A246748 Numbers n such that A242719(n) = (prime(n))^2+1 and A242720(n) - A242719(n) = 2*(prime(n)+1).

3, 52, 104, 209, 343, 373, 398, 473, 628, 2633, 3273, 7538, 8060, 8813, 9025, 10847, 12493, 13768, 14196, 15486, 16865, 17486, 18362, 18613, 18842, 21175, 23522, 31825, 33537, 34507, 38740, 39603, 41802, 41947, 43314, 45479, 47550, 47668, 47787, 50321, 50682

Offset: 1

Author

Vladimir Shevelev, Sep 02 2014

Keywords

• nonn

Comments

If the sequence is infinite, then lim inf(A242719(k)/(prime(k))^2) = 1 and lim inf(A242720(k)/(prime(k))^2) = 1.

In connection with this, one can conjecture that A242719(k) ~ A242720(k) ~ (prime(k))^2, as k goes to infinity (cf. A246819, A246821).

n is in the sequence if and only if prime(n)>=5 and is in the intersection of A001359, A062326, A157468.

Proof. Firstly note that A242719(n) = prime(n)^2 + 1 if and only if prime(n)^2 - 2 is prime. Indeed, let prime(n)^2 + 1 be A242719(n). Then we have lpf(prime(n)^2 - 2) > lpf(prime(n)^2) = prime(n). It is possible only when prime(n)^2 - 2 is prime, i. e., prime(n) is in A062326. Add that prime(n)^2+1 is the smallest value of A242719(n).

Let A242720(n) = A242719(n) + 2*prime(n) + 2 = prime(n)^2 + 2*prime(n) + 3. Then, by the definition of A242720, we have lpf(prime(n)^2 + 2*prime(n) + 2) > lpf(prime(n)*(prime(n)+2)) >= prime(n). Thus prime(n) + 2 is prime, i.e., prime(n) is in A001359. Besides, lpf(prime(n)^2 + 2*prime(n) + 2) > prime(n), or lpf((prime(n)+1)^2 + 1) >= prime(n+1) = prime(n) + 2. So (prime(n)+1)^2+1 is prime, i.e., prime(n) is also in A157468.

Add that, for n>=3, N=prime(n)^2 + 2*prime(n) + 3 is the smallest possible value of A242720(n). Indeed, let prime(n)^2+1 <= N <= prime(n)^2 + 2*prime(n) + 2. Then prime(n)^2-2 <= N - 3 <= prime(n)^2 + 2*prime(n) - 1. Since it should be lpf(N-3) >= prime(n), then there are only two possibilities: N-3 = prime(n)^2 + prime(n) or N-3 = prime(n)^2. However, lpf(prime(n)^2 + prime(n)) = 2, while, although lpf(prime(n)^2) = prime(n), however, in this case, lpf(N-1) = lpf(prime(n)^2+2) = 3, n>=3, and, so the inequalities lpf(N-1) > lpf(N-3) >= prime(n) are impossible in the considered cases for n>=3. - Vladimir Shevelev, Sep 03 2014

Crossrefs

Cf. A001359, A062326, A157468, A242719, A242720, A246819, A246821.

Extensions

More terms from Peter J. C. Moses, Sep 02 2014