A324746 Numbers k with exactly two distinct prime factors and such that phi(k) is square, when k = p^(2s+1) * q^(2t+1) with p < q primes, s,t >= 0.

10, 34, 40, 57, 74, 85, 136, 160, 185, 202, 219, 250, 296, 394, 451, 489, 505, 513, 514, 544, 629, 640, 679, 802, 808, 985, 1000, 1057, 1154, 1184, 1285, 1354, 1387, 1417, 1576, 1717, 1971, 2005, 2047, 2056, 2125, 2176, 2509, 2560, 2594, 2649, 2761, 2885, 3097

Offset

1,1

Comments

An integer belongs to this sequence iff (p-1)*(q-1) = m^2.

This is the first subsequence of A324745, the second one is A324747.

Some values of (k,p,q,m): (10,2,5,2), (34,2,17,4), (40,2,5,4), (57,3,19,4), (74,2,37,6), (85,5,17,8).

The primitive terms of this sequence are the products p * q, with p < q which satisfy (p-1)*(q-1) = m^2; the first few are 10, 34, 57, 74, 85, 185. These primitives form exactly the sequence A247129. Then the integers (p*q) * p^2 and (p*q) * q^2 are new terms of the general sequence.

The number of semiprimes p*q whose totient is a square equal to (2*n)^2 can be found in A306722.

Links

Table of n, a(n) for n=1..49.

Formula

phi(p*q) = (p-1)*(q-1) = m^2 for primitive terms.

phi(k) = (p^s * q^t * m)^2 with k as in the name of this sequence.

Example

629 = 17 * 37 and phi(629) = 16 * 36 = 9^2.
808 = 2^3 * 101 and phi(808) = (2^1 * 101^0 * 10)^2 = 20^2.

Maple

N:= 10^4:
Res:= {}:
p:= 1:
do
  p:= nextprime(p);
  if p^2 >= N then break fi;
  F:= ifactors(p-1)[2];
  dm:= mul(t[1]^ceil(t[2]/2), t=F);
  for j from (p-1)/dm+1 do
    q:= (j*dm)^2/(p-1) + 1;
    if q > N then break fi;
    if isprime(q) then Res:= Res union {seq(seq(
      p^(2*s+1)*q^(2*t+1), t=0..floor((log[q](N/p^(2*s+1))-1)/2)),
      s=0..floor((log[p](N/q)-1)/2))} fi
  od
od:
sort(convert(Res, list)); # Robert Israel, Mar 22 2019

Mathematica

Select[Range[6, 3100], And[PrimeNu@ # == 2, IntegerQ@ Sqrt@ EulerPhi@ #, IntegerQ@ Sqrt[Times @@ (FactorInteger[#][[All, 1]] - 1 )]] &] (* Michael De Vlieger, Mar 24 2019 *)

Prog

(PARI) isok(k) = {if (issquare(eulerphi(k)), my(expo = factor(k)[, 2]); if ((#expo == 2)&& (expo[1]%2) == (expo[2]%2), return (1)); ); } \\ Michel Marcus, Mar 18 2019

Crossrefs

Cf. A039770, A062732, A221285, A054755, A324745, A324747, A306908.

Cf. A306722, A247129 (subsequence of primitives).

Sequence in context: A320565 A177221 A045087 * A119086 A195900 A322412

Adjacent sequences: A324743 A324744 A324745 * A324747 A324748 A324749

Keyword

nonn

Author

Bernard Schott, Mar 12 2019

Status

approved