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A308948 a(n) = A006190(A322907(n)+1) mod n. 1
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 (list; graph; refs; listen; history; text; internal format)
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-Fran├žois 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: A340179 A120394 A016575 * A225246 A116666 A208331

Adjacent sequences:  A308945 A308946 A308947 * A308949 A308950 A308951

KEYWORD

nonn

AUTHOR

Jianing Song, Jul 02 2019

STATUS

approved

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Last modified June 18 18:17 EDT 2021. Contains 345120 sequences. (Running on oeis4.)