Keith Numbers

A Keith Number is an \(n\)-digit integer N with the following property: If a Fibonacci-like sequence (in which each term in the sequence is the sum of the \(n\) previous terms) is formed, with the first \(n\) terms being the decimal digits of the number N, then N itself occurs as a term in the sequence. For example, 197 is a Keith number since it generates the sequence (source)

1, 9, 7, 17, 33, 57, 107, 197, ...

The Keith numbers form OEIS A007629 and the initial members are as follows:

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

Let's take 31331 as another example (permalink):

3, 1, 3, 3, 1, 11, 19, 37, 71, 139, 277, 543, 1067, 2097, 4123, 8107, 15937, 31331

I encountered a variation of Keith numbers when I celebrated being 27223 days old. One of the properties of this number is that it's a member of OEIS A274770:

A274770

Cube analog to Keith numbers.

Here are the OEIS comments regarding this sequence:

Like Keith numbers but starting from \(n^3\) digits to reach \(n\). Consider the digits of the cube of a number \(n\) . Take their sum and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.

The example of \( 776^3 = 467288576 \) is given:

4 + 6 + 7 + 2 + 8 + 8 + 5 + 7 + 6 = 53

6 + 7 + 2 + 8 + 8 + 5 + 7 + 6 + 53 = 102

7 + 2 + 8 + 8 + 5 + 7 + 6 + 53 + 102 = 198

2 + 8 + 8 + 5 + 7 + 6 + 53 + 102 + 198 = 389

8 + 8 + 5 + 7 + 6 + 53 + 102 + 198 + 389 = 776

The initial members of the sequence are:

1, 8, 17, 18, 26, 27, 44, 55, 63, 80, 105, 187, 326, 776, 1095, 2196, 6338, 13031, 13131, 25562, 27223, 70825, 140791, 553076, 632489, 1402680, 1404312, 3183253, 11311424, 50783292, 51231313, 182252596, 255246098, 522599548, 1180697763, 2025114819, 2137581414

In a similar vein, we have the square analogs of Keith numbers. These numbers comprise OEIS A274769: 

A274769

Square analog to Keith numbers.

The OEIS comments state that:

Like Keith numbers but starting from \(n^2\) digits to reach \(n\). Consider the digits of the square of a number \(n\). Take their sum and repeat the process deleting the first addend and adding the previous sum. The sequence lists the numbers that after some iterations reach a sum equal to themselves.

 The example of \( 1264^2 = 1597696 \) is given :

1 + 5 + 9 + 7 + 6 + 9 + 6 = 43

5 + 9 + 7 + 6 + 9 + 6 + 43 = 85

9 + 7 + 6 + 9 + 6 + 43 + 85 = 165

7 + 6 + 9 + 6 + 43 + 85 + 165 = 321

6 + 9 + 6 + 43 + 85 + 165 + 321 = 635

9 + 6 + 43 + 85 + 165 + 321 + 635 = 1264

The initial members of this sequence are:

1, 9, 37, 40, 43, 62, 70, 74, 160, 1264, 1952, 2847, 12799, 16368, 16584, 42696, 83793, 97415, 182011, 352401, 889871, 925356, 1868971, 1881643, 3661621, 7645852, 15033350, 21655382, 63288912, 88192007, 158924174, 381693521, 792090500, 2025078249, 2539401141

Finally, let's return to Keith numbers and include some further comments from Mr Keith himself (source):

There are still a number of unanswered questions about these numbers, such as:

Are there an infinite number of Keith numbers? Heuristic arguments, and the numerical evidence above, both strongly suggest that the answer is "yes" - in fact, we expect to find roughly \(0.9 \log_2{10}\) (about 3) of them between each power of 10. But there is still no proof, constructive or otherwise, that there are an infinite number of them.

Define a cluster of Keith numbers as a set of two or more (all with the same number of digits) in which all the numbers are integer multiples of the smallest one in the set. There are only three known clusters: (14, 28), (1104, 2208), and the remarkable - for having three members - (31331, 62662, 93993). Question: is the number of Keith clusters finite or infinite? Not only do we conjecture it is finite, but we conjecture that the above three clusters are the only ones.  But we have no clue how to prove this.

Is \(n=10\) the only number of digits for which there are no Keith numbers?  (We tentatively think not, but it may be a while before another one is found.)