login
Numbers k such that psi(k^2) = k, psi = A002322; indices of 1 in A341857.
4

%I #42 Aug 28 2024 11:57:48

%S 1,2,4,6,12,20,42,60,84,156,220,420,660,780,1092,1806,1860,2436,3612,

%T 3660,4620,5060,5460,8268,8580,12180,12324,13020,15180,18060,20460,

%U 24180,24492,25620,29820,31668,40260,41340,44220,46956,47580,57876,60060,61620,86268,88620

%N Numbers k such that psi(k^2) = k, psi = A002322; indices of 1 in A341857.

%C For all k we have k divides psi(k^2). This sequence gives those k such that the quotient is 1.

%C Apart from 5 exceptional terms, every term is the product of 4 and distinct odd primes. The exceptional terms are precisely the 5 terms in A014117.

%C Except for k = 1, 2, 6, 42, 1806, k is a term if and only if k = 4*(p_1)*(p_2)*...*(p_m), where p_1 < p_2 < ... < p_m are odd primes, (p_i)-1 | 4*(p_1)*(p_2)*...*(p_(i-1)) for all 1 <= i <= m.

%C The LCM of two terms is again in this sequence.

%C Is this sequence infinite? If this sequence is finite, it means that there exists a term of the form k = 4*(p_1)*(p_2)*...*(p_s), where p_1 < p_2 < ... < p_s are odd primes such that: for every (e_0, e_1, ..., e_s) in {0, 1}^(s+1), 2^((e_0)+1)*(p_1)^(e_1)*(p_2)^(e_2)*...*(p_s)^(e_s)+1 is either composite or equal to some p_i. That term must be divisible by all other terms, since there are no more odd primes q other than p_1, p_2, ..., p_s such that q-1 | k.

%C Numbers k such that b^k == 1 (mod k^2) for every b coprime to k. Proof: these are numbers k such that psi(k^2) divides k, which holds if and only if psi(k^2) = k. Subsequence of A124240 (see my comment there). If k is a term of the sequence and k+1 is prime, then k*(k+1) is also a term. - _Thomas Ordowski_, Jul 26 2024

%H Amiram Eldar, <a href="/A341858/b341858.txt">Table of n, a(n) for n = 1..6034</a> (terms below 10^11; terms 1..249 from Jianing Song)

%H Jianing Song, <a href="/A341858/a341858.txt">Factorization of terms <= 15*10^6 other than 1, 2, 6, 42, 1806</a>

%e 1092 = 4 * 3 * 7 * 13 is a term since 3-1 | 4, 7-1 | 4*3 and 13-1 | 4*3*7. Indeed, we have psi(1092^2) = 1092.

%e 5060 = 4 * 5 * 11 * 23 is a term since 5-1 | 4, 11-1 | 4*5 and 23-1 | 4*5*11.

%t Select[Range[10^5], CarmichaelLambda[#^2] == # &] (* _Paolo Xausa_, Mar 11 2024 *)

%o (PARI) isA341858(n) = (A002322(n^2)==n) \\ See A002322 for its program

%Y Cf. A002322, A341857, A014117.

%Y A229289 gives the set of prime factors of the terms.

%Y Subsequence of A124240.

%K nonn

%O 1,2

%A _Jianing Song_, Feb 21 2021