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

Author

Eric Chen, Mar 02 2015

Keywords

• nonn

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.)

Examples

			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.
		

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)

Crossrefs

Cf. A178764.

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

Programs

Mathematica

a250208[n_] = If[n == 364, 1093, If[n == 1755, 3511, GCD[Cyclotomic[n, 2], n]]]; Table[a250208[n], {n, 0, 200}]

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

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.)