A348037 a(n) = n / gcd(n, A003968(n)), where A003968 is multiplicative with a(p^e) = p*(p+1)^(e-1).
1, 1, 1, 2, 1, 1, 1, 4, 3, 1, 1, 2, 1, 1, 1, 8, 1, 3, 1, 2, 1, 1, 1, 4, 5, 1, 9, 2, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 2, 3, 1, 1, 8, 7, 5, 1, 2, 1, 9, 1, 4, 1, 1, 1, 2, 1, 1, 3, 32, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 5, 2, 1, 1, 1, 8, 27, 1, 1, 2, 1, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 16, 1, 7, 3, 5, 1, 1, 1, 4, 1
Offset: 1
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 1..20000
Crossrefs
Programs
-
Mathematica
f[p_, e_] := p*(p + 1)^(e - 1); a[n_] := n / GCD[n, Times @@ f @@@ FactorInteger[n]]; Array[a, 100] (* Amiram Eldar, Oct 20 2021 *)
-
PARI
A003968(n) = { my(f=factor(n)); for (i=1, #f~, p= f[i, 1]; f[i, 1] = p*(p+1)^(f[i, 2]-1); f[i, 2] = 1); factorback(f); } A348037(n) = (n/gcd(n, A003968(n)));