A270741 a(n) = (Fibonacci(n+2)-1) mod Fibonacci(floor(n/2)).

0, 0, 0, 0, 0, 1, 0, 1, 3, 2, 0, 1, 11, 10, 0, 1, 32, 31, 0, 1, 87, 86, 0, 1, 231, 230, 0, 1, 608, 607, 0, 1, 1595, 1594, 0, 1, 4179, 4178, 0, 1, 10944, 10943, 0, 1, 28655, 28654, 0, 1, 75023, 75022, 0, 1, 196416, 196415, 0, 1, 514227, 514226, 0, 1, 1346267, 1346266, 0, 1, 3524576, 3524575, 0, 1

Offset

2,9

Comments

It appears that a(n)=1 gives same terms as A222945. - Michel Marcus, Mar 23 2016

Links

Table of n, a(n) for n=2..69.

Formula

Empirical g.f.: x^7*(1-x+2*x^2+x^3-3*x^4+3*x^5-6*x^6+5*x^7+x^8-x^9+2*x^10-2*x^11) / ((1-x)*(1+x^2)*(1+x^2-x^4)*(1-x^2-x^4)). - Colin Barker, Mar 23 2016

Example

a(10) = 3 since (1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55) mod 5 = 143 mod 5 = 3.
a(11) = 2 since (1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89) mod 5 = 2.
a(12) = 0 since (1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144) mod 8 = 0.

Mathematica

Table[Mod[Fibonacci[n + 2] - 1, Fibonacci[Floor[n/2]]], {n, 2, 69}] (* Michael De Vlieger, Mar 23 2016 *)

Prog

(PARI) for(n=2, 1e2, print1((fibonacci(n+2)-1) % fibonacci(n\2), ", "));

Crossrefs

Cf. A000045, A000071, A270708.

Sequence in context: A193277 A118972 A171224 * A212220 A193233 A145878

Adjacent sequences: A270738 A270739 A270740 * A270742 A270743 A270744

Keyword

nonn

Author

Altug Alkan, Mar 22 2016

Status

approved