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A337878
a(n) is the smallest m > 0 such that the n-th prime divides Jacobsthal(m).
10
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}.
By definition, a(n) is the multiplicative order of -2 modulo the n-th prime for n > 2. - Jianing Song, Jun 20 2025
LINKS
Eric Weisstein's World of Mathematics, Primitive Prime Factor
FORMULA
A000040(n) == 1 (mod a(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), 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.
Sequence in context: A349372 A386594 A001177 * A053991 A276814 A198617
KEYWORD
nonn
AUTHOR
A.H.M. Smeets, Sep 27 2020
STATUS
approved