## 4. Computation techniques 2, results and higher power cases

In this section, I introduce the advanced technique, written in
"Computational Recreations in Mathematica", Ilan Vardi, Addison-Wesley.

The square roots of 2n-digits, (2n-1)-digits numbers are determined by upper n-digits.
So we can consider the following algorithm.

# Let a be n-digits number consisted of 3 different digits.

Let c be the concatinate of a and (n-1) times 0 (c=a*10n-1).
d=isqrt(c)
Check (d+1)2 is consisted of 3 different digits or not.
In this case, we need to check only lower n digits.
(Because upper n digits is known as consisting of just 3 digits).
If it is just 3 digits, found.

Let c be the concatinate of a and n times 0 (c=a*10n).
d=isqrt(c)
Check (d+1)2 is consisted of 3 different digits or not.
In this case, we need to check only lower n digits.
If it is just 3 digits, found.

In this algorithm, if the digits of a increase, the search range is double-digits.
And computation time of 1 increase of a, just 3 times.
In previous algorithm's case, if the digit of a increase, computation time is getting 10 times.
So using this algorithm, we can get a great advantage.

How to generate the (#) number ?

We need to generate the following number sequencially.

• e.g. 0, 1, 2 case

1 , 2 , 10 , 11 , 12 , 20 , 21 , 22 , 100 , 101 , ...

• e.g. 1, 2, 3 case

1 , 2 , 3 , 11 , 12 , 13 , 21 , 22 , 23 , 31 , 32 , 33 , 111 , ...

The algorithm is the following.
First storing 3 digits in T\$[0], T\$[1], T\$[2] from smaller number.
Let s\$ be the number consisted of 3 different digits.
i is the digits (lengths).

# Noticing the i-th digit.

If i-th digit is T\$[0] then
exchange i-th digits as T\$[1].
(i-1)-digit through 1-digits remain the same.

If i-th digit is T\$[1] then
exchange i-th digits as T\$[2].
(i-1)-digit through 1-digits remain the same.

If i-th digit is T\$[2] then
exchange i-th digits as T\$[0].
i=i-1 : gosub #
If i=0 then
concatinate T\$[0] at the uppermost digit.
If T\$[0]=0 then concatinate T\$[1] at the uppermost digit.

The programs are here.

Up to a ≤ 1023 (b=a2) for patterns including zero, and up to a ≤ 1025 for patterns not including zero has done.
Results are following;

Solutions for 013 and 678 are not found yet.

For higher powers, the results are following.

#### Infinite patterns

power 3 : a=10n+1 (n ≥ 2), b=a3
power 5 : a=10n+1 (n ≥ 2), b=a5

powerab=ap
37343
111331
142744
153375
3646656
62238328
92778688
1735177717
1927077888
2119393931
888700227072
354344474744007
1100111331399339931331
1467963163316636166336
496561
1114641
1665536
341336336
567776
236436343
6646656
836561
465536
113177147
16265536

I couldn't find any solutions for power 7.

At first, the conditions

1. 3 different digits
2. perfect n-th power

are too weak and we will be able to get the many results.
But for higher power cases, the results are only above.

E-mail : kc2h-msm@asahi-net.or.jp
Hisanori Mishima