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

1,1

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

Maple

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

Mathematica

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

Prog

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

Crossrefs

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

Cf. A120326 (partial sums).

Cf. A185934, A217659.

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

Sequence in context: A117929 A306439 A107455 * A025822 A051585 A049115

Adjacent sequences: A039698 A039699 A039700 * A039702 A039703 A039704

Keyword

nonn,easy

Author

Clark Kimberling

Status

approved