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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).
2
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
1,1
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)<c, for some constant c.
LINKS
FORMULA
The sequence consists of the integers k>=2, for which {k*alpha}<beta, where alpha=log_10(phi), beta=log_10(5)/2, {x}=x-floor(x), denotes the fractional part of x, log_10(phi) = A097348, and phi = (1+sqrt(5))/2 = A001622.
EXAMPLE
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.
MATHEMATICA
Select[Range[168], IntegerLength[LucasL[#]]>IntegerLength[Fibonacci[#]]&] (* James C. McMahon, Aug 28 2025 *)
KEYWORD
base,nonn,easy
AUTHOR
Hans J. H. Tuenter, Aug 13 2025
STATUS
approved