OFFSET
1,2
COMMENTS
Conjecture: a(n) > 0 for all n.
If there exists k such that psi(k^2) | k*n and k*n+1 is prime, then A341857(k*(k*n+1)) = n, so a(n) > 0.
For odd n > 1, if a(n) > 0, then a(2n) <= a(n)/2. Proof:
Suppose k is the least integer such that A341857(k) = psi(k^2)/k = n is odd. Note that if m > 1 is odd or m is divisible by 8, then A341857(m) is even. So we must have k = 2t or k = 4t with odd t > 1.
If k = 2t, then n = psi(4t^2)/(2t) = psi(t^2)/(2t), so A341857(t) = psi(t^2)/t = 2n. This gives a(2n) <= t = k/2.
if k = 4t, then n = psi(16t^2)/(4t) = lcm(4, psi(t^2))/(4t). There are two cases: a) if psi(t^2) is divisible by 4, then psi(t^2)/(4t) = n, so A341857(2t) = psi(4t^2)/(2t) = psi(t^2)/(2t) = 2n. This gives a(2n) <= 2t = k/2. b) if psi(t^2) is not divisible by 4, then psi(t^2)/(2t) = n, so A341857(2t) = psi(4t^2)/(2t) = psi(t^2)/(2t) = n. This gives a(n) <= 2t = k/2, contradicting with minimality of k. QED.
The smallest odd n such that a(2n) < a(n)/2 is n = 71, where a(71) = 10236 and a(142) = 2276.
LINKS
Jianing Song, Table of n, a(n) for n = 1..500
EXAMPLE
A341857(116) = psi(116^2)/116 = 812/116 = 7, and there is no k < 116 such that psi(k^2)/k = 7, so a(7) = 116.
MATHEMATICA
A341860[n_] := Block[{k = 0}, While[CarmichaelLambda[++k^2]/k != n]; k];
Array[A341860, 100] (* Paolo Xausa, Mar 11 2024 *)
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Jianing Song, Feb 21 2021
STATUS
approved