A182126 -id:A182126 - cp's OEIS frontend

A333599 a(n) = Fibonacci(n) * Fibonacci(n+1) mod Fibonacci(n+2).

0, 1, 2, 1, 7, 1, 20, 1, 54, 1, 143, 1, 376, 1, 986, 1, 2583, 1, 6764, 1, 17710, 1, 46367, 1, 121392, 1, 317810, 1, 832039, 1, 2178308, 1, 5702886, 1, 14930351, 1, 39088168, 1, 102334154, 1, 267914295, 1, 701408732, 1, 1836311902, 1, 4807526975, 1, 12586269024

Offset: 0

Adnan Baysal, Mar 28 2020

			a(0) = 0*1 mod 1 = 0;
a(1) = 1*1 mod 2 = 1;
a(2) = 1*2 mod 3 = 2;
a(3) = 2*3 mod 5 = 1;
a(4) = 3*5 mod 8 = 7.

Colin Barker, Table of n, a(n) for n = 0..1000
Eric Weisstein's World of Mathematics, d'Ocagne's Identity.
Index entries for linear recurrences with constant coefficients, signature (-1,3,3,-1,-1).

Equals A035508 interleaved with A000012.

Cf. A182126, A000045, A022461, A072565, A022462.

With[{f = Fibonacci}, Table[Mod[f[n] * f[n+1], f[n+2]], {n, 0, 50}]] (* Amiram Eldar, Mar 28 2020 *)

a(n) = if (n % 2, 1, fibonacci(n+2) - 1); \\ Michel Marcus, Mar 29 2020

concat(0, Vec(x*(1 + 3*x - x^3) / ((1 + x)*(1 + x - x^2)*(1 - x - x^2)) + O(x^45))) \\ Colin Barker, Mar 29 2020

def a(n):
    f1 = 0
    f2 = 1
    for i in range(n):
        f = f1 + f2
        f1 = f2
        f2 = f
    return (f1 * f2) % (f1 + f2)

a(2n+1) = 1, and a(2n) = F(2n+2) - 1, and lim(a(2n+2)/a(2n)) = phi^2 by d'Ocagne's identity.
a(n) = F(n) * F(n+1) mod (F(n) + F(n+1)) since F(n+2) := F(n+1) + F(n).
From Colin Barker, Mar 28 2020: (Start)
G.f.: x*(1 + 3*x - x^3) / ((1 + x)*(1 + x - x^2)*(1 - x - x^2)).
a(n) = -a(n-1) + 3*a(n-2) + 3*a(n-3) - a(n-4) - a(n-5) for n>4.
(End)

A288209 Numbers k such that prime(k) * prime(k+1) mod prime(k+2) is odd.

Original entry on oeis.org

1, 2, 5, 7, 10, 14, 15, 23, 29, 46, 61

Offset: 1

Zak Seidov, Jun 06 2017

Finite? Full?
Next term, if it exists, is greater than 1026351685.
From Robert Israel, Jun 19 2017: (Start)
Numbers k such that floor(A001223(k+1)*A031131(k)/prime(k+2)) is odd.
Cramér's conjecture implies the sequence is finite. (End)

			The first five primes are 2, 3, 5, 7, 11.
We see that 2 * 3 = 1 mod 5, so 1 (corresponding to the first prime, 2) is in the sequence.
We see that 3 * 5 = 1 mod 7, so 2 (corresponding to the second prime, 3) is in the sequence.
But 5 * 7 = 2 mod 11, so 3 (corresponding to the third prime, 5) is not in the sequence.

Wikipedia, Cramér's conjecture

Cf. A001223, A006094, A031131, A182126.

P:= select(isprime, [2,seq(i,i=3..10^6,2)]):
select(n -> (P[n]*P[n+1] mod P[n+2])::odd, [$1..nops(P)-2]); # Robert Israel, Jun 19 2017

Select[Range[1000], OddQ[Mod[Prime[#] Prime[# + 1], Prime[# + 2]]] &] (* Alonso del Arte, Jun 06 2017 *)
Position[Partition[Prime[Range[70]],3,1],?(OddQ[Mod[#[[1]]#[[2]], #[[3]]]]&),1,Heads->False]//Flatten (* _Harvey P. Dale, Aug 11 2017 *)

isok(n) = (((prime(n) * prime(n + 1)) % prime(n + 2)) % 2); \\ Michel Marcus, Jun 07 2017

cp's OEIS Frontend

A333599 a(n) = Fibonacci(n) * Fibonacci(n+1) mod Fibonacci(n+2).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

Python

Formula

A288209 Numbers k such that prime(k) * prime(k+1) mod prime(k+2) is odd.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI