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 A337878 a(n) is the smallest m > 0 such that the n-th 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified September 23 09:04 EDT 2023. Contains 365544 sequences. (Running on oeis4.)