;1$ is a natural number. Prove that such an $n$ is prime only if $n=M_p$ for p-prime, where $M_p$ is a Mersennt Prime.   I do not understand what the question is asking me for. Can I get some clarification?
36333
For the recurrence relation:  $f(0)=1$  $f(1)=1$  $f(2)=2$  $f(2n)=f(n)+f(n+1)+n\ \ \ (\forall n&gt;1)$  $f(2n+1)=f(n)+f(n-1)+1\ \ \ (\forall n\ge 1)$  (the first numbers of the sequence are: 1, 1, 2, 3, 7, 4, 13, 6, 15, 11, 22, 12, 25, 18, 28)  Given some number $S$, I would like to find $n$ such that $f(n) = S$, using an approach that runs in $O(\log n)$ time. Any ideas or tips how to solve it? I tried substitution so far but didn't get far.
111147
Prove that there is no arithmetic progression that consists only of primes.  A question that I've been set; I'm guessing it makes use of primes being written in the form 4k+1 and 4k+3? Not sure where to start.  Thanks.
98853
Prove: Given a prime $p$, there are infinitely many $n\in \mathbb{Z}^+$, for   which $pn+1$ is a prime.   This is a simplified version of , so is there any elementary solution to this problem?  A proof by contradiction is recommended. Just like the proof of the existence of infinitely many primes by Euclid.