A242758 Smallest even k such that lpf(k-1) > lpf(k-3) >= prime(n), where lpf=least prime factor (A020639).

6, 8, 14, 14, 20, 20, 32, 32, 32, 44, 44, 44, 62, 62, 62, 62, 74, 74, 74, 104, 104, 104, 104, 104, 104, 110, 110, 140, 140, 140, 140, 140, 152, 152, 182, 182, 182, 182, 182, 182, 194, 194, 200, 200, 230, 230, 230, 230, 242, 242, 242, 272, 272, 272, 272, 272

Offset: 2

Author

Vladimir Shevelev, May 22 2014

Keywords

• nonn

Comments

This is a version of A242720 with the absolute minima of k in the definition. The sequence is nondecreasing. Hypothetically, every pair {a(n)-3, a(n)-1} is a pair of twin primes.

If there exist infinitely many n such that a(n) < A242719(n) < a(n)^2, then from the result in the Shevelev link, it follows that for such n the set of numbers {even k: lpf(k-1) > lpf(k-3) >= prime(n)} either attains the absolute minimum of a(n) only in the case when {a(n)-3, a(n)-1} are twin primes, or does not attain it at all. Therefore, if there is only a finite number of twin primes, we have a contradiction. Thus the above condition is sufficient for infinity of twin primes.

Note also that, if there is only a finite number of twin primes, then after the last pair of them, this sequence will coincide with A242720. Then, in order to avoid a contradiction (again according to the Shevelev link), we should accept that there exists a number N_0 such that, for every n >= N_0, the following inequality holds: max(A242719(n),A242720(n)) > (min(A242719(n),A242720(n)))^2. - Vladimir Shevelev, May 24 2014

It is easy to prove that min(A242719(n), A242720(n)) >= prime(n)^2+1, while we conjecture that max(A242719(n), A242720(n)) <= prime(n)^4. Thus this conjecture implies there are infinitely many twin primes. - Vladimir Shevelev, Jun 01 2014

Links

• Peter J. C. Moses, Table of n, a(n) for n = 2..10001
• V. Shevelev, Theorems on twin primes-dual case, arXiv:0912.4006 [math.GM], 2009-2014.

Crossrefs

Cf. A001359, A006512, A242489, A242490, A242720.

Programs

Mathematica

lpf[k_] := FactorInteger[k][[1, 1]];
a[n_] := a[n] = For[k = If[n == 2, 2, a[n-1]], True, k = k+2, If[lpf[k-1] > lpf[k-3] >= Prime[n], Return[k]]];
Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Nov 03 2018 *)

PARI

lpf(k) = factorint(k)[1,1];
vector(100, n, k=6; while(lpf(k-1)<=lpf(k-3) || lpf(k-3)Colin Barker, Jun 01 2014