A341857 a(n) = psi(n^2)/n, psi = A002322.

1, 1, 2, 1, 4, 1, 6, 2, 6, 2, 10, 1, 12, 3, 4, 4, 16, 3, 18, 1, 2, 5, 22, 2, 20, 6, 18, 3, 28, 2, 30, 8, 10, 8, 12, 3, 36, 9, 4, 2, 40, 1, 42, 5, 12, 11, 46, 4, 42, 10, 16, 3, 52, 9, 4, 6, 6, 14, 58, 1, 60, 15, 6, 16, 12, 5, 66, 4, 22, 6, 70, 6, 72, 18, 20, 9

Offset: 1

Author

Jianing Song, Feb 21 2021

Keywords

• nonn

Comments

It is easy to show that n divides psi(n) for all n.

For k >= 2, n^k divides psi(n^(k+1)) if and only if n is not of the form 2*(p_1)^(e_1)*(p_2)^(e_2)*...*(p_m)^(e_m), where p_i are distinct odd primes not congruent to 1 modulo 2^k.

It seems that every positive integer occurs in this sequence. The first occurrence of each k is given by A341860.

Examples

			psi(220^2) = psi(2^4 * 5^2 * 11^2) = lcm(psi(2^4), psi(5^2), psi(11^2)) = lcm(4, 20, 110) = 220, so a(220) = psi(220^2)/220 = 220/220 = 1.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A002322, A341860.

Indices of 1 are given by A341858.

Programs

Mathematica

Array[CarmichaelLambda[#^2]/# &, 100] (* Paolo Xausa, Mar 11 2024 *)

PARI

a(n) = A002322(n^2)/n \\ See A002322 for its program

Formula

For odd n > 1, a(2n) = a(n)/2.