# Four is the number of letters in the ...

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Four is the number of letters in the ...
You are encouraged to solve this task according to the task description, using any language you may know.

The     Four is ...     sequence is based on the counting of the number of letters in the words of the (never─ending) sentence:

```  Four is the number of letters in the first word of this sentence, two in the second,
three in the third, six in the fourth, two in the fifth, seven in the sixth, ···
```

Definitions and directives
•   English is to be used in spelling numbers.
•   Letters   are defined as the upper─ and lowercase letters in the Latin alphabet   (A──►Z   and   a──►z).
•   Commas are not counted,   nor are hyphens (dashes or minus signs).
•   twenty─three   has eleven letters.
•   twenty─three   is considered one word   (which is hyphenated).
•   no   and   words are to be used when spelling a (English) word for a number.
•   The American version of numbers will be used here in this task   (as opposed to the British version).
```         2,000,000,000   is two billion,   not   two milliard.
```

•   Write a driver (invoking routine) and a function (subroutine/routine···) that returns the sequence (for any positive integer) of the number of letters in the first   N   words in the never─ending sentence.   For instance, the portion of the never─ending sentence shown above (2nd sentence of this task's preamble),   the sequence would be:
```         4  2  3  6  2  7
```
•   Only construct as much as is needed for the never─ending sentence.
•   Write a driver (invoking routine) to show the number of letters in the   Nth   word,   as well as   showing the   Nth   word itself.
•   After each test case, show the total number of characters   (including blanks, commas, and punctuation)   of the sentence that was constructed.
•   Show all output here.

Test cases
``` Display the first  201  numbers in the sequence   (and the total number of characters in the sentence).
Display the number of letters  (and the word itself)  of the       1,000th  word.
Display the number of letters  (and the word itself)  of the      10,000th  word.
Display the number of letters  (and the word itself)  of the     100,000th  word.
Display the number of letters  (and the word itself)  of the   1,000,000th  word.
Display the number of letters  (and the word itself)  of the  10,000,000th  word  (optional).
```

