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Numbers n such that there are exactly 3 primes p satisfying the inequality n < p < n + tau(n)^2 where tau(n) = A000005(n).
4

%I #24 Jan 05 2018 22:02:42

%S 4,9,21,51,55,62,74,77,82,86,87,91,106,122,123,129,134,142,143,145,

%T 146,155,158,159,161,177,183,214,215,217,237,249,254,259,265,274,278,

%U 298,299,301,309,334,335,339,341,343,358,365,371,377,382,386,394,395,407

%N Numbers n such that there are exactly 3 primes p satisfying the inequality n < p < n + tau(n)^2 where tau(n) = A000005(n).

%H Robert Israel, <a href="/A069231/b069231.txt">Table of n, a(n) for n = 1..10000</a>

%p filter:= n -> nops(select(isprime, [$(n+1) .. (n+numtheory:-tau(n)^2-1)]))=3:

%p select(filter, [$1..1000]); # _Robert Israel_, Jan 05 2018

%t fQ[n_] := Block[{r = Range[n, n + DivisorSigma[0, n]^2]}, If[ PrimeQ@ n, r = Rest@ r]; If[ PrimeQ[ r[[-1]]], r = Most@ r]; Length@ Select[r, PrimeQ] == 3]; Select[Range@410, fQ] (* _Robert G. Wilson v_, Jan 05 2018 *)

%o (PARI) isok(n) = #select(x->isprime(x), vector(numdiv(n)^2-1, k, k+n)) == 3; \\ _Michel Marcus_, Jun 18 2017

%Y Cf. A000005, A049591, A069230, A069232, A069233.

%K easy,nonn

%O 1,1

%A _Benoit Cloitre_, Apr 13 2002