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

%I A253247 #47 Jun 28 2025 09:25:57
%S A253247 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,
%T A253247 29,5,31,32,33,17,35,36,37,19,39,40,41,7,43,11,45,23,47,16,49,25,51,
%U A253247 26,53,27,55,14,57,29,59,10,61,31,63,64,65,11,67,17,69,35,71,72
%N A253247 Pisano period of A006190(n^2) divided by Pisano period of A006190(n).
%C A253247 For all n, a(n)|n.
%C A253247 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.)
%C A253247 If there exists any prime p such that A175182(p^2) = A175182(p), then the conjecture fails.
%C A253247 For any prime p, these three statements are equivalent:
%C A253247 (1) A175182(p^2) = A175182(p).
%C A253247 (2) A006190(p-(p|13)) = 3 (mod p^2).
%C A253247 (3) A006497(p) = 1 (mod p^2).
%C A253247 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.
%C A253247 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.
%H A253247 Eric Chen, <a href="/A253247/b253247.txt">Table of n, a(n) for n = 1..1000</a>
%F A253247 a(n) = A175182(n^2) / A175182(n).
%p A253247 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:
%p A253247 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:
%p A253247 k := 3 ; seq( Pper(k, m^2) div Pper(k, m), m=1..300) ;
%t A253247 A006190[n_] := Fibonacci[n, 3];
%t A253247 A175182[n_] := Module[{k=1}, While[Mod[A006190[k], n] != 0 || Mod[A006190[k+1]-1, n] != 0, k++]; k];
%t A253247 Table[A175182[n^2] / A175182[n], {n, 72}] (* corrected by _Jason Yuen_, Jun 28 2025 *)
%o A253247 (PARI)
%o A253247 fibmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
%o A253247 entry_p(p)=my(k=1, c=Mod(1, p), o); while(c, [o, c]=[c, 3*c+o]; k++); k
%o A253247 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)
%o A253247 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)))
%o A253247 a(n)=per(n^2)/per(n)
%Y A253247 Cf. A237517, A175182.
%K A253247 nonn
%O A253247 1,2
%A A253247 _Eric Chen_, Apr 09 2015