A250208 Ratio of the primitive part of 2^n-1 to the product of primitive prime factors of 2^n-1.

1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

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

Eric Chen, Table of n, a(n) for n = 1..990

Eric Chen, Table of n, a(n) for n = 1..1280 status (with unknown terms but all are conjectured to be 1)

Formula

a(n) = A019320(n) / A064078(n) while Phi_n(2) is squarefree.

a(n) = GCD(Phi_n(2), n) while Phi_n(2) is squarefree.

Notice: a(364) = 1093, a(1755) = 3511. (See A001220.)

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

Cf. A178764.

Cf. A001220, A237043, A093106, A049094, A049093, A014491, A072226, A161508.

Sequence in context: A354997 A327537 A120263 * A325825 A030580 A030579

Adjacent sequences: A250205 A250206 A250207 * A250209 A250210 A250211

Keyword

nonn

Author

Eric Chen, Mar 02 2015

Status

approved