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Added base 10 tetration score
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Riker
Riker
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AWK -, 100 bytes --, Score ≈ 10^(5e80) ≈ 10↑↑2.280320629

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

AWK - 100 bytes -- Score ≈ 10^(5e80) ≈ 10↑↑2.280320629

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

AWK, 100 bytes, Score ≈ 10^(5e80) ≈ 10↑↑2.280320629

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

Added base 10 tetration score
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edit approved May 19, 2017 at 0:27
Simply Beautiful Art
edit approved May 19, 2017 at 0:27
Simply Beautiful Art
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AWK - 100 bytes -- Score ~≈ 10^(5e80) ≈ 10↑↑2.280320629

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

AWK - 100 bytes -- Score ~ 10^(5e80)

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

AWK - 100 bytes -- Score ≈ 10^(5e80) ≈ 10↑↑2.280320629

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

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Robert Benson
Robert Benson
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AWK - 100 bytes -- Score ~ 10^(5e80)

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.