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%I #27 Oct 05 2020 01:05:42
%S 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,
%T 102,53,36,28,14,65,68,69,148,30,52,81,166,172,89,180,190,96,196,198,
%U 105,74,113,76,58,238,24,25,16,262,268,270,92,35,47,292,51
%N a(n) is the smallest m > 0 such that the n-th prime divides Jacobsthal(m).
%C All positive Jacobsthal numbers are odd, so the index starts at n = 2.
%C The set of primitive prime factors of J_k is given by {A000040(j) | a(j) = k}.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimitivePrimeFactor.html">Primitive Prime Factor</a>
%F a(n) = 1 (mod A000040(n)) for n > 2.
%e 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.
%t 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 *)
%o (Python)
%o n = 1
%o while n < 63:
%o n, J0, J1, a = n+1, 3, 1, 3
%o p = A000040(n)
%o J0 = J0%p
%o while J0 != 0:
%o J0, J1, a = (J0+2*J1)%p, J0, a+1
%o print(n,a)
%o (PARI) J(n) = (2^n - (-1)^n)/3; \\ A001045
%o a(n) = {my(k=1, p=prime(n)); while (J(k) % p, k++); k;} \\ _Michel Marcus_, Sep 29 2020
%Y Cf. A000040 (primes), A001045 (Jacobsthal numbers), A001602 (similar for Fibonacci numbers), A129738.
%K nonn
%O 2,1
%A _A.H.M. Smeets_, Sep 27 2020
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