A092330 -id:A092330 - cp's OEIS frontend

A296240 Pisano quotients: a(n) = (p-1)/k(p) if p == +- 1 mod 5, = (2*p+2)/k(p) if p == +- 2 mod 5, where p = prime(n) and k(p) = Pisano period(p).

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 1, 3, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 9, 5, 1, 1, 2, 9, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 7, 1, 1, 1, 3, 1, 3, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 5, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 2, 20, 1, 6, 1, 9, 3, 1, 1, 1, 1, 1, 1

Offset: 4

Jonathan Sondow, Dec 09 2017

nonn

Wall (1960) in Theorems 6 and 7 proved that a(n) is an integer for n >= 4. Jarden (1946) proved that the sequence is unbounded. See Elsenhans and Jahnel (2010), pp. 1-2.

A. Elsenhans and J. Jahnel, The Fibonacci sequence modulo p^2 -- An investigation by computer for p < 10^14, arXiv 1006.0824 [math.NT], 2010.
D. Jarden, Two theorems on Fibonacci's sequence, Amer. Math. Monthly, 53 (1946), 425-427.
D. D. Wall, Fibonacci series modulo m, Amer. Math. Monthly, 67 (1960), 525-532.
Wikipedia, Wall-Sun-Sun prime

Cf. A001175, A092330, A060305, A222361, A222413.

With[{p = Prime[n]}, T = Table[a = {1, 0}; a0 = a; k = 0; While[k++; s = Mod[Plus @@ a, p]; a = RotateLeft[a]; a[[2]] = s; a != a0]; k, {n, 1, 130}]; Table[L = KroneckerSymbol[p, 5]; (3 - L)/2 (p - L)/T[[n]], {n, 4, 130}]] (* after T. D. Noe *)

a(n) = (3 - L(p))/2 * (p - L(p)) / k(p), where p = prime(n), L(p) = Legendre(p|5), and k(p) = Pisano period(p) = A001175(p).

a(n) > 1 if and only if prime(n) is in A222413.

A222361 Fibonacci-Legendre quotients: (Fibonacci(p) - L(p/5)) / p, where p = prime(n) and L(p/5) is the Legendre symbol.

Original entry on oeis.org

1, 1, 1, 2, 8, 18, 94, 220, 1246, 17732, 43428, 652914, 4038540, 10081266, 63217342, 1005967758, 16215627560, 41061160360, 670829406162, 4338894664368, 11048157986978, 183194101578180, 1195118711985006, 19999768719154092, 862073644225241474, 5674731128849674100, 14568160545698020226, 96118885585174929102, 247025215671874138312, 1633201998168434481118

Offset: 1

Jonathan Sondow, Feb 23 2013

nonn

Fibonacci(p) == L(p/5) mod p, where the Legendre symbol L(p/5) equals 0, +1, -1 according as p = 5, 5*k+-1, 5*k+-2 for some k.

Not to be confused with Fibonacci(p - L(p,5)) / p, which is A092330.

			Prime(4) = 7, so a(4) = (Fibonacci(7)-L(7/5))/7 = (13-(-1))/7 = 14/7 = 2.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Wikipedia, Prime divisors of Fibonacci numbers

Cf. A000045, A092330.

Table[p = Prime[n]; (Fibonacci[p] - JacobiSymbol[p, 5])/p, {n, 1, 30}]

For n>=4, a(n) = (Fibonacci(prime(n)) +/- 1)/prime(n), where '+' is chosen if prime(n)== 2 or 3 (mod 5), '-' is chosen otherwise. For n>=2, a(n) = round(Fibonacci(prime(n))/prime(n)). - Vladimir Shevelev, Mar 12 2014

A176951 Let p = prime(n). Then a(n) = Fibonacci(p+1)/p if this is an integer, otherwise a(n) = Fibonacci(p-1)/p if this is an integer, and fall back to a(n)=0 if both are non-integer.

Original entry on oeis.org

1, 1, 0, 3, 5, 29, 152, 136, 2016, 10959, 26840, 1056437, 2495955, 16311831, 102287808, 1627690024, 10021808981, 25377192720, 1085424779823, 2681584376185, 17876295136009, 113220181313816, 1933742696582736

Offset: 1

Roger L. Bagula, Apr 29 2010

nonn

This differs only trivially from the better-defined A092330. - John Blythe Dobson, Sep 20 2014

Harvey P. Dale, Table of n, a(n) for n = 1..647

Cf. A092330.

A176951aux := proc(n)
        if n = 0 then
                0;
        elif combinat[fibonacci](n+1) mod n = 0 then
                combinat[fibonacci](n+1)/n ;
        elif combinat[fibonacci](n-1) mod n = 0 then
                combinat[fibonacci](n-1)/n ;
        else
                0 ;
        end if;
end proc:
A176951 := proc(n)
        A176951aux(ithprime(n)) ;
end proc:
seq(A176951(n),n=1..20) ; # R. J. Mathar, Oct 29 2011

f[n_] = If[n == 0, 0, If[Mod[Fibonacci[n + 1], n] == 0, Fibonacci[n + 1]/n, If[Mod[Fibonacci[n - 1], n] == 0, Fibonacci[n - 1]/n, 0]]];
Table[f[Prime[n + 1]], {n, 0, 50}]
Table[With[{f1=Fibonacci[p+1],f2=Fibonacci[p-1]},Which[IntegerQ[f1/p],f1/p,IntegerQ[f2/p],f2/p,True,0]],{p,Prime[Range[30]]}] (* Harvey P. Dale, Jun 09 2025 *)

a(n)=my(p=prime(n),t);t=fibonacci(p+1);if(t%p==0,t/p,t=fibonacci(p-1);if(t%p==0,t/p,0)) \\ Charles R Greathouse IV, Oct 29 2011

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A296240 Pisano quotients: a(n) = (p-1)/k(p) if p == +- 1 mod 5, = (2*p+2)/k(p) if p == +- 2 mod 5, where p = prime(n) and k(p) = Pisano period(p).

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A222361 Fibonacci-Legendre quotients: (Fibonacci(p) - L(p/5)) / p, where p = prime(n) and L(p/5) is the Legendre symbol.

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A176951 Let p = prime(n). Then a(n) = Fibonacci(p+1)/p if this is an integer, otherwise a(n) = Fibonacci(p-1)/p if this is an integer, and fall back to a(n)=0 if both are non-integer.

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