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

Author

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

Keywords

• nonn
• easy

Examples

			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.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 3..1000
• R. Fröberg, C. Gottlieb and R. Häggkvist, On numerical semigroups, Semigroup Forum, 35 (1987), 63-83 (for definition of Frobenius number).

Crossrefs

Cf. A000045.

Programs

Magma

[Fibonacci(n+1)*Fibonacci(n+2)-Fibonacci(n+3): n in [2..30]]; // Vincenzo Librandi, Mar 06 2016

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 *)
FrobeniusNumber/@Partition[Fibonacci[Range[3,40]],2,1] (* Harvey P. Dale, Feb 07 2025 *)

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

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

Extensions

Corrected by T. D. Noe, Nov 27 2006