A039701 a(n) = n-th prime modulo 3.

2, 0, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

If n > 2 and prime(n) is a Mersenne prime then a(n) = 1. Proof: prime(n) = 2^p - 1 for some odd prime p, so prime(n) = 2*4^((p-1)/2) - 1 == 2 - 1 = 1 (mod 3). - Santi Spadaro, May 03 2002; corrected and simplified by Dean Hickerson, Apr 20 2003

Except for n = 2, a(n) is the smallest number k > 0 such that 3 divides prime(n)^k - 1. - T. D. Noe, Apr 17 2003

a(n) <> 0 for n <> 2; a(A049084(A003627(n))) = 2; a(A049084(A002476(n))) = 1; A134323(n) = (1 - 0^a(n)) * (-1)^(a(n)+1). - Reinhard Zumkeller, Oct 21 2007

Probability of finding 1 (or 2) in this sequence is 1/2. This follows from the Prime Number Theorem in arithmetic progressions. Examples: There are 4995 1's in terms (10^9 +1) to (10^9+10^4); there are 10^9/2-1926 1's in the first 10^9 terms. - Jerzy R Borysowicz, Mar 06 2022

Links

• Nathaniel Johnston, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A091178 (indices of 1's), A091177 (indices of 2's).

Cf. A120326 (partial sums).

Cf. A185934, A217659.

Cf. A010872.

Other moduli: A039702-A039706, A038194, A007652, A039709-A039715.

Programs

Haskell

a039701 = (`mod` 3) . a000040
a039701_list = map (`mod` 3) a000040_list
-- Reinhard Zumkeller, Nov 16 2012

Magma

[p mod(3): p in PrimesUpTo(500)]; // Vincenzo Librandi, May 06 2014

Maple

seq(ithprime(n) mod 3, n=1..105); # Nathaniel Johnston, Jun 29 2011

Mathematica

Table[Mod[Prime[n], 3], {n, 100}]

PARI

primes(100)%3 \\ Charles R Greathouse IV, May 06 2014

Formula

Sum_k={1..n} a(k) ~ (3/2)*n. - Amiram Eldar, Dec 11 2024