A130245 Number of Lucas numbers (A000032) <= n.

0, 1, 2, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 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

Offset: 0

Author

Hieronymus Fischer, May 19 2007, Jul 02 2007

Keywords

• nonn

Comments

Partial sums of the Lucas indicator sequence A102460.

For n>=2, we have a(A000032(n)) = n + 1.

Examples

			a(9)=5 because there are 5 Lucas numbers <=9 (2,1,3,4 and 7).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..64079
• Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Tųrcąs, On some new results for the generalised Lucas sequences, An. Şt. Univ. Ovidius Constanţa (Romania, 2021) Vol. 29, No. 1, 17-36.

Crossrefs

Partial sums of A102460.

For partial sums of this sequence, see A130246. Other related sequences: A000032, A130241, A130242, A130247, A130249, A130253, A130255, A130259.

For Fibonacci inverse, see A130233 - A130240, A104162, A108852.

Programs

Magma

[0] cat [1+Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [1..100]]; // G. C. Greubel, Sep 09 2018

Mathematica

Join[{0}, Table[1+Floor[Log[GoldenRatio, (2*n+1)/2]], {n,1,100}]] (* G. C. Greubel, Sep 09 2018 *)

PARI

A102460(n) = { my(u1=1,u2=3,old_u1); if(n<=2,sign(n),while(n>u2,old_u1=u1;u1=u2;u2=old_u1+u2);(u2==n)); };
A130245(n) = if(!n,n,A102460(n)+A130245(n-1));
\\ Or just as:
c=0; for(n=0,123,c += A102460(n); print1(c,", ")); \\ Antti Karttunen, May 13 2018

Python

from itertools import count, islice
def A130245_gen(): # generator of terms
    yield from (0, 1, 2)
    a, b = 3,4
    for i in count(3):
        yield from (i,)*(b-a)
        a, b = b, a+b
A130245_list = list(islice(A130245_gen(),40)) # Chai Wah Wu, Jun 08 2022

Formula

a(n) = 1 +floor(log_phi((n+sqrt(n^2+4))/2)) = 1 +floor(arcsinh(n/2)/log(phi)) for n>=2, where phi = (1+sqrt(5))/2.

a(n) = A130241(n)+1 = A130242(n+1) for n>=2.

G.f.: g(x) = 1/(1-x)*sum{k>=0, x^Lucas(k)}.

a(n) = 1 +floor(log_phi(n+1/2)) for n>=1, where phi is the golden ratio.

Sum_{n>=1} (-1)^(n+1)/a(n) = 3/2 - Pi/(6*sqrt(3)) - log(3)/2. - Amiram Eldar, Jul 25 2025