A247011 Numbers n for which A242719(n) = (prime(n) + 2)^2 + 1.

5, 7, 13, 17, 26, 33, 64, 81, 98, 140, 171, 176, 190, 201, 215, 225, 318, 332, 336, 444, 469, 475, 495, 551, 558, 563, 577, 601, 636, 671, 828, 849, 862, 870, 948, 1004, 1064, 1074, 1189, 1198, 1230, 1238, 1305, 1328, 1445, 1449, 1528, 1618, 1634, 1642, 1679

Offset: 1

Author

Vladimir Shevelev, Sep 09 2014

Keywords

• nonn

Comments

(prime(n) + 2)^2 + 1 is the second minimal possible value of A242719(n) after prime(n)^2 + 1. Indeed, by the definition lpf(A242719(n) - 3) > lpf(A242719(n) - 1) >= prime(n), thus after prime(n)^2 + 1 we should consider prime(n)*(prime(n) + 2) + 1. Then prime(n) should be lesser number of twin primes, but then prime(n) + 1 == 0 (mod 3). So, prime(n)*(prime(n) + 2) - 2 == 0 (mod 3). Analogously one can prove that prime(n)*(prime(n) + 4) - 2 == 0 (mod 3).

Note that for the sequence prime(n+1) is in intersection of A006512 and A062326, but prime(n) is not in A062326.

Links

• François Marques, Table of n, a(n) for n = 1..109

Crossrefs

Cf. A242719, A246748.

Formula

If prime(n) is not in A062326, then A242719(n) >= (prime(n)+2)^2 + 1.

Intersection of A247011 and A246824 forms sequence 81, 215, 828, 1189, 1634, ... For these values of n we have A242719(n) - A242720(n) = 2*(prime(n) + 1).

Extensions

More terms from Peter J. C. Moses, Sep 09 2014