A337878 a(n) is the smallest m > 0 such that the n-th prime divides Jacobsthal(m).

3, 4, 6, 5, 12, 8, 9, 22, 28, 10, 36, 20, 7, 46, 52, 29, 60, 33, 70, 18, 78, 41, 22, 48, 100, 102, 53, 36, 28, 14, 65, 68, 69, 148, 30, 52, 81, 166, 172, 89, 180, 190, 96, 196, 198, 105, 74, 113, 76, 58, 238, 24, 25, 16, 262, 268, 270, 92, 35, 47, 292, 51

Offset: 2

Author

A.H.M. Smeets, Sep 27 2020

Keywords

• nonn

Comments

All positive Jacobsthal numbers are odd, so the index starts at n = 2.

The set of primitive prime factors of J_k is given by {A000040(j) | a(j) = k}.

By definition, a(n) is the multiplicative order of -2 modulo the n-th prime for n > 2. - Jianing Song, Jun 20 2025

Examples

			The 4th prime number is 7, and 7 divides 21 which is Jacobsthal(6), so a(4) = 6. The second prime number, 3, divides Jacobsthal(6) as well, but it divides also the smaller Jacobsthal(3), i.e., a(2) = 3.
		

Links

• Jianing Song, Table of n, a(n) for n = 2..10000
• Eric Weisstein's World of Mathematics, Primitive Prime Factor

Crossrefs

Cf. A000040 (primes), A001045 (Jacobsthal numbers), A001602 (similar for Fibonacci numbers), A105874 (primes having primitive root -2), A129738.

Cf. multiplicative orders of 2..10: A014664, A062117, A082654, A211241, A211242, A211243, A211244, A211245, A002371.

Cf. multiplicative orders of -2..-10: this sequence (if first term 1), A380482, A380531, A380532, A380533, A380540, A380541, A380542, A385222.

Programs

Mathematica

m = 300; j = LinearRecurrence[{1, 2}, {3, 5}, m]; s = {}; p = 3; While[(ind = Select[Range[m], Divisible[j[[#]], p] &, 1]) != {}, AppendTo[s, ind[[1]] + 2]; p = NextPrime[p]]; s (* Amiram Eldar, Sep 28 2020 *)

PARI

J(n) = (2^n - (-1)^n)/3; \\ A001045
a(n) = {my(k=1, p=prime(n)); while (J(k) % p, k++); k;} \\ Michel Marcus, Sep 29 2020

Python

n = 1
while n < 63:
    n, J0, J1, a = n+1, 3, 1, 3
    p = A000040(n)
    J0 = J0%p
    while J0 != 0:
        J0, J1, a = (J0+2*J1)%p, J0, a+1
    print(n,a)

Formula

A000040(n) == 1 (mod a(n)) for n > 2.