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

Author

Eric Chen, Apr 09 2015

Keywords

• nonn

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

Crossrefs

Cf. A237517, A175182.

Programs

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_] := Module[{k=1}, While[Mod[A006190[k], n] != 0 || Mod[A006190[k+1]-1, n] != 0, k++]; k];
Table[A175182[n^2] / A175182[n], {n, 72}] (* corrected by Jason Yuen, Jun 28 2025 *)

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<
				

Formula

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