# Undulating numbers

Undulating numbers
You are encouraged to solve this task according to the task description, using any language you may know.

An Undulating number in some base is a number which has the digit form ABABAB... where A and B are digits in that base.

For the purposes of this task, we shall only consider a number to be undulating if it has at least 3 digits in a given base and A != B.

Examples

The numbers: 101 and 9898 are undulating numbers in base 10.

The numbers: 50 and 2350 are undulating numbers in base 7 because their base 7 representations are: 101 and 6565 respectively.

1. All three digit undulating numbers.

2. All four digit undulating numbers.

3. All three digit undulating numbers which are primes.

4. The 600th undulating number.

5. How many undulating numbers are less than 2^53 and the largest such number.

Bonus

Do the same for base 7 expressing the results in base 10 apart from 4. and 5. which should be expressed in base 7 also.

Note that undulating numbers with a given number of digits in base 7 may, of course, have less digits when expressed in base 10 and are unlikely to be still undulating. However, 121 (or 232 in base 7) is an exception to this.

References

## ALGOL 68

As undulating numbers are relatively sparse, this counts the numbers by enumerating them.

```BEGIN # show some Undulating numbers - numbers whose digits form the pattern #
#                                ABAB...                               #

LONG INT max number = LONG 2 ^ 53;     # maximum number we will consider #
STRING   max name   = "2^53";          # "name" of max number            #
# returns TRUE if n is prime, FALSE otherwise - uses trial division      #
PROC is prime = ( LONG INT n )BOOL:
IF   n < 3       THEN n = 2
ELIF n MOD 3 = 0 THEN n = 3
ELIF NOT ODD n   THEN FALSE
ELSE
BOOL is a prime := TRUE;
INT  f          := 5;
INT  f2         := 25;
INT  to next    := 24;
WHILE f2 <= n AND is a prime DO
is a prime := n MOD f /= 0;
f         +:= 2;
f2        +:= to next;
to next   +:= 8
OD;
is a prime
FI # is prime # ;
# returns s with comma separators                                        #
PROC commatise = ( STRING s )STRING:
BEGIN
STRING result := "";
INT r pos     := -1;
FOR s pos FROM UPB s BY -1 TO LWB s DO
IF ( r pos +:= 1 ) = 3 THEN
r pos := 0;
"," +=: result
FI;
s[ s pos ] +=: result
OD;
result
END # commatise # ;
# returns a string representation of the undulating number in base b     #
#         with first digit f and second digit s with d digits in total   #
PROC undulating string = ( INT d, f, s, b )STRING:
BEGIN
STRING un    := "";
CHAR   fc     = REPR ( f + ABS "0" );
CHAR   sc     = REPR ( s + ABS "0" );
FOR digit FROM d BY -1 TO 1 DO
IF ODD digit THEN fc ELSE sc FI +=: un
OD;
commatise( un )
END # undulating string # ;
# shows various Undulating numbers in base b                             #
PROC print undulation = ( INT b )VOID:
BEGIN
LONG INT un          := 0;
INT      un count    := 0;
INT      p count     := 0;
INT      last d      := 0;
INT      last f      := 0;
INT      last s      := 0;
LONG INT last un     := 0;
[ 1 : ( b - 1 ) * ( b - 1 ) ]LONG INT prime un;
FOR digits FROM 3 WHILE un < max number DO
IF digits = 3 OR digits = 4 THEN
print( ( whole( digits, 0 ), " digit Undulating Numbers in base " ) );
print( ( whole( b, 0 ) ) );
IF b /= 10 THEN print( ( " (shown in base 10)" ) ) FI;
print( ( ":", newline ) )
FI;
FOR f TO b - 1 WHILE  un < max number DO
FOR s FROM 0 TO b - 1 WHILE  un < max number DO
IF f /= s THEN
un        := 0;
FOR d TO digits DO
un *:= b
+:= IF ODD d THEN f ELSE s FI
OD;
IF un < max number THEN
un count +:= 1;
last un   := un;
last d    := digits;
last f    := f;
last s    := s;
IF digits = 3 OR digits = 4 THEN
print( ( " ", whole( un, -digits ) ) );
IF digits = 3 THEN
IF is prime( un ) THEN
prime un[ p count +:= 1 ] := un
FI
FI
ELIF un count = 600 THEN
print( ( "Undulating number 600 in base ", whole( b, 0 ), ": " ) );
print( ( undulating string( digits, f, s, b ) ) );
IF b /= 10 THEN
print( ( newline, "        which is: " ) );
print( ( commatise( whole( un, 0 ) ) ) );
print( ( " in base 10", newline ) )
FI;
print( ( newline ) )
FI
FI
FI
OD;
IF digits = 3 OR digits = 4 THEN print( ( newline ) ) FI
OD;
IF digits = 3 OR digits = 4 THEN
print( ( newline ) );
IF digits = 4 THEN
print( ( "Prime 3 digit Undulating Numbers in base " ) );
print( ( whole( b, 0 ) ) );
IF b /= 10 THEN print( ( " (shown in base 10)" ) ) FI;
print( ( ":", newline ) );
FOR i TO p count DO
print( ( whole( prime un[ i ], -4 ) ) );
IF i MOD b = ( b - 1 ) THEN print( ( newline ) ) FI
OD;
IF p count MOD b /= ( b - 1 ) THEN print( ( newline ) ) FI;
print( ( newline ) )
FI
FI
OD;
print( ( "There are ", whole( un count, 0 ) ) );
print( ( " Undulating nnumbers in base ", whole( b, 0 ), " before ", max name, newline ) );
print( ( "The last is: ", undulating string( last d, last f, last s, b ), newline ) );
IF b /= 10 THEN
print( ( "   which is: ", commatise( whole( last un, 0 ) ), " in base 10", newline ) )
FI
END # print undulation # ;

print undulation( 10 ); # base 10 undulating numbers                #
print( ( newline ) );
print undulation(  7 )  # base  7 undulating numbers                #

END```
Output:
```3 digit Undulating Numbers in base 10:
101 121 131 141 151 161 171 181 191
202 212 232 242 252 262 272 282 292
303 313 323 343 353 363 373 383 393
404 414 424 434 454 464 474 484 494
505 515 525 535 545 565 575 585 595
606 616 626 636 646 656 676 686 696
707 717 727 737 747 757 767 787 797
808 818 828 838 848 858 868 878 898
909 919 929 939 949 959 969 979 989

4 digit Undulating Numbers in base 10:
1010 1212 1313 1414 1515 1616 1717 1818 1919
2020 2121 2323 2424 2525 2626 2727 2828 2929
3030 3131 3232 3434 3535 3636 3737 3838 3939
4040 4141 4242 4343 4545 4646 4747 4848 4949
5050 5151 5252 5353 5454 5656 5757 5858 5959
6060 6161 6262 6363 6464 6565 6767 6868 6969
7070 7171 7272 7373 7474 7575 7676 7878 7979
8080 8181 8282 8383 8484 8585 8686 8787 8989
9090 9191 9292 9393 9494 9595 9696 9797 9898

Prime 3 digit Undulating Numbers in base 10:
101 131 151 181 191 313 353 373 383
727 757 787 797 919 929

Undulating number 600 in base 10: 4,646,464,646
There are 1125 Undulating nnumbers in base 10 before 2^53
The last is: 8,989,898,989,898,989

3 digit Undulating Numbers in base 7 (shown in base 10):
50  64  71  78  85  92
100 107 121 128 135 142
150 157 164 178 185 192
200 207 214 221 235 242
250 257 264 271 278 292
300 307 314 321 328 335

