A386760 Numbers k such that the number of decimal digits of the Lucas number L(k) is greater than the number of decimal digits of the Fibonacci number F(k).

5, 6, 10, 11, 15, 16, 20, 24, 25, 29, 30, 34, 35, 39, 44, 48, 49, 53, 54, 58, 59, 63, 67, 68, 72, 73, 77, 78, 82, 83, 87, 91, 92, 96, 97, 101, 102, 106, 111, 115, 116, 120, 121, 125, 126, 130, 134, 135, 139, 140, 144, 145, 149, 150, 154, 158, 159, 163, 164, 168

Offset: 1

Author

Hans J. H. Tuenter, Aug 13 2025

Keywords

• base
• nonn
• easy
• new

Comments

The difference in the number of decimal digits, A055642(L(k))-A055642(F(k)) = A060384(k)-A386758(k) is either zero or one. In fact, this difference is ceiling(beta-{k*alpha}), with alpha and beta as defined in the Formula section. This implies that, asymptotically, a fraction of beta=0.349485... of the Lucas numbers has one more decimal digit than the corresponding Fibonacci number. This gives the asymptotic behavior of the sequence as a(n)~n/beta. Conjecture: abs(a(n)-n/beta)

Examples

			5 is a term since F(5)=5 has length 1 decimal digit, but L(5)=11 has length 2 decimal digits which is greater.
		

Links

• Hans J. H. Tuenter, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A000032, A000045, A001622, A055642, A060384, A097348, A386758.

Programs

Mathematica

Select[Range[168],IntegerLength[LucasL[#]]>IntegerLength[Fibonacci[#]]&] (* James C. McMahon, Aug 28 2025 *)

Formula

The sequence consists of the integers k>=2, for which {k*alpha}A097348, and phi = (1+sqrt(5))/2 = A001622.

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