A306198 Multiplicative with a(p^e) = p^(e-1)*(p^2 - p - 1).

1, 1, 5, 2, 19, 5, 41, 4, 15, 19, 109, 10, 155, 41, 95, 8, 271, 15, 341, 38, 205, 109, 505, 20, 95, 155, 45, 82, 811, 95, 929, 16, 545, 271, 779, 30, 1331, 341, 775, 76, 1639, 205, 1805, 218, 285, 505, 2161, 40, 287, 95, 1355, 310, 2755, 45, 2071, 164, 1705, 811

Offset: 1

Author

Jianing Song, Jan 28 2019

Keywords

• easy
• nonn
• mult

Comments

For any positive integer n and any m coprime to n, define R(n,m) = Product_{primes p divides n} (p - [m == 1 (mod p)]), where [] is an Iverson branket. Then we have the following conjecture: (Start)

Let k == 2, 3 (mod 4) be a squarefree number, b be any positive integer such that k*b^2 is not a perfect power and not equal to -1, n be either coprime to or divisible by 4*k. Define Q(N,k*b^2,n,m) = # {primes p <= N : p == m (mod n), k*b^2 is a primitive modulo p}, then:

(a) if gcd(n, 4*k) = 1, then Q(N,k*b^2,n,m)/(C*PrimePi(N)) ~ R(n,m)/a(n);

(b) if 4*k divides n, then Q(N,k*b^2,n,m)/(C*PrimePi(N)) ~ 2*R(n,m)/a(n) if Jacobi(k/m) = -1 and 0 if Jacobi(k/m) = +1,

Where C is the Artin's constant = A005596, PrimePi = A000720. (End)

(Note that Sum_{m=1..n, gcd(m,n)=1} R(n,m) = a(n).)

For example, let N = 10^6:

k*b^2 | n | m | Q(N,k*b^2,n,m) | Q(N,k*b^2,n,m)/(C*PrimePi(N))

2 | 8 | 3 | 14642 | 0.498794... approx = 2/4

3 | 5 | 1 | 6192 | 0.210936... approx = 4/19

-2 | 48 | 13 | 2933 | 0.099915... approx = 4/40

-5 | 9 | 5 | 5933 | 0.202113... approx = 3/15

Links

• Jianing Song, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Artin's constant.
• Wikipedia, Artin's conjecture on primitive roots.

Crossrefs

Cf. A000720 (PrimePi), A005596 (Artin's constant), A086463.

Programs

Maple

P := (p, e) -> p^(e-1)*(p^2 - p - 1):
a := n -> mul(P(f[1], f[2]), f in ifactors(n)[2]):
seq(a(n), n=1..58); # Peter Luschny, Feb 13 2019

Mathematica

a[n_] := Product[{p, e} = pe; p^(e-1) (p^2-p-1), {pe, FactorInteger[n]}]; a[1] = 1; Array[a, 58] (* Jean-François Alcover, Jul 22 2019 *)

PARI

a(n) = my(f=factor(n)); prod(j=1, #f[, 1], my(p=f[j, 1], e=f[j, 2]); (p^2 - p - 1)*p^(e-1))

Formula

Sum_{k=1..n} a(k) ~ c * n^3, where c = (Pi^2/18) * Product_{p prime} (1 - 3/p^2 + 1/p^3 + 1/p^4) = 0.1314639252... . - Amiram Eldar, Dec 01 2022