4 digit Undulating Numbers in base 7 (shown in base 10):
350  450  500  550  600  650
700  750  850  900  950 1000
1050 1100 1150 1250 1300 1350
1400 1450 1500 1550 1650 1700
1750 1800 1850 1900 1950 2050
2100 2150 2200 2250 2300 2350

Prime 3 digit Undulating Numbers in base 7 (shown in base 10):
71 107 157 257 271 307

Undulating number 600 in base 7: 4,646,464,646,464,646,464
which is: 8,074,217,422,972,642 in base 10

There are 603 Undulating nnumbers in base 7 before 2^53
The last is: 5,252,525,252,525,252,525
which is: 8,786,648,372,058,464 in base 10
```

## C++

```#include <algorithm>
#include <cassert>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <vector>

bool is_prime(uint64_t n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
for (uint64_t p = 5; p * p <= n; p += 4) {
if (n % p == 0)
return false;
p += 2;
if (n % p == 0)
return false;
}
return true;
}

class undulating_number_generator {
public:
explicit undulating_number_generator(int base) : base_(base) {}

uint64_t next() {
uint64_t n = 0;
for (int d = 0; d < digits_; ++d)
n = n * base_ + (d % 2 == 0 ? a_ : b_);
++b_;
if (a_ == b_)
++b_;
if (b_ == base_) {
++a_;
b_ = 0;
if (a_ == base_) {
a_ = 1;
++digits_;
}
}
return n;
}

private:
int base_;
int a_ = 1;
int b_ = 0;
int digits_ = 3;
};

std::string to_string(uint64_t n, int base) {
assert(base > 1 && base <= 16);
const char digits[] = "0123456789ABCDEF";
std::string str;
for (; n > 0; n /= base)
str += digits[n % base];
reverse(str.begin(), str.end());
return str;
}

void undulating(int base) {
undulating_number_generator gen(base);
uint64_t n = gen.next();
int i = 1;
uint64_t limit = base * base * base;
std::vector<uint64_t> primes;
std::cout << "3-digit undulating numbers in base " << base << ":\n";
for (; n < limit; ++i) {
std::cout << std::setw(3) << n << (i % 9 == 0 ? '\n' : ' ');
if (is_prime(n))
primes.push_back(n);
n = gen.next();
}
limit *= base;
std::cout << "\n4-digit undulating numbers in base " << base << ":\n";
for (; n < limit; ++i) {
std::cout << std::setw(4) << n << (i % 9 == 0 ? '\n' : ' ');
n = gen.next();
}
std::cout << "\n3-digit undulating numbers in base " << base
<< " which are prime:\n";
for (auto prime : primes)
std::cout << prime << ' ';
std::cout << '\n';
for (; i != 600; ++i)
n = gen.next();
std::cout << "\nThe 600th undulating number in base " << base << " is "
<< n;
if (base != 10) {
std::cout << "\nor expressed in base " << base << ": "
<< to_string(n, base);
}
std::cout << ".\n";
for (;; ++i) {
uint64_t next = gen.next();
if (next >= (1ULL << 53))
break;
n = next;
}
std::cout << "\nTotal number of undulating numbers < 2^53 in base " << base
<< ": " << i << "\nof which the largest is " << n;
if (base != 10) {
std::cout << "\nor expressed in base " << base << ": "
<< to_string(n, base);
}
std::cout << ".\n";
}

int main() {
undulating(10);
std::cout << '\n';
undulating(7);
}
```
Output:
```3-digit undulating numbers in base 10:
101 121 131 141 151 161 171 181 191
202 212 232 242 252 262 272 282 292
303 313 323 343 353 363 373 383 393
404 414 424 434 454 464 474 484 494
505 515 525 535 545 565 575 585 595
606 616 626 636 646 656 676 686 696
707 717 727 737 747 757 767 787 797
808 818 828 838 848 858 868 878 898
909 919 929 939 949 959 969 979 989

4-digit undulating numbers in base 10:
1010 1212 1313 1414 1515 1616 1717 1818 1919
2020 2121 2323 2424 2525 2626 2727 2828 2929
3030 3131 3232 3434 3535 3636 3737 3838 3939
4040 4141 4242 4343 4545 4646 4747 4848 4949
5050 5151 5252 5353 5454 5656 5757 5858 5959
6060 6161 6262 6363 6464 6565 6767 6868 6969
7070 7171 7272 7373 7474 7575 7676 7878 7979
8080 8181 8282 8383 8484 8585 8686 8787 8989
9090 9191 9292 9393 9494 9595 9696 9797 9898

3-digit undulating numbers in base 10 which are prime:
101 131 151 181 191 313 353 373 383 727 757 787 797 919 929

The 600th undulating number in base 10 is 4646464646.

Total number of undulating numbers < 2^53 in base 10: 1125
of which the largest is 8989898989898989.

3-digit undulating numbers in base 7:
50  64  71  78  85  92 100 107 121
128 135 142 150 157 164 178 185 192
200 207 214 221 235 242 250 257 264
271 278 292 300 307 314 321 328 335

4-digit undulating numbers in base 7:
350  450  500  550  600  650  700  750  850
900  950 1000 1050 1100 1150 1250 1300 1350
1400 1450 1500 1550 1650 1700 1750 1800 1850
1900 1950 2050 2100 2150 2200 2250 2300 2350

3-digit undulating numbers in base 7 which are prime:
71 107 157 257 271 307

The 600th undulating number in base 7 is 8074217422972642
or expressed in base 7: 4646464646464646464.

Total number of undulating numbers < 2^53 in base 7: 603
of which the largest is 8786648372058464
or expressed in base 7: 5252525252525252525.
```

## EasyLang

Translation of: C++
```func isprim num .
i = 2
while i <= sqrt num
if num mod i = 0
return 0
.
i += 1
.
return 1
.
subr init
a_ = 1
b_ = 0
digits_ = 3
max_ = pow 2 53 - 1
.
func nxtundul .
for d = 1 to digits_
if d mod 2 = 1
n = n * 10 + a_
else
n = n * 10 + b_
.
if n > max_
return 0
.
.
b_ += 1
if a_ = b_
b_ += 1
.
if b_ = 10
a_ += 1
b_ = 0
if a_ = 10
a_ = 1
digits_ += 1
.
.
return n
.
init
while digits_ = 3
write nxtundul & " "
.
print "\n"
while digits_ = 4
write nxtundul & " "
.
print "\n"
init
while digits_ = 3
h = nxtundul
if isprim h = 1
write h & " "
.
.
print "\n"
init
for i to 600
h = nxtundul
.
print h
print ""
init
repeat
last = h
h = nxtundul
until h = 0
cnt += 1
.
print cnt & " " & last```
Output:
```101 121 131 141 151 161 171 181 191 202 212 232 242 252 262 272 282 292 303 313 323 343 353 363 373 383 393 404 414 424 434 454 464 474 484 494 505 515 525 535 545 565 575 585 595 606 616 626 636 646 656 676 686 696 707 717 727 737 747 757 767 787 797 808 818 828 838 848 858 868 878 898 909 919 929 939 949 959 969 979 989

1010 1212 1313 1414 1515 1616 1717 1818 1919 2020 2121 2323 2424 2525 2626 2727 2828 2929 3030 3131 3232 3434 3535 3636 3737 3838 3939 4040 4141 4242 4343 4545 4646 4747 4848 4949 5050 5151 5252 5353 5454 5656 5757 5858 5959 6060 6161 6262 6363 6464 6565 6767 6868 6969 7070 7171 7272 7373 7474 7575 7676 7878 7979 8080 8181 8282 8383 8484 8585 8686 8787 8989 9090 9191 9292 9393 9494 9595 9696 9797 9898

101 131 151 181 191 313 353 373 383 727 757 787 797 919 929

4646464646

1125 8989898989898989
```

## J

Inspection shows that there are 81 undulating numbers (base 10) for any given digit count, and 36 for base 7. (In other words two times the number of two digit combinations of the base minus the number of non-zero digits (or: the square of one less than the base).)

