|
|
A341858
|
|
Numbers k such that psi(k^2) = k, psi = A002322; indices of 1 in A341857.
|
|
3
|
|
|
1, 2, 4, 6, 12, 20, 42, 60, 84, 156, 220, 420, 660, 780, 1092, 1806, 1860, 2436, 3612, 3660, 4620, 5060, 5460, 8268, 8580, 12180, 12324, 13020, 15180, 18060, 20460, 24180, 24492, 25620, 29820, 31668, 40260, 41340, 44220, 46956, 47580, 57876, 60060, 61620, 86268, 88620
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
For all k we have k divides psi(k^2). This sequence gives those k such that the quotient is 1.
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.
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.
The LCM of two terms is again in this sequence.
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.
|
|
LINKS
|
|
|
EXAMPLE
|
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.
5060 = 4 * 5 * 11 * 23 is a term since 5-1 | 4, 11-1 | 4*5 and 23-1 | 4*5*11.
|
|
MATHEMATICA
|
Select[Range[10^5], CarmichaelLambda[#^2] == # &] (* Paolo Xausa, Mar 11 2024 *)
|
|
PROG
|
(PARI) isA341858(n) = (A002322(n^2)==n) \\ See A002322 for its program
|
|
CROSSREFS
|
A229289 gives the set of prime factors of the terms.
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|