A247011 - OEIS

%I #25 Feb 08 2021 14:00:34

%S 5,7,13,17,26,33,64,81,98,140,171,176,190,201,215,225,318,332,336,444,

%T 469,475,495,551,558,563,577,601,636,671,828,849,862,870,948,1004,

%U 1064,1074,1189,1198,1230,1238,1305,1328,1445,1449,1528,1618,1634,1642,1679

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

%C (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).

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

%H François Marques, <a href="/A247011/b247011.txt">Table of n, a(n) for n = 1..109</a>

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

%F 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).

%Y Cf. A242719, A246748.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Sep 09 2014

%E More terms from _Peter J. C. Moses_, Sep 09 2014