A308947 a(n) = A000129(A214028(n)+1) mod n.

0, 1, 2, 1, 2, 5, 1, 1, 8, 9, 10, 5, 5, 1, 11, 1, 16, 17, 18, 1, 8, 21, 1, 1, 7, 25, 26, 1, 12, 11, 1, 1, 32, 33, 29, 17, 31, 37, 14, 1, 1, 29, 42, 21, 26, 1, 1, 1, 1, 49, 16, 1, 30, 53, 21, 1, 56, 57, 58, 41, 50, 1, 8, 1, 8, 65, 66, 33, 47, 29, 1, 1, 72, 73

Offset

1,3

Comments

A214028(n) is the smallest k > 0 such that n divides A000129(k).

Let M = [{2, 1}, {1, 0}], I = [{1, 0}, {0, 1}] is the 2 X 2 identity matrix, then A214028(n) is the smallest k > 0 such that M^k == r*I (mod n) for some r such that 0 <= r < n, and a(n) gives the value r.

A214027(n) is the multiplicative order of a(n) modulo n, which can only take value 1, 2 or 4.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

Also a(n) = A000129(A214028(n)-1) mod n.

a(2^e) = 1; a(p^e) = a(p)^(p^(e-1)) mod p^e for odd primes p.

For odd primes p, a(p^e) = 1 if and only if A214028(p) == 2 (mod 4); a(p^e) = p^e - 1 if and only if 4 divides A214028(p).

Example

For n = 7, {A000129(n) mod 7 : n > 0} = {1, 2, 5, 5, 1, 0, 1, ...}, so a(7) = 1. Also, A214028(7) = 6, and M^6 mod 7 = [{1, 0}, {0, 1}], so a(7) = 1.

Mathematica

a[n_] := For[k = 1, True, k++, If[Divisible[Fibonacci[k, 2], n], Return[ Mod[ Fibonacci[k+1, 2], n]]]];
Array[a, 100] (* Jean-François Alcover, Jul 05 2019 *)

Prog

(PARI) a(n) = my(M=[2, 1; 1, 0]); for(k=1, 4*n/3, if((Mod(M, n)^k)[2, 1]==0, return(lift((Mod(M, n)^k)[1, 1]))))

Crossrefs

Cf. A000129, A214027, A214028.

Similar sequences: A217036, A308948.

Sequence in context: A245841 A011404 A002211 * A175011 A211700 A171840

Adjacent sequences: A308944 A308945 A308946 * A308948 A308949 A308950

Keyword

nonn

Author

Jianing Song, Jul 02 2019

Status

approved