Problems & Puzzles: Puzzles

 Puzzle 1134 F(n)=|(2^n-1)/(2n+1)| By Sebastián Martín Ruiz. Consider the integer values of the function F(n)=(2^n-1)/(2n+1). We have if F(n) is prime then n and 2n+1 are also prime numbers. Q1) Prove it or find a counterexample. I have found the following values expressed in MATHEMATICA F[n_]:=(2^n-1)/(2n+1) Do[If[PrimeQ[F[n]],Print[F[n]," ",n," ",2n+1," ",PrimeQ[F[n]]," ",PrimeQ[n]," ",PrimeQ[2n+1]]],{n,1,15000}] 89 11 23 True True True 178481 23 47 True True True 57912614113275649087721 83 167 True True True 10350794431055162386718619237468234569 131 263 True True True 2150061062571132232545030150231494321261931502937914161854451735782815972183153772 9658958459122860204118390753258481506847174729117738689892562247720853011571496235 529484213513789047439494933924925933540771001858448005515782538708941691223325271 4054247018216597994795059161567922302450277281351135838393171424038832688432240078 3612641615239043555390859277387539681570185502584761638520908267561579157052834132 2600081615171254383858106628160065027869053471937111299739319072106813684059679052 5950480851370510277560248341182341805553054000587378384785994695875587394905226703 1496056898302577682299877147709491924833025835697991410798675970511901340787180117 3050848254256728441883811900056344398559369122120306013704764871309550287777529006 2907208508727269017130916691676838817452529964938878349395785642430571852241837461 6041363744484437301750818895020562905124977171775774927365557840817319985657655985 1810482251652034030170103412392676747278466577677948062882162827968765173619854133 0802238405154786248043073 3359 6719 True True True Q2) Find more prime values of F(n) , n and 2n+1

During the week 10-16 JUne 2023, contributoons came from Michael Branicky, Alessandro Casini, Giorgos Kalogeropoulos, Oscar Volpatti, Emmanuel Vantieghem

Michael wrote:

This problem is partially studied in OEIS A239638, so besides n =  11, 23, 83, 131, and 3359 above, we have n = 130439 and 406583 are solutions with n and 2n+1 prime.

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Alessandro wrote:

Two related OEIS sequence are A085724 and A002326.

Q1)  I could neither find a counterexample nor prove the statement. However, here is the proof of a (very) little step towards it. Therefore the search for counterexamples is limited to composite with 2n+1 composite only, that are nevertheless a lot.

Q2) No other N < 10^5

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Giorgos wrote:

Q2: We know two more terms from A239638
n=130439 (prime) and 2n+1=260879 (prime)
n=406583 (prime) and 2n+1=813167 (prime)
F(n) is also prime in both cases because it is stated that 2^n-1 is semiprime

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Oscar Volpatti wrote:

About Q1, the statement by Sebastián Martín Ruiz is true.
I'll prove it in two steps.
Step 1: if F(n) is a prime number, then n is prime too.
Step 2: if F(n) is an integer and n is a prime number, then 2*n+1 is prime too.

STEP 1.
Given the Mersenne number M(n)=2^n-1 (exponent not notessarily prime), let A(n)=2*n+1 be its desired factor, so that F(n)=M(n)/A(n).
Also define functions B(n)=2^sqrt(n)-1 and C(n)=2^(n/2)-1, which will be used later.
Consider the following inequality chain:
n < A(n) < B(n) < C(n) < F(n) < M(n).
It necessarily holds for n large enough, because the given functions have different, increasing growth rates.
More precisely, the chain actually holds for every integer n>40; the bottleneck is inequality A(n)<B(n).

Assume that F(n) is a prime number but n is composite, so that n=x*y with 2<=x<=y.
As F(n) is prime, M(n) has no divisor d belonging to interval A(n)<d<F(n):
if d is coprime with F(n), then d<= A(n);
otherwise d must be a positive multiple of F(n), therefore d>=F(n).
But, as n has the proper factor y, M(n) has the proper factor M(y)=2^y-1, belonging to the forbidden interval:
n=x*y<=y^2, so y>=sqrt(n), so M(y)>=B(n)>A(n);
x>=2, so y=n/x<=n/2, so M(y)<=C(n)<F(n).

Exaustive search for 0<=n<=40.
F(n) is an integer for n in {0,3,8,11,15,20,23,35,36,39}.
F(n) is a prime for n in {11,23}; in both cases, n is prime too.

STEP 2
The value F(2)=3/5 is not an integer, so let n be an odd prime. A(n) divides M(n), so:
2^n == 1  mod  A(n),
4^n = (2^n)^2 == 1*1 == 1  mod  A(n).
By Miguel Pineda Martin's proof of puzzle 1133, A(n) is prime.

About Q2, next prime value of sequence F(n) is F(130439),
I tested prime arguments n<=200000; A(n) divides M(n) for about one thousand primes.
I checked if M(n) has more small factors of the form q=2*n*k+1, with k<=10^6, excluding about 50% of candidates.
Then I checked if F(n) is a Fermat 3-PRP, obtaining a positive answer for n=130439.
According to factordb.com, F(130439) is actually a proven prime, with certificate uploaded by Alan Mock in February 2023.

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Emmanuel wrote:

Q2
The sequence  11, 23, 83, 131, 3359  is  A239638  in the OEIS.
There we find two extra terms : 130439  and  406583.
Also, in the comment lines we find that the next term (if existing) is > 500000

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