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

2,1

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}.

Links

Table of n, a(n) for n=2..63.

Eric Weisstein's World of Mathematics, Primitive Prime Factor

Formula

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

Example

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.

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 *)

Prog

(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)
(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

Crossrefs

Cf. A000040 (primes), A001045 (Jacobsthal numbers), A001602 (similar for Fibonacci numbers), A129738.

Sequence in context: A058838 A349372 A001177 * A053991 A276814 A198617

Adjacent sequences: A337875 A337876 A337877 * A337879 A337880 A337881

Keyword

nonn

Author

A.H.M. Smeets, Sep 27 2020

Status

approved