A130234 Minimal index k of a Fibonacci number such that Fibonacci(k) >= n (the 'upper' Fibonacci Inverse).

0, 1, 3, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11

Offset: 0

Author

Hieronymus Fischer, May 17 2007

Keywords

• nonn
• easy

Comments

Inverse of the Fibonacci sequence (A000045), nearly, since a(Fibonacci(n)) = n except for n = 2 (see A130233 for another version). a(n+1) is equal to the partial sum of the Fibonacci indicator sequence (see A104162).

Examples

			a(10) = 7, since Fibonacci(7) = 13 >= 10 but Fibonacci(6) = 8 < 10.
		

Crossrefs

Partial sums: A130236.

Other related sequences: A000045, A130233, A130256, A130260, A104162, A108852.

Lucas inverse: A130241 - A130248.

Programs

Maple

A130234 := proc(n)
    local i;
    for i from 0 do
        if A000045(i) >= n then
            return i;
        end if;
    end do:
end proc: # R. J. Mathar, Jan 31 2015

Mathematica

a[n_] := For[i = 0, True, i++, If[Fibonacci[i] >= n, Return[i]]];
a /@ Range[0, 80] (* Jean-François Alcover, Apr 13 2020 *)

PARI

a(n)=my(k);while(fibonacci(k)Charles R Greathouse IV, Feb 03 2014, corrected by M. F. Hasler, Apr 07 2021

Formula

a(n) = ceiling(log_phi((sqrt(5)*n + sqrt(5*n^2-4))/2)) = ceiling(arccosh(sqrt(5)*n/2)/log(phi)) where phi = (1+sqrt(5))/2, the golden ratio, for n > 0.

a(n) = A130233(n-1) + 1 for n > 0.

G.f.: x/(1-x) * Sum_{k >= 0} x^Fibonacci(k).

a(n) = ceiling(log_phi(sqrt(5)*n - 1)) for n > 0, where phi is the golden ratio. - Hieronymus Fischer, Jul 02 2007

a(n) = A108852(n-1). - R. J. Mathar, Jan 31 2015