cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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).

Original entry on oeis.org

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

Views

Author

Jonathan Sondow, Dec 09 2017

Keywords

Comments

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.

Crossrefs

Programs

  • Mathematica
    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 *)

Formula

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.

A092330 Fibonacci quotients: Fibonacci(p - Legendre(p|5))/p where p runs through the primes.

Original entry on oeis.org

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

Views

Author

Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 17 2004

Keywords

Comments

If p is prime then p divides fibonacci(p - Legendre(p|5)).
The result is known as the Fibonacci Quotient. - John Blythe Dobson, Sep 20 2014
Legendre(p|5) = 1 if prime p == 1 or 4 mod 5, -1 if p == 2 or 3 mod 5, 0 if p = 5. - Robert Israel, Sep 21 2014
Not to be confused with (Fibonacci(p) - Legendre(p|5))/p, which is A222361. - Jonathan Sondow, Dec 08 2017

Crossrefs

Programs

  • Maple
    f:= proc(n) local p; p:= ithprime(n); combinat:-fibonacci(p - numtheory:-legendre(p,5))/p end proc:
    seq(f(n),n=1..30); # Robert Israel, Sep 21 2014
  • Mathematica
    a[n_] := With[{p = Prime[n]}, Fibonacci[p - KroneckerSymbol[p, 5]]/p];
    Array[a, 23] (* Jean-François Alcover, Nov 25 2017 *)
  • PARI
    forprime (i=1,150,print1(fibonacci(i-kronecker(i,5))/i,","))

Extensions

Offset corrected by Jonathan Sondow, Dec 11 2017
Showing 1-2 of 2 results.