So, given:

```require'stats'
undul=: {{ /:~ m #. y\$"1(<:m)}.(,|."1)2 comb m }}
und10=: 10 undul
und7=: 7 undul

fmt7=: 7{{' '-.~":m&#.inv y}}
```

As an aside, undul could have been made more efficient, if speed was a concern, precalculating and sorting the digit pairs when the base was choosen:

```undul=: {{ m {{ m #. y\$"1 n}} (/:~ (<:m)}.(,|."1)2 comb m) }}
```

We get:

```   9 9\$und10 3
101 121 131 141 151 161 171 181 191
202 212 232 242 252 262 272 282 292
303 313 323 343 353 363 373 383 393
404 414 424 434 454 464 474 484 494
505 515 525 535 545 565 575 585 595
606 616 626 636 646 656 676 686 696
707 717 727 737 747 757 767 787 797
808 818 828 838 848 858 868 878 898
909 919 929 939 949 959 969 979 989
9 9\$und10 4
1010 1212 1313 1414 1515 1616 1717 1818 1919
2020 2121 2323 2424 2525 2626 2727 2828 2929
3030 3131 3232 3434 3535 3636 3737 3838 3939
4040 4141 4242 4343 4545 4646 4747 4848 4949
5050 5151 5252 5353 5454 5656 5757 5858 5959
6060 6161 6262 6363 6464 6565 6767 6868 6969
7070 7171 7272 7373 7474 7575 7676 7878 7979
8080 8181 8282 8383 8484 8585 8686 8787 8989
9090 9191 9292 9393 9494 9595 9696 9797 9898
(#~ 1 p:])und10 3
101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
(<:600){;und10&.>3+i.>.600%81
4646464646
(#,{:)(#~ (2^53)>:]);und10&.>3+i.>.10^.2^53
1125 8989898989898989

6 7\$ und7 3
50  64  71  78  85  92 100
107 121 128 135 142 150 157
164 178 185 192 200 207 214
221 235 242 250 257 264 271
278 292 300 307 314 321 328
335  50  64  71  78  85  92
6 7\$ und7 4
350  450  500  550  600  650  700
750  850  900  950 1000 1050 1100
1150 1250 1300 1350 1400 1450 1500
1550 1650 1700 1750 1800 1850 1900
1950 2050 2100 2150 2200 2250 2300
2350  350  450  500  550  600  650
(#~ 1 p:])und7 3
71 107 157 257 271 307
fmt7(<:600){;und7&.>3+i.>.600%36
4646464646464646464
(#,&":' ',fmt7@{:)(#~ (2^53)>:]);und7&.>3+i.>.7^.2^53
603 5252525252525252525
```

## Java

```public final class UndulatingNumbers {

public static void main(String[] args) {
UndulatingNumberIterator iterator = new UndulatingNumberIterator(3, 3, 10);
System.out.println("Three digit undulating numbers in base 10:");
int count = 0;
while ( iterator.hasNext() ) {
count += 1;
System.out.print(String.format("%3d%s", iterator.next(), ( ( count % 9 == 0 ) ? "\n" : " " )));
}
System.out.println();

iterator = new UndulatingNumberIterator(4, 4, 10);
System.out.println("Four digit undulating numbers in base 10:");
count = 0;
while ( iterator.hasNext() ) {
count += 1;
System.out.print(String.format("%3d%s", iterator.next(), ( ( count % 9 == 0 ) ? "\n" : " " )));
}
System.out.println();

iterator = new UndulatingNumberIterator(3, 3, 10);
System.out.println("Three digit undulating numbers in base 10 which are prime numbers:");
while ( iterator.hasNext() ) {
long undulatingNumber = iterator.next();
if ( isPrime(undulatingNumber) ) {
System.out.print(undulatingNumber + " ");
}
}
System.out.println(System.lineSeparator());

iterator = new UndulatingNumberIterator(3, 100, 10);
count = 0;
while ( count < 599 && iterator.hasNext() ) {
count += 1;
iterator.next();
}
System.out.println("The 600th undulating number in base 10 is " + iterator.next());
System.out.println();

final long TWO_POWER_53 = (long) Math.pow(2, 53);
final int maxDigitsBase10 = String.valueOf(TWO_POWER_53).length();
long number = 0;
long largest = 0;
count = 0;
iterator = new UndulatingNumberIterator(3, maxDigitsBase10, 10);
while ( iterator.hasNext() && ( number = iterator.next() ) < TWO_POWER_53 ) {
count += 1;
largest = number;
}
System.out.println("The number of undulating numbers in base 10 less than 2^53 is " + count);
System.out.println("The last undulating number in base 10 less than 2^53 is " + largest);
System.out.println();

// Bonus - Part 1
System.out.println("Three digit numbers, written in base 10, which are undulating in base 7:");
iterator = new UndulatingNumberIterator(3, 3, 7);
count = 0;
while ( iterator.hasNext() ) {
count += 1;
System.out.print(String.format("%3d%s", iterator.next(), ( ( count % 9 == 0 ) ? "\n" : " " )));
}
System.out.println();

// Bonus - Part 2
System.out.println("Four digit numbers, written in base 10, which are undulating in base 7:");
iterator = new UndulatingNumberIterator(4, 4, 7);
count = 0;
while ( iterator.hasNext() ) {
count += 1;
System.out.print(String.format("%3d%s", iterator.next(), ( ( count % 9 == 0 ) ? "\n" : " " )));
}
System.out.println();

// Bonus - Part 3
iterator = new UndulatingNumberIterator(3, 3, 7);
System.out.println("Three digit prime numbers, written in base 10, which are undulating in base 7:");
while ( iterator.hasNext() ) {
long undulatingNumber = iterator.next();
if ( isPrime(undulatingNumber) ) {
System.out.print(undulatingNumber + " ");
}
}
System.out.println(System.lineSeparator());

// Bonus - Part 4
iterator = new UndulatingNumberIterator(3, 100, 7);
count = 0;
while ( count < 599 && iterator.hasNext() ) {
count += 1;
iterator.next();
}
final long undulatingNumber = iterator.next();
System.out.println("The 600th undulating number in base 7 is " + convertToBase(7, undulatingNumber));
System.out.println("which is " + undulatingNumber + " written in base 10");
System.out.println();

final int maxDigitsBase7 = convertToBase(7, TWO_POWER_53).length();
number = 0;
largest = 0;
count = 0;
iterator = new UndulatingNumberIterator(3, maxDigitsBase7, 7);
while ( iterator.hasNext() && ( number = iterator.next() ) < TWO_POWER_53 ) {
count += 1;
largest = number;
}
System.out.println("The number of undulating numbers in base 7 less than 2^53 is " + count);
System.out.println("The last undulating number in base 7 less than 2^53 is " + convertToBase(7, largest));
System.out.println("which is " + largest+ " written in base 10");
System.out.println();
}

private static final class UndulatingNumberIterator {

public UndulatingNumberIterator(int aMinDigits, int aMaxDigits, int aBase) {
minDigits = aMinDigits;
maxDigits = aMaxDigits;
base = aBase;
}

public boolean hasNext() {
return minDigits <= maxDigits;
}

public long next() {
long result = 0;
for ( int digit = 0; digit < minDigits; digit++ ) {
result = result * base + ( digit % 2 == 0 ? a : b );
}

b += 1;
if ( a == b ) {
b += 1;
}
if ( b == base ) {
b = 0;
a += 1;
if ( a == base ) {
a = 1;
minDigits += 1;
}
}
return result;
}

private int minDigits;
private int a = 1;
private int b = 0;

private final int base;
private final int maxDigits;

}

private static String convertToBase(int base, long number) {
if ( base < 2 || base > 10 ) {
throw new AssertionError("Base should be in the range: 2 << base << 10");
}

if ( number == 0 ) {
return "0";
}

StringBuilder result = new StringBuilder();
while ( number != 0 ) {
result.append(number % base);
number /= base;
}

return result.reverse().toString();
}

private static boolean isPrime(long number) {
if ( number < 2 ) {
return false;
}
if ( number % 2 == 0 ) {
return number == 2;
}
if ( number % 3 == 0 ) {
return number == 3;
}

for ( long p = 5; p * p <= number; p += 4 ) {
if ( number % p == 0 ) {
return false;
}
p += 2;
if ( number % p == 0 ) {
return false;
}
}

return true;
}

}
```
Output:
```Three digit undulating numbers in base 10:
101 121 131 141 151 161 171 181 191
202 212 232 242 252 262 272 282 292
303 313 323 343 353 363 373 383 393
404 414 424 434 454 464 474 484 494
505 515 525 535 545 565 575 585 595
606 616 626 636 646 656 676 686 696
707 717 727 737 747 757 767 787 797
808 818 828 838 848 858 868 878 898
909 919 929 939 949 959 969 979 989

Four digit undulating numbers in base 10:
1010 1212 1313 1414 1515 1616 1717 1818 1919
2020 2121 2323 2424 2525 2626 2727 2828 2929
3030 3131 3232 3434 3535 3636 3737 3838 3939
4040 4141 4242 4343 4545 4646 4747 4848 4949
5050 5151 5252 5353 5454 5656 5757 5858 5959
6060 6161 6262 6363 6464 6565 6767 6868 6969
7070 7171 7272 7373 7474 7575 7676 7878 7979
8080 8181 8282 8383 8484 8585 8686 8787 8989
9090 9191 9292 9393 9494 9595 9696 9797 9898

Three digit undulating numbers in base 10 which are prime numbers:
101 131 151 181 191 313 353 373 383 727 757 787 797 919 929

The 600th undulating number in base 10 is 4646464646

The number of undulating numbers in base 10 less than 2^53 is 1125
The last undulating number in base 10 less than 2^53 is 8989898989898989

Three digit numbers, written in base 10, which are undulating in base 7:
50  64  71  78  85  92 100 107 121
128 135 142 150 157 164 178 185 192
200 207 214 221 235 242 250 257 264
271 278 292 300 307 314 321 328 335

Four digit numbers, written in base 10, which are undulating in base 7:
350 450 500 550 600 650 700 750 850
900 950 1000 1050 1100 1150 1250 1300 1350
1400 1450 1500 1550 1650 1700 1750 1800 1850
1900 1950 2050 2100 2150 2200 2250 2300 2350

Three digit prime numbers, written in base 10, which are undulating in base 7:
71 107 157 257 271 307

The 600th undulating number in base 7 is 4646464646464646464
which is 8074217422972642 written in base 10

The number of undulating numbers in base 7 less than 2^53 is 603
The last undulating number in base 7 less than 2^53 is 5252525252525252525
which is 8786648372058464 written in base 10
```

