A059769 Frobenius number of the subsemigroup of the natural numbers generated by successive pairs of Fibonacci numbers.

1, 7, 27, 83, 239, 659, 1781, 4751, 12583, 33175, 87231, 228983, 600473, 1573655, 4122467, 10796939, 28273519, 74031979, 193835949, 507497759, 1328692751, 3478637807, 9107313407, 23843452463, 62423286769, 163426800679, 427857750891, 1120147480451

Offset

3,2

Links

Table of n, a(n) for n=3..30.

R. Fröberg, C. Gottlieb and R. Häggkvist, On numerical semigroups, Semigroup Forum, 35 (1987), 63-83 (for definition of Frobenius number).

Formula

a(n) = (F(n)-1)*(F(n+1)-1)-1 where F(n) is the n-th Fibonacci number.

G.f.: x^3*(1+4*x+5*x^2-x^4)/((1+x)*(1-3*x+x^2)*(1-x-x^2)). [Colin Barker, Feb 17 2012]

a(n) = F(n)*F(n+1) - F(n+2). - Clark Kimberling, Mar 05 2016

Example

a(3)=1 because the 3rd and 4th Fibonacci numbers are 2 and 3, so a(3)=(2-1)(3-1)-1=1. Or, a(3)=1 because 1 is the largest positive integer that is not a nonnegative linear combination of 2 and 3.

Mathematica

Table[(Fibonacci[n]-1)(Fibonacci[n+1]-1)-1, {n, 3, 28}] (* T. D. Noe, Nov 27 2006 *)
f[n_]:=Fibonacci[n]; Table[f[n+1]f[n+2]-f[n+3], {n, 2, 40}] (* Clark Kimberling, Mar 05 2016 *)

Prog

(PARI) x='x+O('x^100); Vec(x^3*(1+4*x+5*x^2-x^4)/(1+x)/(1-3*x+x^2)/(1-x-x^2)) \\ Altug Alkan, Mar 05 2016
(Magma) [Fibonacci(n+1)*Fibonacci(n+2)-Fibonacci(n+3): n in [2..30]]; // Vincenzo Librandi, Mar 06 2016

Crossrefs

Cf. A000045.

Sequence in context: A038092 A059823 A339771 * A135914 A213588 A305653

Adjacent sequences: A059766 A059767 A059768 * A059770 A059771 A059772

Keyword

nonn,easy

Author

Victoria A Sapko (vsapko(AT)math.unl.edu), Feb 21 2001

Extensions

Corrected by T. D. Noe, Nov 27 2006

Status

approved