OFFSET
1,2
COMMENTS
Conjecture: There are no 1's after the initial term. Remark: If there were some k = x^2 > 1, for which a(x) = 1, then sigma(k) would be a divisor of A003961(k). In other words, d = A350073(k) = A064989(sigma(k)) would be a divisor of k. Then, if that divisor were also a unitary divisor [with gcd(d,k/d) = 1], it would need to satisfy the equation sigma(k) = sigma(d) * sigma(k/d) = sigma(A064989(sigma(k))) * sigma(k/A064989(sigma(k))), because sigma is a multiplicative function. (Minor correction by Antti Karttunen, Jul 11 2023)
Note that if d = A064989(sigma(k)) were a unitary divisor of a square k, then sigma(k) would also be a square, the cases which are quite rare (see A008848 and A336547). Also compare to A349756. - Antti Karttunen, Jul 24 2022
LINKS
MATHEMATICA
f1[p_, e_] := (p^(2*e + 1) - 1)/(p - 1); f2[p_, e_] := NextPrime[p]^(2*e); a[1] = 1; a[n_] := (s = Times @@ f1 @@@ (f = FactorInteger[n])) / GCD[s, Times @@ f2 @@@ f]; Array[a, 60] (* Amiram Eldar, Dec 12 2021 *)
PROG
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Dec 12 2021
STATUS
approved