## jq

Works with: jq

Also works with gojq, the Go implementation of jq.

```## For the sake of gojq:
def _nwise(\$n):
def n: if length <= \$n then . else .[0:\$n] , (.[\$n:] | n) end;
n;

## Generic functions
def lpad(\$len): tostring | (\$len - length) as \$l | (" " * \$l) + .;

def tobase(\$b):
def digit: "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[.:.+1];
def mod: . % \$b;
def div: ((. - mod) / \$b);
def digits: recurse( select(. > 0) | div) | mod ;
# For jq it would be wise to protect against `infinite` as input, but using `isinfinite` confuses gojq
select( (tostring|test("^[0-9]+\$")) and 2 <= \$b and \$b <= 36)
| if . == 0 then "0"
else [digits | digit] | reverse[1:] | add
end;

# tabular print
def tprint(columns; wide):
reduce _nwise(columns) as \$row ("";
. + (\$row|map(lpad(wide)) | join(" ")) + "\n" );

def is_prime:
. as \$n
| if (\$n < 2)         then false
elif (\$n % 2 == 0)  then \$n == 2
elif (\$n % 3 == 0)  then \$n == 3
elif (\$n % 5 == 0)  then \$n == 5
elif (\$n % 7 == 0)  then \$n == 7
elif (\$n % 11 == 0) then \$n == 11
elif (\$n % 13 == 0) then \$n == 13
elif (\$n % 17 == 0) then \$n == 17
elif (\$n % 19 == 0) then \$n == 19
else sqrt as \$s
| 23
| until( . > \$s or (\$n % . == 0); . + 2)
| . > \$s
end;

## Undulating numbers
def undulating(\$base; \$n):
53 as \$mpow
| (pow(2;\$mpow) - 1) as \$limit
| (\$base * \$base) as \$bsquare
| { u3: [], u4: [] }
| reduce range(1; \$base) as \$a (.;
reduce range(0; \$base) as \$b (.;
if \$b == \$a then .
else (\$a * \$bsquare + \$b * \$base + \$a) as \$u
| .u3 += [\$u]
| (\$a * \$base + \$b) as \$v
| .u4 += [\$v * \$bsquare + \$v]
end) )
| "All 3-digit undulating numbers in base \(\$base):",
(.u3 | tprint(9; 4)),
"All 4-digit undulating numbers in base \(\$base):",
(.u4 | tprint(9; 5)),
"All 3-digit undulating numbers which are primes in base \(\$base):",
( .primes = []
| reduce .u3[] as \$u (.;
if \$u % 2 == 1 and \$u % 5 != 0 and (\$u | is_prime)
then .primes += [\$u]
else .
end)
| (.primes | tprint(10; 4))),
( .un = (.u3 + .u4)
| (.un|length) as \$unc
| .j = 0
| .i = 0
| .done = false
| until(.done;
.i = 0
| until(.i >= \$unc;
(.un[.j * \$unc + .i] * \$bsquare + (.un[.j * \$unc + .i] % \$bsquare)) as \$u
| if \$u > \$limit then .done = true | .i = \$unc
else .un += [\$u]
| .i += 1
end )
| .j += 1 )
| "\nThe \(\$n)th undulating number in base \(\$base) is: \(.un[\$n-1])",
(select(\$base != 10) | "or expressed in base \(\$base): \(.un[\$n-1] | tobase(\$base))"),
"\nTotal number of undulating numbers in base \(\$base) < 2^\(\$mpow) = \(.un|length)",
"of which the largest is: \(.un[-1])",
(select(\$base != 10) | "or expressed in base \(\$base): \(.un[-1]| tobase(\$base))")
) ;

(10, 7) as \$base
| undulating(\$base; 600), ""```
Output:

Essentially as for Wren.

## Julia

```using Primes

""" An undulating number is an integer which has the digit form ABABAB... """
struct UndulatingInteger
ubase::Int
min_digits::Int
end

