Parasitic number


An n-parasitic number is a positive natural number which can be multiplied by n by moving the rightmost digit of its decimal representation to the front. Here n is itself a single-digit positive natural number. In other words, the decimal representation undergoes a right circular shift by one place. For example, 4•128205=512820, so 128205 is 4-parasitic. Most authors do not allow leading zeros to be used, and this article follows that convention. So even though 4•025641=102564, the number 025641 is not 4-parasitic.

Derivation

An n-parasitic number can be derived by starting with a digit k in the rightmost place, and working up one digit at a time.
For example, for n = 4 and k = 7
So 179487 is a 4-parasitic number with units digit 7. Others are 179487179487, 179487179487179487, etc.
Notice that the repeating decimal
Thus
In general, an n-parasitic number can be found as follows. Pick a one digit integer k such that, and take the period of the repeating decimal k/.
This will be
where m is the length of the period; i.e. the multiplicative order of 10 modulo.
For another example, if n = 2, then 10n − 1 = 19 and the repeating decimal for 1/19 is
So that for 2/19 is double that:
The length m of this period is 18, the same as the order of 10 modulo 19, so = 105263157894736842.
105263157894736842 × 2 = 210526315789473684, which is the result of moving the last digit of 105263157894736842 to the front.

Additional information

The step-by-step derivation algorithm depicted above is a great core technique but will not find all n-parasitic numbers. It literally gets stuck in an infinite loop with the derived number equaling the derivation source. An example of this occurs when n = 5 and k = 5. The 42-digit n-parasitic number to be derived is 102040816326530612244897959183673469387755. Check the steps in Table One below. The algorithm begins building from right to left until it reaches step 15—then the infinite loop occurs. Lines 16 and 17 are pictured to show that nothing changes. There is a fix for this problem, and when applied, the algorithm will not only find all n-parasitic numbers in base ten, it will find them in base 8 and base 16 as well. Look at line 15 in Table Two. The fix, when this condition is identified and the n-parasitic number has not been found, is simply to not shift the product from the multiplication, but use it as is, and append n to the end. After 42 steps, the proper parasitic number will be found.

Table One

1. 5 × 5 = 25 − Shift = 55
2. 5 × 55 = 275 − Shift = 755
3. 5 × 755 = 3775 − Shift = 7755
4. 5 × 7755 = 38775 − Shift = 87755
5. 5 × 87755 = 438775 − Shift = 387755
6. 5 × 387755 = 1938775 − Shift = 9387755
7. 5 × 9387755 = 46938775 − Shift = 69387755
8. 5 × 69387755 = 346938775 − Shift = 469387755
9. 5 × 469387755 = 2346938775 − Shift = 3469387755
10. 5 × 3469387755 = 17346938775 − Shift = 73469387755
11. 5 × 73469387755 = 367346938775 − Shift = 673469387755
12. 5 × 673469387755 = 3367346938775 − Shift = 3673469387755
13. 5 × 3673469387755 = 18367346938775 − Shift = 83673469387755
14. 5 × 83673469387755 = 418367346938775 − Shift = 183673469387755
15. 5 × 183673469387755 = 918367346938775 − Shift = 183673469387755
16. 5 × 183673469387755 = 918367346938775 − Shift = 183673469387755
17. 5 × 183673469387755 = 918367346938775 − Shift = 183673469387755

Table Two

1. 5 × 5 = 25 − Shift = 55
2. 5 × 55 = 275 − Shift = 755
3. 5 × 755 = 3775 − Shift = 7755
4. 5 × 7755 = 38775 − Shift = 87755
5. 5 × 87755 = 438775 − Shift = 387755
6. 5 × 387755 = 1938775 − Shift = 9387755
7. 5 × 9387755 = 46938775 − Shift = 69387755
8. 5 × 69387755 = 346938775 − Shift = 469387755
9. 5 × 469387755 = 2346938775 − Shift = 3469387755
10. 5 × 3469387755 = 17346938775 − Shift = 73469387755
11. 5 × 73469387755 = 367346938775 − Shift = 673469387755
12. 5 × 673469387755 = 3367346938775 − Shift = 3673469387755
13. 5 × 3673469387755 = 18367346938775 − Shift = 83673469387755
14. 5 × 83673469387755 = 418367346938775 − Shift = 183673469387755
15. 5 × 183673469387755 = 918367346938775 − Shift = 9183673469387755
16. 5 × 9183673469387755 = 45918367346938775 − Shift = 59183673469387755
17. 5 × 59183673469387755 = 295918367346938775 − Shift = 959183673469387755

There is one more condition to be aware of when working with this algorithm, leading zeros must not be lost. When the shift number is created it may contain a leading zero which is positionally important and must be carried into and through the next step. Calculators and computer math methods will remove leading zeros. Look at Table Three below displaying the derivation steps for n = 4 and k = 4. The Shift number created in step 4, 02564, has a leading zero which is fed into step 5 creating a leading zero product. The resulting Shift is fed into Step 6 which displays a product proving the 4-parasitic number ending in 4 is 102564.

Table Three

1. 4 × 4 = 16 − Shift = 64
2. 4 × 64 = 256 − Shift = 564
3. 4 × 564 = 2256 − Shift = 2564
4. 4 × 2564 = 10256 − Shift = 02564
5. 4 × 02564 = 010256 − Shift = 102564
6. 4 × 102564 = 410256 − Shift = 102564

Smallest ''n''-parasitic numbers

The smallest n-parasitic numbers are also known as Dyson numbers, after a puzzle concerning these numbers posed by Freeman Dyson. They are:
nSmallest n-parasitic numberDigitsPeriod of
1111/9
2105263157894736842182/19
31034482758620689655172413793283/29
410256464/39
5102040816326530612244897959183673469387755425/49
61016949152542372881355932203389830508474576271186440677966586/59
71014492753623188405797227/69
81012658227848138/79
910112359550561797752808988764044943820224719449/89

General note

In general, if we relax the rules to allow a leading zero, then there are 9 n-parasitic numbers for each n. Otherwise only if k ≥ n then the numbers do not start with zero and hence fit the actual definition.
Other n-parasitic integers can be built by concatenation. For example, since 179487 is a 4-parasitic number, so are 179487179487, 179487179487179487 etc.

Other bases

In duodecimal system, the smallest n-parasitic numbers are:
nSmallest n-parasitic numberDigitsPeriod of
1111/Ɛ
210631694842Ɛ2/1Ɛ
3249747/2Ɛ = 1/5
410309236ᘔ882061647195441Ɛ4/3Ɛ
51025355ᘔ9433073ᘔ458409919Ɛ715255/4Ɛ
61020408142854ᘔ997732650ᘔ183469163062Ɛ6/5Ɛ
7101899Ɛ864406Ɛ33ᘔᘔ15423913745949305255Ɛ17357/6Ɛ
8131ᘔ8ᘔ6ᘔ/7Ɛ = 2/17
9101419648634459Ɛ9384Ɛ26Ɛ533040547216ᘔ1155Ɛ3Ɛ12978ᘔ399459/8Ɛ
ᘔ14Ɛ36429ᘔ70857921412/9Ɛ = 2/15
Ɛ1011235930336ᘔ53909ᘔ873Ɛ325819Ɛ9975055Ɛ54ᘔ3145ᘔ42694157078404491Ɛ55Ɛ/ᘔƐ

Strict definition

In strict definition, least number m beginning with 1 such that the quotient m/n is obtained merely by shifting the leftmost digit 1 of m to the right end are
They are the period of n/, also the period of the decadic integer -n/.
Number of digits of them are