A323916 - OEIS

%I #41 Jan 06 2022 12:50:36

%S 6,21,24,28,54,68,69,96,112,124,133,141,189,216,237,272,284,301,384,

%T 388,448,481,486,496,501,508,589,621,657,669,781,796,864,964,1025,

%U 1029,1077,1088,1136,1141,1269,1317,1348,1357,1372,1417,1536,1537,1552,1701,1792,1796

%N Numbers k with exactly two distinct prime divisors and such that cototient(k) is a square.

%C The integers with only one prime factor and whose cototient is a square are in A246551.

%C This sequence is the intersection of A007774 and A063752.

%C There are exactly two different families of integers which realize a partition of this sequence (A323917 and A323918); there is also another family with the even perfect numbers of A000396 which is a subsequence of this sequence.

%C See the file "Subfamilies of terms" (& II) in A063752 for more details, proofs with data, comments, formulas and examples.

%H Harvey P. Dale, <a href="/A323916/b323916.txt">Table of n, a(n) for n = 1..1000</a>

%F 1st family (A323917): if k = p^(2s+1) * q^(2t+1) with s,t >=0, p, q primes and p + q - 1 = M^2, then cototient(k) = (p^r * q^s * M)^2. The primitive terms are p*q with cototient(p*q) = p+q-1 = M^2

%F 2nd family (A323918): if k = p^(2s) * q^(2t+1) with s>=1, t>=0, p, q primes, p < q and such that p*(p+q-1)= M^2, then cototient(k) = (p^(s-1) * q^t * M)^2. The primitive terms are p^2 *q with cototient(p^2 * q) = p * (p+q-1) = M^2

%F 3rd family (A000396): the even perfect Numbers, if 2^p - 1 is a Mersenne prime, then cototient(2^(p-1) * (2^p - 1)) = (2^(p-1))^2.

%e 1st family: 189 = 3^3 * 7 and cototient(189) = 9^2;

%e 2nd family: 272 = 2^4 * 17 and cototient(272) = 12^2;

%e 3rd family: 8128 = 2^6 * 127 and cototient(8128) = 64^2.

%t Select[Range[1800], 2 == Length@ FactorInteger@ # == 2 && IntegerQ@ Sqrt[# - EulerPhi@ #] &] (* _Giovanni Resta_, Feb 27 2019 *)

%t Select[Range[2000],PrimeNu[#]==2&&IntegerQ[Sqrt[#-EulerPhi[#]]]&] (* _Harvey P. Dale_, Jan 06 2022 *)

%o (PARI) isok(n) = (omega(n)==2) && issquare(n - eulerphi(n)); \\ _Michel Marcus_, Feb 10 2019

%o (Sage) [n for n in (1..2500) if len([1 for d in divisors(n) if is_prime(d)])==2 and is_square(n - euler_phi(n))] # _G. C. Greubel_, Mar 01 2019

%Y Subsequence of A063752.

%Y Cf. A051953.

%Y Cf. A000396, A323917, A323918

%K nonn

%O 1,1

%A _Bernard Schott_, Feb 09 2019