""" Iterate undulating numbers """
function Base.iterate(u::UndulatingInteger, state = (1, 0, u.min_digits))
a, b, n = state
i = foldl((i, j) -> u.ubase * i + (iseven(j) ? b : a), 1:n, init = 0)
b += 1
if b == a
b += 1
end
if b >= u.ubase
b = 0
a += 1
if a >= u.ubase
a = 1
n += 1
end
end
return i, (a, b, n)
end

""" Run tests on the sequence in a given base `ubase` """
function test_undulating(ubase)
println("Three digit undulating numbers in base \$ubase:")
for (i, n) in enumerate(UndulatingInteger(ubase, 3))
n >= ubase^3 - 1 && break
print(lpad(n, 5), i % 9 == 0 ? "\n" : " ")
end
println("\nFour digit undulating numbers in base \$ubase:")
for (i, n) in enumerate(UndulatingInteger(ubase, 4))
n >= ubase^4 - 1 && break
print(lpad(n, 5), i % 9 == 0 ? "\n" : " ")
end
println("\nThree digit undulating numbers in base \$ubase which are primes:")
for (i, n) in enumerate(Iterators.filter(isprime, UndulatingInteger(ubase, 3)))
n >= ubase^3 - 1 && break
print(n, i % 20 == 0 ? "\n" : " ")
end

lastn = 0
for (i, n) in enumerate(UndulatingInteger(ubase, 3))
if i == 600
print("\n\nThe 600th undulating number in base \$ubase is \$n.")
elseif n > 2^53
print("\n\nNumber of undulating numbers in base \$ubase less than 2^53 is ",
i - 1, "\n   with the largest such number \$lastn.\n\n")
break
end
lastn = n
end
end

test_undulating(10)
test_undulating(7)
```
Output:
```Three digit undulating numbers in base 10:
101   121   131   141   151   161   171   181   191
202   212   232   242   252   262   272   282   292
303   313   323   343   353   363   373   383   393
404   414   424   434   454   464   474   484   494
505   515   525   535   545   565   575   585   595
606   616   626   636   646   656   676   686   696
707   717   727   737   747   757   767   787   797
808   818   828   838   848   858   868   878   898
909   919   929   939   949   959   969   979   989

Four digit undulating numbers in base 10:
1010  1212  1313  1414  1515  1616  1717  1818  1919
2020  2121  2323  2424  2525  2626  2727  2828  2929
3030  3131  3232  3434  3535  3636  3737  3838  3939
4040  4141  4242  4343  4545  4646  4747  4848  4949
5050  5151  5252  5353  5454  5656  5757  5858  5959
6060  6161  6262  6363  6464  6565  6767  6868  6969
7070  7171  7272  7373  7474  7575  7676  7878  7979
8080  8181  8282  8383  8484  8585  8686  8787  8989
9090  9191  9292  9393  9494  9595  9696  9797  9898

Three digit undulating numbers in base 10 which are primes:
101 131 151 181 191 313 353 373 383 727 757 787 797 919 929

The 600th undulating number in base 10 is 4646464646.

Number of undulating numbers in base 10 less than 2^53 is 1125
with the largest such number 8989898989898989.

Three digit undulating numbers in base 7:
50    64    71    78    85    92   100   107   121
128   135   142   150   157   164   178   185   192
200   207   214   221   235   242   250   257   264
271   278   292   300   307   314   321   328   335

Four digit undulating numbers in base 7:
350   450   500   550   600   650   700   750   850
900   950  1000  1050  1100  1150  1250  1300  1350
1400  1450  1500  1550  1650  1700  1750  1800  1850
1900  1950  2050  2100  2150  2200  2250  2300  2350

Three digit undulating numbers in base 7 which are primes:
71 107 157 257 271 307

The 600th undulating number in base 7 is 8074217422972642.

Number of undulating numbers in base 7 less than 2^53 is 603
with the largest such number 8786648372058464.
```

## Nim

```import std/[algorithm, math, strutils]

func isPrime(n: Natural): bool =
## Return true if "n" is prime.
if n < 2: return false
if (n and 1) == 0: return n == 2
if n mod 3 == 0: return n == 3
var k = 5
var delta = 2
while k * k <= n:
if n mod k == 0: return false
inc k, delta
delta = 6 - delta
result = true

func toBase(n, base: Natural): string =
## Return the string representation of "n" in given base.
assert base in 2..10
var n = n
var s: seq[int]
if n == 0: return "0"
while n != 0:
n = n div base
result = reversed(s).join()

func buildNumber (a, b, n, base: int): int =
## Build an undulating number of length "n" in
## given base, returning its value in base 10.
var digits = [a, b]
var idx = 0
for i in 1..n:
result = base * result + digits[idx]
idx = 1 - idx

iterator undulatingNumbers(nStart, nEnd: Positive; base = 10): int =
## Yield the successive undulating numbers in given base (expressed in
## base 10), starting with "nStart" digits and ending with "nEnd" digits.
assert nStart >= 3, "need at least three digits."
var n = nStart
while n <= nEnd :
for a in 1..<base:
for b in 0..<base:
if b == a: continue
yield buildNumber(a, b, n, base)
inc n

### Task - Part 1 ###

echo "Three digits undulating numbers in base 10:"
var count = 0
for unum in undulatingNumbers(3, 3):
inc count
stdout.write unum
stdout.write if count mod 9 == 0: '\n' else: ' '

### Task - Part 2 ###

echo "\nFour digits undulating numbers in base 10:"
count = 0
for unum in undulatingNumbers(4, 4):
inc count
stdout.write unum
stdout.write if count mod 9 == 0: '\n' else: ' '

### Task - Part 3 ###

echo "\nThree digits undulating numbers in base 10 which are primes:"
count = 0
for unum in undulatingNumbers(3, 3):
if unum.isPrime:
inc count
stdout.write unum
stdout.write if count == 15: '\n' else: ' '

### Task - Part 4 ###

count = 0
for unum in undulatingNumbers(3, 10):
inc count
if count == 600:
echo "\nThe 600th undulating number in base 10 is ", unum
break

### Task - Part 5 ###

const N = 2^53
count = (D10 - 3) * 81    # For each number of digits, there are 9 × 9 undulating numbers.
count = (D10 - 3) * 81
var last: int
for unum in undulatingNumbers(D10, D10):
if unum < N:
inc count
last = unum
else: break
echo "\nNumber of undulating numbers in base 10 less than 2^53: ", count
echo "The last undulating number in base 10 less than 2^53 is ", last

### Bonus - Part 1 ###
echo "\nThree digits undulating numbers in base 7 (shown in base 10):"
count = 0
for unum in undulatingNumbers(3, 3, 7):
inc count
stdout.write align(\$unum, 3)
stdout.write if count mod 9 == 0: '\n' else: ' '

### Bonus - Part 2 ###
echo "\nFour digits undulating numbers in base 7 (shown in base 10):"
count = 0
for unum in undulatingNumbers(4, 4, 7):
inc count
stdout.write align(\$unum, 4)
stdout.write if count mod 9 == 0: '\n' else: ' '

### Bonus - Part 3 ###

echo "\nThree digits undulating numbers in base 7 which are primes:"
count = 0
for unum in undulatingNumbers(3, 3, 7):
if unum.isPrime:
inc count
stdout.write unum
stdout.write if count == 6: '\n' else: ' '

### Bonus - Part 4 ###

count = 0
for unum in undulatingNumbers(3, 19, 7):
inc count
if count == 600:
echo "\nThe 600th undulating number in base 7 is ", unum.toBase(7)
echo "which is ", unum, " in base 10."
break

