A308948 a(n) = A006190(A322907(n)+1) mod n.

0, 1, 1, 1, 3, 1, 6, 5, 1, 3, 10, 1, 8, 13, 4, 9, 16, 1, 18, 9, 13, 21, 1, 13, 18, 5, 1, 13, 12, 19, 30, 17, 10, 33, 6, 1, 31, 37, 25, 29, 32, 13, 1, 21, 19, 1, 46, 25, 48, 43, 16, 25, 1, 1, 21, 41, 37, 17, 58, 49, 1, 61, 55, 33, 18, 43, 66, 33, 1, 41, 70, 37

Offset

1,5

Comments

A322907(n) is the smallest k > 0 such that n divides A006190(k).

Let M = [{3, 1}, {1, 0}], I = [{1, 0}, {0, 1}] is the 2 X 2 identity matrix, then A322907(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.

A322906(n) is the multiplicative order of a(n) modulo n, which can only take value 1, 2 or 4.

Links

Table of n, a(n) for n=1..72.

Formula

Also a(n) = A006190(A322907(n)-1) mod n.

a(2^e) = 1 if e = 1, 2, 2^(e-1) + 1 if e >= 3; 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 A322907(p) == 2 (mod 4); a(p^e) = p^e - 1 if and only if 4 divides A322907(p).

Example

For n = 7, {A006190(n) mod 7 : n > 0} = {1, 3, 3, 5, 4, 3, 6, 0, 6, ...}, so a(7) = 6. Also, A322907(7) = 8, and M^8 mod 7 = [{6, 0}, {0, 6}], so a(7) = 6.

Mathematica

a[n_] := For[k = 1, True, k++, If[Divisible[Fibonacci[k, 3], n], Return[ Mod[Fibonacci[k + 1, 3], n]]]];
Array[a, 100] (* Jean-François Alcover, Jul 05 2019 *)

Prog

(PARI) a(n) = my(M=[3, 1; 1, 0]); for(k=1, 12*n/7, if((Mod(M, n)^k)[2, 1]==0, return(lift((Mod(M, n)^k)[1, 1]))))

Crossrefs

Cf. A006190, A322906, A322907.

Similar sequences: A217036, A308947.

Sequence in context: A120394 A371667 A016575 * A225246 A116666 A208331

Adjacent sequences: A308945 A308946 A308947 * A308949 A308950 A308951

Keyword

nonn

Author

Jianing Song, Jul 02 2019

Status

approved