Also see

## C

Library: GLib
```#include <ctype.h>
#include <locale.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdint.h>
#include <glib.h>

typedef uint64_t integer;

typedef struct number_names_tag {
const char* cardinal;
const char* ordinal;
} number_names;

const number_names small[] = {
{ "zero", "zeroth" }, { "one", "first" }, { "two", "second" },
{ "three", "third" }, { "four", "fourth" }, { "five", "fifth" },
{ "six", "sixth" }, { "seven", "seventh" }, { "eight", "eighth" },
{ "nine", "ninth" }, { "ten", "tenth" }, { "eleven", "eleventh" },
{ "twelve", "twelfth" }, { "thirteen", "thirteenth" },
{ "fourteen", "fourteenth" }, { "fifteen", "fifteenth" },
{ "sixteen", "sixteenth" }, { "seventeen", "seventeenth" },
{ "eighteen", "eighteenth" }, { "nineteen", "nineteenth" }
};

const number_names tens[] = {
{ "twenty", "twentieth" }, { "thirty", "thirtieth" },
{ "forty", "fortieth" }, { "fifty", "fiftieth" },
{ "sixty", "sixtieth" }, { "seventy", "seventieth" },
{ "eighty", "eightieth" }, { "ninety", "ninetieth" }
};

typedef struct named_number_tag {
const char* cardinal;
const char* ordinal;
integer number;
} named_number;

const named_number named_numbers[] = {
{ "hundred", "hundredth", 100 },
{ "thousand", "thousandth", 1000 },
{ "million", "millionth", 1000000 },
{ "billion", "biliionth", 1000000000 },
{ "trillion", "trillionth", 1000000000000 },
{ "quintillion", "quintillionth", 1000000000000000000ULL }
};

const char* get_small_name(const number_names* n, bool ordinal) {
return ordinal ? n->ordinal : n->cardinal;
}

const char* get_big_name(const named_number* n, bool ordinal) {
return ordinal ? n->ordinal : n->cardinal;
}

const named_number* get_named_number(integer n) {
const size_t names_len = sizeof(named_numbers)/sizeof(named_numbers[0]);
for (size_t i = 0; i + 1 < names_len; ++i) {
if (n < named_numbers[i + 1].number)
return &named_numbers[i];
}
return &named_numbers[names_len - 1];
}

typedef struct word_tag {
size_t offset;
size_t length;
} word_t;

typedef struct word_list_tag {
GArray* words;
GString* str;
} word_list;

void word_list_create(word_list* words) {
words->words = g_array_new(FALSE, FALSE, sizeof(word_t));
words->str = g_string_new(NULL);
}

void word_list_destroy(word_list* words) {
g_string_free(words->str, TRUE);
g_array_free(words->words, TRUE);
}

void word_list_clear(word_list* words) {
g_string_truncate(words->str, 0);
g_array_set_size(words->words, 0);
}

void word_list_append(word_list* words, const char* str) {
size_t offset = words->str->len;
size_t len = strlen(str);
g_string_append_len(words->str, str, len);
word_t word;
word.offset = offset;
word.length = len;
g_array_append_val(words->words, word);
}

word_t* word_list_get(word_list* words, size_t index) {
return &g_array_index(words->words, word_t, index);
}

void word_list_extend(word_list* words, const char* str) {
word_t* word = word_list_get(words, words->words->len - 1);
size_t len = strlen(str);
word->length += len;
g_string_append_len(words->str, str, len);
}

size_t append_number_name(word_list* words, integer n, bool ordinal) {
size_t count = 0;
if (n < 20) {
word_list_append(words, get_small_name(&small[n], ordinal));
count = 1;
} else if (n < 100) {
if (n % 10 == 0) {
word_list_append(words, get_small_name(&tens[n/10 - 2], ordinal));
} else {
word_list_append(words, get_small_name(&tens[n/10 - 2], false));
word_list_extend(words, "-");
word_list_extend(words, get_small_name(&small[n % 10], ordinal));
}
count = 1;
} else {
const named_number* num = get_named_number(n);
integer p = num->number;
count += append_number_name(words, n/p, false);
if (n % p == 0) {
word_list_append(words, get_big_name(num, ordinal));
++count;
} else {
word_list_append(words, get_big_name(num, false));
++count;
count += append_number_name(words, n % p, ordinal);
}
}
return count;
}

size_t count_letters(word_list* words, size_t index) {
const word_t* word = word_list_get(words, index);
size_t letters = 0;
const char* s = words->str->str + word->offset;
for (size_t i = 0, n = word->length; i < n; ++i) {
if (isalpha((unsigned char)s[i]))
++letters;
}
return letters;
}

void sentence(word_list* result, size_t count) {
static const char* words[] = {
"Four", "is", "the", "number", "of", "letters", "in", "the",
"first", "word", "of", "this", "sentence,"
};
word_list_clear(result);
size_t n = sizeof(words)/sizeof(words[0]);
for (size_t i = 0; i < n; ++i)
word_list_append(result, words[i]);
for (size_t i = 1; count > n; ++i) {
n += append_number_name(result, count_letters(result, i), false);
word_list_append(result, "in");
word_list_append(result, "the");
n += 2;
n += append_number_name(result, i + 1, true);
// Append a comma to the final word
word_list_extend(result, ",");
}
}

size_t sentence_length(const word_list* words) {
size_t n = words->words->len;
if (n == 0)
return 0;
return words->str->len + n - 1;
}

int main() {
setlocale(LC_ALL, "");
size_t n = 201;
word_list result = { 0 };
word_list_create(&result);
sentence(&result, n);
printf("Number of letters in first %'lu words in the sequence:\n", n);
for (size_t i = 0; i < n; ++i) {
if (i != 0)
printf("%c", i % 25 == 0 ? '\n' : ' ');
printf("%'2lu", count_letters(&result, i));
}
printf("\nSentence length: %'lu\n", sentence_length(&result));
for (n = 1000; n <= 10000000; n *= 10) {
sentence(&result, n);
const word_t* word = word_list_get(&result, n - 1);
const char* s = result.str->str + word->offset;
printf("The %'luth word is '%.*s' and has %lu letters. ", n,
(int)word->length, s, count_letters(&result, n - 1));
printf("Sentence length: %'lu\n" , sentence_length(&result));
}
word_list_destroy(&result);
return 0;
}
```
Output:
```Number of letters in first 201 words in the sequence:
4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
11
Sentence length: 1,203
The 1,000th word is 'in' and has 2 letters. Sentence length: 6,279
The 10,000th word is 'in' and has 2 letters. Sentence length: 64,140
The 100,000th word is 'one' and has 3 letters. Sentence length: 659,474
The 1,000,000th word is 'the' and has 3 letters. Sentence length: 7,113,621
The 10,000,000th word is 'thousand' and has 8 letters. Sentence length: 70,995,756
```

