A102460 a(n) = 1 if n is a Lucas number, else a(n) = 0.

0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Author

Casey Mongoven, Apr 18 2005

Keywords

• nonn

Comments

From Hieronymus Fischer, Jul 02 2007: (Start)

a(n) is the number of nonnegative integer solutions to 25*x^4-10*n^2*x^2+n^4-16=0.

a(n)=1 if and only if there is an integer m such that x=n is a root of p(x)=x^4-10*m^2*x^2+25*m^4-16.

For n>=3: a(n)=1 iff floor(log_phi(n+1/2))=ceiling(log_phi(n-1/2)). (End)

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537
• Casey Mongoven, Lucas Binary no. 1; electronic music created with this sequence.
• Index entries for characteristic functions

Crossrefs

Cf. A000032, A130241, A130242, A130247, A123927.

Cf. partial sums A130245.

Cf. also A010056, A104162, A105348, A147612, A294878.

Programs

Mathematica

{0}~Join~ReplacePart[ConstantArray[0, Last@ #], Map[# -> 1 &, #]] &@ Array[LucasL, 11, 0] (* Michael De Vlieger, Nov 22 2017 *)
With[{nn=130,lc=LucasL[Range[0,20]]},Table[If[MemberQ[lc,n],1,0],{n,0,nn}]] (* Harvey P. Dale, Jul 03 2022 *)

PARI

a(n)=my(f=factor(25*'x^4-10*n^2*'x^2+n^4-16)[,1]); sum(i=1,#f, poldegree(f[i])==1 && polcoeff(f[i],0)<=0) \\ Charles R Greathouse IV, Nov 06 2014

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)); }; \\ Antti Karttunen, Nov 22 2017

Python

from sympy.ntheory.primetest import is_square
def A102460(n): return int(is_square(m:=5*(n**2-4)) or is_square(m+40)) # Chai Wah Wu, Jun 13 2024

Formula

From Hieronymus Fischer, Jul 02 2007: (Start)

G.f.: g(x) = Sum_{k>=0} x^A000032(k).

a(n) = 1+floor(arcsinh(n/2)/log(phi))-ceiling(arccosh(n/2)/log(phi)) for n>=3, where phi=(1+sqrt(5))/2.

a(n) = 1+A130241(n)-A130242(n) for n>=3.

a(n) = 1+A130247(n)-A130242(n) for n=>2.

a(n) = A130245(n)-A130245(n-1) for n>=1.

For n>=3: a(n)=1 iff A130241(n)=A130242(n). (End)

Extensions

Data section extended up to a(123) by Antti Karttunen, Nov 22 2017