### Bonus - Part 5 ###

const D7 = int(ln(N.toFloat) / ln(7.0)) + 1
count = (D7 - 3) * 36   # For each number of digits, there are 6 × 6 undulating numbers.
last = 0
for unum in undulatingNumbers(D7, D7, 7):
if unum < N:
inc count
last = unum
else: break
echo "\nNumber of undulating numbers in base 7 less than 2^53: ", count
echo "The last undulating number in base 7 less than 2^53 is ", last.toBase(7)
echo "which is ", last, " in base 10."
```
Output:
```Three digits undulating numbers in base 10:
101 121 131 141 151 161 171 181 191
202 212 232 242 252 262 272 282 292
303 313 323 343 353 363 373 383 393
404 414 424 434 454 464 474 484 494
505 515 525 535 545 565 575 585 595
606 616 626 636 646 656 676 686 696
707 717 727 737 747 757 767 787 797
808 818 828 838 848 858 868 878 898
909 919 929 939 949 959 969 979 989

Four digits undulating numbers in base 10:
1010 1212 1313 1414 1515 1616 1717 1818 1919
2020 2121 2323 2424 2525 2626 2727 2828 2929
3030 3131 3232 3434 3535 3636 3737 3838 3939
4040 4141 4242 4343 4545 4646 4747 4848 4949
5050 5151 5252 5353 5454 5656 5757 5858 5959
6060 6161 6262 6363 6464 6565 6767 6868 6969
7070 7171 7272 7373 7474 7575 7676 7878 7979
8080 8181 8282 8383 8484 8585 8686 8787 8989
9090 9191 9292 9393 9494 9595 9696 9797 9898

Three digits undulating numbers in base 10 which are primes:
101 131 151 181 191 313 353 373 383 727 757 787 797 919 929

The 600th undulating number in base 10 is 4646464646

Number of undulating numbers in base 10 less than 2^53: 1125
The last undulating number in base 10 less than 2^53 is 8989898989898989

Three digits undulating numbers in base 7 (shown in base 10):
50  64  71  78  85  92 100 107 121
128 135 142 150 157 164 178 185 192
200 207 214 221 235 242 250 257 264
271 278 292 300 307 314 321 328 335

Four digits undulating numbers in base 7 (shown in base 10):
350  450  500  550  600  650  700  750  850
900  950 1000 1050 1100 1150 1250 1300 1350
1400 1450 1500 1550 1650 1700 1750 1800 1850
1900 1950 2050 2100 2150 2200 2250 2300 2350

Three digits undulating numbers in base 7 which are primes:
71 107 157 257 271 307

The 600th undulating number in base 7 is 4646464646464646464
which is 8074217422972642 in base 10.

Number of undulating numbers in base 7 less than 2^53: 603
The last undulating number in base 7 less than 2^53 is 5252525252525252525
which is 8786648372058464 in base 10.
```

## Perl

Library: ntheory
```use v5.36;
use bigint;
use experimental <builtin for_list>;
use ntheory <is_prime vecfirstidx>;

sub X     (\$a, \$b) { my @c; for my \$aa (0..\$a) { for my \$bb (0..\$b) { push @c, \$aa, \$bb } } @c }
sub table (\$c, @V) { my \$t = \$c * (my \$w = 6); ( sprintf( ('%'.\$w.'d')x@V, @V) ) =~ s/.{1,\$t}\K/\n/gr }

my \$max = 2**53;

my(@pairs,@U);
for my(\$i,\$j) ( X(9,9) ) { push @pairs, \$i . \$j unless \$i == 0 || \$i == \$j }
for my \$l (3..length \$max) {
if (0 == \$l%2) { push @U, "\$_"x( \$l   /2) for @pairs }
else           { push @U, "\$_"x((\$l+1)/2) and chop \$U[-1] for @pairs }
}

say "All 3 digit undulating numbers:"; say table 9, grep { 3 == length \$_ } @U;
say "All 3 digit undulating primes:";  say table 9, grep { 3 == length \$_ and is_prime \$_ } @U;
say "All 4 digit undulating numbers:"; say table 9, grep { 4 == length \$_ } @U;

my \$fmt = "%34s: %d\n";
printf \$fmt, 'The 600th undulating number is', \$U[599];
printf \$fmt, 'Undulating numbers less than 2**53', (my \$i = vecfirstidx { \$_ >= \$max } @U);
printf \$fmt, 'That number is', \$U[\$i-1];
```
Output:
```All 3 digit undulating numbers:
101   121   131   141   151   161   171   181   191
202   212   232   242   252   262   272   282   292
303   313   323   343   353   363   373   383   393
404   414   424   434   454   464   474   484   494
505   515   525   535   545   565   575   585   595
606   616   626   636   646   656   676   686   696
707   717   727   737   747   757   767   787   797
808   818   828   838   848   858   868   878   898
909   919   929   939   949   959   969   979   989

All 3 digit undulating primes:
101   131   151   181   191   313   353   373   383
727   757   787   797   919   929

All 4 digit undulating numbers:
1010  1212  1313  1414  1515  1616  1717  1818  1919
2020  2121  2323  2424  2525  2626  2727  2828  2929
3030  3131  3232  3434  3535  3636  3737  3838  3939
4040  4141  4242  4343  4545  4646  4747  4848  4949
5050  5151  5252  5353  5454  5656  5757  5858  5959
6060  6161  6262  6363  6464  6565  6767  6868  6969
7070  7171  7272  7373  7474  7575  7676  7878  7979
8080  8181  8282  8383  8484  8585  8686  8787  8989
9090  9191  9292  9393  9494  9595  9696  9797  9898

The 600th undulating number is: 4646464646
Undulating numbers less than 2**53: 1125
That number is: 8989898989898989
```

## Phix

Translation of: Wren
```with javascript_semantics
for base in {10,7} do
integer n = 600, mpow = 53, bsquare = base * base
atom limit = power(2,53)-1
sequence u3 = {}, u4 = {}
for a=1 to base-1 do
for b=0 to base-1 do
if b!=a then
integer u = a * base + b,
v = u * base + a,
w = v * base + b
u3 &= v
u4 &= w
end if
end for
end for
for d,u in {{u3,"%3d"},{u4,"%4d"}} do
printf(1,"All %d digit undulating numbers in base %d:\n%s\n",
{d+2,base,join_by(u[1],1,9,"  ",fmt:=u[2])})
end for
printf(1,"All 3 digit undulating numbers which are primes in base %d:\n%s\n",
{base,join_by(filter(u3,is_prime),1,10,"  ",fmt:="%3d")})
sequence un = u3 & u4
integer start = 1, done = false
while not done do
integer finish = length(un)
for i=start to finish do
atom u = un[i] * bsquare + remainder(un[i],bsquare)
if u>limit then done = true; exit end if
un &= u
end for
start = finish+1
end while
printf(1,"The %,d%s undulating number in base %d is: %,d\n", {n,ord(n),base,un[n]})
if base!=10 then printf(1,"or expressed in base %d : %,a\n", {base,{base,un[n]}}) end if
printf(1,"\nTotal number of undulating numbers in base %d < 2^%d = %,d\n",{base,mpow,length(un)})
printf(1,"of which the largest is: %,d\n", un[\$])
if base!=10 then printf(1,"or expressed in base %d : %,a\n", {base,{base,un[\$]}}) end if
printf(1,"\n")
end for
```

Output same as Wren, except for "⁵³"->"^53". Note that comma-fill on %a under p2js needs a trivial fix ("df".indexOf->"aAdf".indexOf)/1.0.3 or later.

## Raku

Translation of: Wren
```# 20230602 Raku programming solution

