OFFSET
1,6
COMMENTS
As with A178764, it can be shown that all terms are either 1 or prime.
a(2*3^n) = 3 (n>=1).
a(4*5^n) = 5 (n>=1).
a(3*7^n) = 7 (n>=1).
a(10*11^n) = 11 (n>=1).
a(12*13^n) = 13 (n>=1).
a(8*17^n) = 17 (n>=1).
a(18*19^n) = 19 (n>=1).
...
a(A014664(k)*prime(k)^n) = prime(k).
For other n (while Phi_n(2) is squarefree), a(n) = 1.
a(n) != 1 for n = {6, 18, 20, 21, 54, 100, 110, 136, 147, 155, 156, 162, ...}.
At least, a(A049093(n)) = 1. (In fact, since Phi_n(2) is not completely factored for n = 991, 1207, 1213, 1217, 1219, 1229, 1231, 1237, 1243, 1249, ..., so it is unknown whether they are squarefree or not, but it is likely that Phi_n(2) is squarefree for all n except 364 and 1755 (because it is likely 1093 and 3511 are the only two Wieferich primes), so a(991), a(1207), a(1213), ..., are likely to be 1.)
LINKS
FORMULA
EXAMPLE
a(11) = 1 since Phi_11(2) = (2^11-1)/(2-1) = 2047, and the primitive prime factors of 2^11-1 are 23 and 89, so a(11) = 2047/(23*89) = 1.
a(18) = 3 since Phi_18(2) = 2^6 - 2^3 + 1 = 57, and the only primitive prime factor of 2^18-1 is 19, so a(18) = 57/19 = 3.
MATHEMATICA
a250208[n_] = If[n == 364, 1093, If[n == 1755, 3511, GCD[Cyclotomic[n, 2], n]]]; Table[a250208[n], {n, 0, 200}]
PROG
(PARI) a(n) = if (n==364, 1093, if (n==1755, 3511, gcd(polcyclo(n, 2), n)));
(PARI) isprimitive(p, n) = {for (r=1, n-1, if (((2^r-1) % p) == 0, return (0)); ); return (1); }
ppf(n) = {my(pf = factor(2^n-1)[, 1]); prod(k=1, #pf, if (isprimitive(pf[k], n), pf[k], 1)); }
a(n) = if (issquarefree(m=polcyclo(n, 2)), gcd(m, n), m/ppf(n)); \\ Michel Marcus, Mar 06 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Eric Chen, Mar 02 2015
STATUS
approved