A345379 Number of terms m <= n, where m is a term in the bisection of Lucas numbers (A005248).

0, 0, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset: 0

Author

Ovidiu Bagdasar, Jun 16 2021

Keywords

• nonn
• easy

Examples

			a(0)=a(1)=0, since the least term in A005248 is 2.
a(2)=1 since A005248(0) = 2 is followed in that sequence by 3.
a(k)=3 for 3 <= k <= 6 since the first terms of A005248 are {0, 2, 3, 7}.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10000
• 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. See Section 5.4, pp. 33-34, Table 4.

Crossrefs

Cf. A005248, A108852 (Fibonacci), A130245 (Lucas), A130260.

Programs

Mathematica

Block[{a = 3, b = 1, nn = 105, u, v = {}}, u = {2, a}; Do[AppendTo[u, Total[{-b, a} u[[-2 ;; -1]]]]; AppendTo[v, Count[u, _?(# <= i &)]], {i, nn}]; {Boole[First[u] <= 0]}~Join~v] (* or *)
{0}~Join~Accumulate@ ReplacePart[ConstantArray[0, Last[#]], Map[# -> 1 &, #]] &@ LucasL@ Range[0, 10, 2] (* Michael De Vlieger, Jun 16 2021 *)