sub undulating (\$base, \n) {
my \limit = 2**(my \mpow = 53) - 1;
my (\bsquare,@u3,@u4) = \$base*\$base;
for 1..^\$base X 0..^\$base -> (\a,\b) {
next if b == a;
@u3.push(a * bsquare + b * \$base + a);
@u4.push((my \v = a * \$base + b) * bsquare + v)
}
say "\nAll 3 digit undulating numbers in base \$base:";
.fmt('%3d').say for @u3.rotor: 9;
say "\nAll 4 digit undulating numbers in base \$base:";
.fmt('%4d').say for @u4.rotor: 9;
say "\nAll 3 digit undulating numbers which are primes in base \$base:";
my @primes = @u3.grep: *.is-prime;
.fmt('%3d').say for @primes.rotor: 10, :partial;
my \unc = (my @un = @u3.append: @u4).elems;
my (\$j, \$done) = 0, False;
loop {
for 0..^unc {
my  \u = @un[\$j * unc + \$_] * bsquare + @un[\$j * unc + \$_] % bsquare;
u > limit ?? ( \$done = True and last ) !! ( @un.push: u );
}
\$done ?? ( last ) !! \$j++
}
say "\nThe {n} undulating number in \$base \$base is: @un[n-1]";
say "or expressed in base \$base : {@un[n-1].base(\$base)}" unless \$base == 10;
say "\nTotal number of undulating numbers in base \$base < 2**{mpow} = {+@un}";
say "of which the largest is: ", @un[*-1];
say "or expressed in base \$base : {@un[*-1].base(\$base)}" unless \$base == 10;
}

undulating \$_, 600 for 10, 7;
```

You may Try it online!

## RPL

Once found the direct formula, the code is fairly simple to write.

Works with: HP version 49
```« 1 - → n
« n 81 MOD 9 / FLOOR 1 +    @ A = mod(N-1,81)//9 + 1
n 9 MOD DUP2 ≤ +          @ B = mod(N-1,9) + (A ≤ mod(N-1,9))
→STR +
n 81 / FLOOR 3 + SWAP
WHILE DUP2 SIZE > REPEAT DUP + END
1 ROT SUB STR→            @ reduce # of digits to N//81 + 3
» » 'N→UND' STO

« « n N→UND » 'n' 1 81 1 SEQ
« n N→UND » 'n' 82 162 1 SEQ
{ }
1 4 PICK SIZE FOR j
PICK3 j GET
IF DUP ISPRIME? THEN + ELSE DROP END
NEXT
625 N→UND
2 53 ^ 0
DO 1 + UNTIL DUP2 N→UND < END
1 - NIP DUP N→UND
```
Output:
```6: { 101 121 131 141 151 161 171 181 191 202 212 232 242 252 262 272 282 292 303 313 323 343 353 363 373 383 393 404 414 424 434 454 464 474 484 494 505 515 525 535 545 565 575 585 595 606 616 626 636 646 656 676 686 696 707 717 727 737 747 757 767 787 797 808 818 828 838 848 858 868 878 898 909 919 929 939 949 959 969 979 989 }
5: { 1010 1212 1313 1414 1515 1616 1717 1818 1919 2020 2121 2323 2424 2525 2626 2727 2828 2929 3030 3131 3232 3434 3535 3636 3737 3838 3939 4040 4141 4242 4343 4545 4646 4747 4848 4949 5050 5151 5252 5353 5454 5656 5757 5858 5959 6060 6161 6262 6363 6464 6565 6767 6868 6969 7070 7171 7272 7373 7474 7575 7676 7878 7979 8080 8181 8282 8383 8484 8585 8686 8787 8989 9090 9191 9292 9393 9494 9595 9696 9797 9898 }
4: { 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 }
3: 4646464646
2: 1125
1: 8989898989898989
```

## Rust

```// [dependencies]

fn is_prime(n: u64) -> bool {
if n < 2 {
return false;
}
if n % 2 == 0 {
return n == 2;
}
if n % 3 == 0 {
return n == 3;
}
let mut p: u64 = 5;
while p * p <= n {
if n % p == 0 {
return false;
}
p += 2;
if n % p == 0 {
return false;
}
p += 4;
}
true
}

fn undulating_numbers(base: u64) -> impl std::iter::Iterator<Item = u64> {
let mut a = 1;
let mut b = 0;
let mut digits = 3;
std::iter::from_fn(move || {
let mut n = 0;
for d in 0..digits {
n = n * base + if d % 2 == 0 { a } else { b };
}
b += 1;
if a == b {
b += 1;
}
if b == base {
a += 1;
b = 0;
if a == base {
a = 1;
digits += 1;
}
}
Some(n)
})
}

fn undulating(base: u64) {
let mut count = 0;
let limit3 = base * base * base;
let limit4 = base * limit3;
let mut u3 = Vec::new();
let mut u4 = Vec::new();
let mut umax = 0;
let mut u600 = 0;

for n in undulating_numbers(base).take_while(|x| *x < 1u64 << 53) {
if n < limit3 {
u3.push(n);
} else if n < limit4 {
u4.push(n);
}
count += 1;
umax = n;
if count == 600 {
u600 = n;
}
}

println!("3-digit undulating numbers in base {}:", base);
for (i, n) in u3.iter().enumerate() {
print!("{:3}{}", n, if (i + 1) % 9 == 0 { '\n' } else { ' ' });
}

println!("\n4-digit undulating numbers in base {}:", base);
for (i, n) in u4.iter().enumerate() {
print!("{:4}{}", n, if (i + 1) % 9 == 0 { '\n' } else { ' ' });
}

println!(
"\n3-digit undulating numbers in base {} which are prime:",
base
);
for n in u3 {
if is_prime(n) {
print!("{} ", n);
}
}
println!();

print!("\nThe 600th undulating number in base {} is {}", base, u600);
if base != 10 {
print!(
"\nor expressed in base {}: {}",
base,
);
}
println!(".");

print!(
"\nTotal number of undulating numbers < 2^53 in base {}: {}\nof which the largest is {}",
base, count, umax
);
if base != 10 {
print!(
"\nor expressed in base {}: {}",
base,
);
}
println!(".");
}

fn main() {
undulating(10);
println!();
undulating(7);
}
```
Output:
```3-digit undulating numbers in base 10:
101 121 131 141 151 161 171 181 191
202 212 232 242 252 262 272 282 292
303 313 323 343 353 363 373 383 393
404 414 424 434 454 464 474 484 494
505 515 525 535 545 565 575 585 595
606 616 626 636 646 656 676 686 696
707 717 727 737 747 757 767 787 797
808 818 828 838 848 858 868 878 898
909 919 929 939 949 959 969 979 989

4-digit undulating numbers in base 10:
1010 1212 1313 1414 1515 1616 1717 1818 1919
2020 2121 2323 2424 2525 2626 2727 2828 2929
3030 3131 3232 3434 3535 3636 3737 3838 3939
4040 4141 4242 4343 4545 4646 4747 4848 4949
5050 5151 5252 5353 5454 5656 5757 5858 5959
6060 6161 6262 6363 6464 6565 6767 6868 6969
7070 7171 7272 7373 7474 7575 7676 7878 7979
8080 8181 8282 8383 8484 8585 8686 8787 8989
9090 9191 9292 9393 9494 9595 9696 9797 9898

3-digit undulating numbers in base 10 which are prime:
101 131 151 181 191 313 353 373 383 727 757 787 797 919 929

The 600th undulating number in base 10 is 4646464646.

Total number of undulating numbers < 2^53 in base 10: 1125
of which the largest is 8989898989898989.

