

A337878


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


0



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
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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



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
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



KEYWORD

nonn


AUTHOR



STATUS

approved



