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5 votes
0 answers
161 views

Consecutive integers each of which has a large prime factor

There are many results about consecutive integers all having small prime factors. But what about consecutive integers each of which has a large prime factor? More precisely, let $P(n)$ be the ...
Penchez's user avatar
  • 341
3 votes
3 answers
696 views

For any prime $p$, is there $C$ such that if $x\ge C$, then all but one integer among $x+1, x+2, \dots, x+p$ has Greatest Prime Factor $> p$

I apologize if this is a naive question about greatest prime factors (gpf). I was thinking about the sequence of integers where $\mathrm{gpf}(x) \le p$ where $p$ is any prime. Clearly, as $x$ ...
Larry Freeman's user avatar
1 vote
1 answer
181 views

On the sequence $a(n)=\gcd(2^n-1,\phi(2^n-1))$

For natural $n$, define the sequence $$ a(n)=\gcd(2^n-1,\phi(2^n-1)) $$ It doesn't appear to be in OEIS and starts $1,1,1,1,9,1,1,1,3,1,9,1,3,1,1,1,27,1,75,49$ Q1 Can we unconditionally prove $a(n)=1$...
joro's user avatar
  • 25.4k
1 vote
0 answers
60 views

On the parity of $(2^{\varphi(n)}-1) \bmod{n^2}$

For odd integer $n$ define the function $$ J(n)=(2^{\varphi(n)}-1) \bmod{n^2}$$ $J(n)$ is integer in $[0,n^2-1]$ and it is divisible by $n$. Integer $n$ is Wieferich number iff $J(n)=0$ and if $n$ is ...
joro's user avatar
  • 25.4k
1 vote
0 answers
70 views

Is this factorization problem in EXP?

Factorization is not known to have a polynomial time algorithm. Traditionally the input length is number of bits in representation of the integer to be factored. However now consider integers of form $...
Turbo's user avatar
  • 13.9k