3-digit undulating numbers in base 7:
50  64  71  78  85  92 100 107 121
128 135 142 150 157 164 178 185 192
200 207 214 221 235 242 250 257 264
271 278 292 300 307 314 321 328 335

4-digit undulating numbers in base 7:
350  450  500  550  600  650  700  750  850
900  950 1000 1050 1100 1150 1250 1300 1350
1400 1450 1500 1550 1650 1700 1750 1800 1850
1900 1950 2050 2100 2150 2200 2250 2300 2350

3-digit undulating numbers in base 7 which are prime:
71 107 157 257 271 307

The 600th undulating number in base 7 is 8074217422972642
or expressed in base 7: 4646464646464646464.

Total number of undulating numbers < 2^53 in base 7: 603
of which the largest is 8786648372058464
or expressed in base 7: 5252525252525252525.
```

## Uiua

Works with: Uiua version 0.10.0-dev.1

More of a script than a program, and only for base 10 as there's no built-in support for other radixes in Uiua yet.

```# Generate all distinct two digit numbers as strings.
≡(°⋕)▽≡(≠0◿11).↘10⇡100
&p≡(⋕⊂⟜⊢) . &pf "Three digits: "
&p≡(⋕⊂.) . &pf "Four digits: "
≡(⋕⊂⟜⊢) . &pf "Three digit primes: "
# Primes by sieve.
⇌◌⍢(▽≠0◿⊃⊢(.↘1)⟜(⊂⊢)|>0⧻)↘2⇡1000[]
&p▽:⟜(/+⍉⊞⌕):
# Collect all 3 and 4 digit numbers.
⊂⊃≡(⋕⊂⟜⊢)≡(⋕⊂.)
# Double-up last two digits of each and remove the ensuing duplicates.
F ← ◴⊂⟜≡(⋕⊂⟜(↙¯2)°⋕)
# Repeat until length is okay.
⍢(F|<600⧻)
&p⊡599 . &pf "Six hundredth: "
# Repeat until last is larger than target.
⍢(F|<ⁿ53 2⊡¯1)
&p⧻.▽<ⁿ53 2. &pf "Count less than 2^53: "
&p⊡¯1⊏⍏. &pf "Last one less than 2^53: "```
Output:
```Three digits: [101 121 131 141 151 161 171 181 191 202 212 232 242 252 262 272 282 292 303 313 323 343 353 363 373 383 393 404 414 424 434 454 464 474 484 494 505 515 525 535 545 565 575 585 595 606 616 626 636 646 656 676 686 696 707 717 727 737 747 757 767 787 797 808 818 828 838 848 858 868 878 898 909 919 929 939 949 959 969 979 989]
Four digits: [1010 1212 1313 1414 1515 1616 1717 1818 1919 2020 2121 2323 2424 2525 2626 2727 2828 2929 3030 3131 3232 3434 3535 3636 3737 3838 3939 4040 4141 4242 4343 4545 4646 4747 4848 4949 5050 5151 5252 5353 5454 5656 5757 5858 5959 6060 6161 6262 6363 6464 6565 6767 6868 6969 7070 7171 7272 7373 7474 7575 7676 7878 7979 8080 8181 8282 8383 8484 8585 8686 8787 8989 9090 9191 9292 9393 9494 9595 9696 9797 9898]
Three digit primes: [101 131 151 181 191 313 353 373 383 727 757 787 797 919 929]
Six hundredth: 4646464646
Count less than 2^53: 1125
Last one less than 2^53: 8989898989898989
```

## Wren

Library: Wren-fmt
Library: Wren-math
```import "./fmt" for Fmt, Conv
import "./math" for Int

var undulating = Fn.new { |base, n|
var mpow = 53
var limit = 2.pow(mpow) - 1
var u3 = []
var u4 = []
var bsquare = base * base
for (a in 1...base) {
for (b in 0...base) {
if (b == a) continue
var u = a * bsquare + b * base + a
var v = a * base + b
}
}
System.print("All 3 digit undulating numbers in base %(base):")
Fmt.tprint("\$3d ", u3, 9)
System.print("\nAll 4 digit undulating numbers in base %(base):")
Fmt.tprint("\$4d ", u4, 9)
System.print("\nAll 3 digit undulating numbers which are primes in base %(base):")
var primes = []
for (u in u3) {
if (u % 2 == 1 && u % 5 != 0 && Int.isPrime(u)) primes.add(u)
}
Fmt.tprint("\$3d ", primes, 10)
var un = u3 + u4
var unc = un.count
var j = 0
var done = false
while (true) {
for (i in 0...unc) {
var u = un[j * unc + i] * bsquare + un[j * unc + i] % bsquare
if (u > limit) {
done = true
break
}
}
if (done) break
j = j + 1
}
Fmt.print("\nThe \$,r undulating number in base \$d is: \$,d", n, base, un[n-1])
if (base != 10) Fmt.print("or expressed in base \$d : \$,s", base, Conv.itoa(un[n-1], base))
Fmt.print("\nTotal number of undulating numbers in base \$d < 2\$S = \$,d", base, mpow, un.count)
Fmt.print("of which the largest is: \$,d", un[-1])
if (base != 10) Fmt.print("or expressed in base \$d : \$,s", base, Conv.itoa(un[-1], base))
System.print()
}

for (base in [10, 7]) undulating.call(base, 600)
```
Output:
```All 3 digit undulating numbers in base 10:
101  121  131  141  151  161  171  181  191
202  212  232  242  252  262  272  282  292
303  313  323  343  353  363  373  383  393
404  414  424  434  454  464  474  484  494
505  515  525  535  545  565  575  585  595
606  616  626  636  646  656  676  686  696
707  717  727  737  747  757  767  787  797
808  818  828  838  848  858  868  878  898
909  919  929  939  949  959  969  979  989

All 4 digit undulating numbers in base 10:
1010  1212  1313  1414  1515  1616  1717  1818  1919
2020  2121  2323  2424  2525  2626  2727  2828  2929
3030  3131  3232  3434  3535  3636  3737  3838  3939
4040  4141  4242  4343  4545  4646  4747  4848  4949
5050  5151  5252  5353  5454  5656  5757  5858  5959
6060  6161  6262  6363  6464  6565  6767  6868  6969
7070  7171  7272  7373  7474  7575  7676  7878  7979
8080  8181  8282  8383  8484  8585  8686  8787  8989
9090  9191  9292  9393  9494  9595  9696  9797  9898

All 3 digit undulating numbers which are primes in base 10:
101  131  151  181  191  313  353  373  383  727
757  787  797  919  929

The 600th undulating number in base 10 is: 4,646,464,646

Total number of undulating numbers in base 10 < 2⁵³ = 1,125
of which the largest is: 8,989,898,989,898,989

All 3 digit undulating numbers in base 7:
50   64   71   78   85   92  100  107  121
128  135  142  150  157  164  178  185  192
200  207  214  221  235  242  250  257  264
271  278  292  300  307  314  321  328  335

All 4 digit undulating numbers in base 7:
350   450   500   550   600   650   700   750   850
900   950  1000  1050  1100  1150  1250  1300  1350
1400  1450  1500  1550  1650  1700  1750  1800  1850
1900  1950  2050  2100  2150  2200  2250  2300  2350

All 3 digit undulating numbers which are primes in base 7:
71  107  157  257  271  307

The 600th undulating number in base 7 is: 8,074,217,422,972,642
or expressed in base 7 : 4,646,464,646,464,646,464

Total number of undulating numbers in base 7 < 2⁵³ = 603
of which the largest is: 8,786,648,372,058,464
or expressed in base 7 : 5,252,525,252,525,252,525
```