In number theory, a Keith number or repfigit number (short for repetitive Fibonacci-like digit) is a natural number in a given number base with digits such that when a sequence is created such that the first terms are the digits of and each subsequent term is the sum of the previous terms, is part of the sequence. Keith numbers were introduced by Mike Keith in 1987. They are computationally very challenging to find, with only about 100 known.
Let be a natural number, let be the number of digits in the number in base , and let
be the value of each digit of the number.
We define a linear recurrence relation. such that for ,
If there exists an such that , then is said to be a Keith number.
For example, 88 is a Keith number in base 6, as
and the entire sequence
Whether or not there are infinitely many Keith numbers in a particular base is currently a matter of speculation. Keith numbers are rare and hard to find. They can be found by exhaustive search, and no more efficient algorithm is known. According to Keith, in base 10, on average Keith numbers are expected between successive powers of 10. Known results seem to support this.
14, 19, 28, 47, 61, 75, 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385, 7647, 7909, 31331, 34285, 34348, 55604, 62662, 86935, 93993, 120284, 129106, 147640, 156146, 174680, 183186, 298320, 355419, 694280, 925993, 1084051, 7913837, 11436171, 33445755, 44121607, 129572008, 251133297, ...
In base 2, there exists a method to construct all Keith numbers.
The Keith numbers in base 12, written in base 12, are
A Keith cluster is a related set of Keith numbers such that one is a multiple of another. For example, in base 10, , , and are all Keith clusters. These are possibly the only three examples of a Keith cluster in base 10.
The example below implements the sequence defined above in Python to determine if a number in a particular base is a Keith number:
def is_repfigit(x: int, b: int) -> bool: """Determine if a number in a particular base is a Keith number.""" if x == 0: return True sequence =  y = x while y > 0: sequence.append(y % b) y = y // b digit_count = len(sequence) sequence.reverse() while sequence[len(sequence) - 1] < x: n = 0 for i in range(0, digit_count): n = n + sequence[len(sequence) - digit_count + i] sequence.append(n) return (sequence[len(sequence) - 1] == x)