A253247 Pisano period of A006190(n^2) divided by Pisano period of A006190(n).

1, 2, 3, 4, 5, 1, 7, 8, 9, 5, 11, 4, 13, 7, 5, 16, 17, 9, 19, 10, 21, 11, 23, 8, 25, 13, 27, 7, 29, 5, 31, 32, 33, 17, 35, 36, 37, 19, 39, 40, 41, 7, 43, 11, 45, 23, 47, 16, 49, 25, 51, 26, 53, 27, 55, 14, 57, 29, 59, 10, 61, 31, 63, 64, 65, 11, 67, 17, 69, 35, 71, 72

Offset

1,2

Comments

For all n, a(n)|n.

Conjecture: a(n) = 1 only for n = 1 and 6. (This conjecture is true if and only if the generalized Wall's conjecture to A006190 is true.)

If there exists any prime p such that A175182(p^2) = A175182(p), then the conjecture fails.

For any prime p, these three statements are equivalent:

(1) A175182(p^2) = A175182(p).

(2) A006190(p-(p|13)) = 3 (mod p^2).

(3) A006497(p) = 1 (mod p^2).

Since A175182(241^2) = A175182(241) = 484, so the prime 241 is a Wall-Sun-Sun prime to A006190 (Lucas (P, Q) = (3, -1)) and no others < 10^8, so the conjecture is true for all primes < 10^8 except 241.

All of Wall's theorems are true for A175182. For example, let P(n) = A175182(n), p and q are primes, then P(pq) = lcm(P(p), P(q)), and for every prime p, P(p)|(p-1) if (p|13) = 1, P(p)|(2p+2) if (p|13) = -1 (P(13) = 52, which if divisible by 13), while (p|13) is the Legendre symbol, and the fixed points of A175182 are 1, 6, and 12*13^k, k>0.

Links

Eric Chen, Table of n, a(n) for n = 1..1000

Formula

a(n) = A175182(n^2) / A175182(n).

Maple

F := proc(k, n) option remember; if n <= 1 then n; else k*procname(k, n-1)+procname(k, n-2) ; end if; end proc:
Pper := proc(k, m) local cha, zer, n, fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k, n) mod m ; cha := [op(cha), fmodm] ; if fmodm = 0 then zer := [op(zer), n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2], cha) ] = [ op(zer[2]+1..zer[3], cha) ] and [op(1..zer[2], cha)] = [ op(zer[3]+1..zer[4], cha) ] and [op(1..zer[2], cha)] = [ op(zer[4]+1..zer[5], cha) ] then return zer[2] ; elif [op(1..zer[3], cha) ] = [ op(zer[3]+1..zer[5], cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 3 ; seq( Pper(k, m^2) div Pper(k, m), m=1..300) ;

Mathematica

A006190[n_] = Fibonacci[n, 3];
A175182[n_2 = For[k=1, while[Mod[A006190[k], n] != 0 || Mod[A006190[k+1], n] != 1, k++]; k];
Table[A175182[n^2] / A175182[n], {n, 300}]

Prog

(PARI)
fibmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
entry_p(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, 3*c+o]; k++); k
entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e8 && f[i, 1] != 241, entry_p(f[i, 1]^f[i, 2]), entry_p(f[i, 1])*f[i, 1]^(f[i, 2] - 1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<<max(f[1, 2]-2, 1)); lcm(v)
per(n)=if(n==1, return(1)); my(k=entry(n)); forstep(i=k, n^2, k, if(fibmod(i-1, n)==1, return(i)))
a(n)=per(n^2)/per(n)

Crossrefs

Cf. A237517, A175182.

Sequence in context: A237517 A332883 A017666 * A201059 A325959 A325976

Adjacent sequences: A253244 A253245 A253246 * A253248 A253249 A253250

Keyword

nonn

Author

Eric Chen, Apr 09 2015

Status

approved