## C++

```#include <cctype>
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <string>
#include <vector>

struct number_names {
const char* cardinal;
const char* ordinal;
};

const number_names small[] = {
{ "zero", "zeroth" }, { "one", "first" }, { "two", "second" },
{ "three", "third" }, { "four", "fourth" }, { "five", "fifth" },
{ "six", "sixth" }, { "seven", "seventh" }, { "eight", "eighth" },
{ "nine", "ninth" }, { "ten", "tenth" }, { "eleven", "eleventh" },
{ "twelve", "twelfth" }, { "thirteen", "thirteenth" },
{ "fourteen", "fourteenth" }, { "fifteen", "fifteenth" },
{ "sixteen", "sixteenth" }, { "seventeen", "seventeenth" },
{ "eighteen", "eighteenth" }, { "nineteen", "nineteenth" }
};

const number_names tens[] = {
{ "twenty", "twentieth" }, { "thirty", "thirtieth" },
{ "forty", "fortieth" }, { "fifty", "fiftieth" },
{ "sixty", "sixtieth" }, { "seventy", "seventieth" },
{ "eighty", "eightieth" }, { "ninety", "ninetieth" }
};

struct named_number {
const char* cardinal;
const char* ordinal;
uint64_t number;
};

const named_number named_numbers[] = {
{ "hundred", "hundredth", 100 },
{ "thousand", "thousandth", 1000 },
{ "million", "millionth", 1000000 },
{ "billion", "biliionth", 1000000000 },
{ "trillion", "trillionth", 1000000000000 },
{ "quintillion", "quintillionth", 1000000000000000000ULL }
};

const char* get_name(const number_names& n, bool ordinal) {
return ordinal ? n.ordinal : n.cardinal;
}

const char* get_name(const named_number& n, bool ordinal) {
return ordinal ? n.ordinal : n.cardinal;
}

const named_number& get_named_number(uint64_t n) {
constexpr size_t names_len = std::size(named_numbers);
for (size_t i = 0; i + 1 < names_len; ++i) {
if (n < named_numbers[i + 1].number)
return named_numbers[i];
}
return named_numbers[names_len - 1];
}

size_t append_number_name(std::vector<std::string>& result, uint64_t n, bool ordinal) {
size_t count = 0;
if (n < 20) {
result.push_back(get_name(small[n], ordinal));
count = 1;
}
else if (n < 100) {
if (n % 10 == 0) {
result.push_back(get_name(tens[n/10 - 2], ordinal));
} else {
std::string name(get_name(tens[n/10 - 2], false));
name += "-";
name += get_name(small[n % 10], ordinal);
result.push_back(name);
}
count = 1;
} else {
const named_number& num = get_named_number(n);
uint64_t p = num.number;
count += append_number_name(result, n/p, false);
if (n % p == 0) {
result.push_back(get_name(num, ordinal));
++count;
} else {
result.push_back(get_name(num, false));
++count;
count += append_number_name(result, n % p, ordinal);
}
}
return count;
}

size_t count_letters(const std::string& str) {
size_t letters = 0;
for (size_t i = 0, n = str.size(); i < n; ++i) {
if (isalpha(static_cast<unsigned char>(str[i])))
++letters;
}
return letters;
}

std::vector<std::string> sentence(size_t count) {
static const char* words[] = {
"Four", "is", "the", "number", "of", "letters", "in", "the",
"first", "word", "of", "this", "sentence,"
};
std::vector<std::string> result;
result.reserve(count + 10);
size_t n = std::size(words);
for (size_t i = 0; i < n && i < count; ++i) {
result.push_back(words[i]);
}
for (size_t i = 1; count > n; ++i) {
n += append_number_name(result, count_letters(result[i]), false);
result.push_back("in");
result.push_back("the");
n += 2;
n += append_number_name(result, i + 1, true);
result.back() += ',';
}
return result;
}

size_t sentence_length(const std::vector<std::string>& words) {
size_t n = words.size();
if (n == 0)
return 0;
size_t length = n - 1;
for (size_t i = 0; i < n; ++i)
length += words[i].size();
return length;
}

int main() {
std::cout.imbue(std::locale(""));
size_t n = 201;
auto result = sentence(n);
std::cout << "Number of letters in first " << n << " words in the sequence:\n";
for (size_t i = 0; i < n; ++i) {
if (i != 0)
std::cout << (i % 25 == 0 ? '\n' : ' ');
std::cout << std::setw(2) << count_letters(result[i]);
}
std::cout << '\n';
std::cout << "Sentence length: " << sentence_length(result) << '\n';
for (n = 1000; n <= 10000000; n *= 10) {
result = sentence(n);
const std::string& word = result[n - 1];
std::cout << "The " << n << "th word is '" << word << "' and has "
<< count_letters(word) << " letters. ";
std::cout << "Sentence length: " << sentence_length(result) << '\n';
}
return 0;
}
```
Output:
```Number of letters in first 201 words in the sequence:
4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
11
Sentence length: 1,203
The 1,000th word is 'in' and has 2 letters. Sentence length: 6,279
The 10,000th word is 'in' and has 2 letters. Sentence length: 64,140
The 100,000th word is 'one' and has 3 letters. Sentence length: 659,474
The 1,000,000th word is 'the' and has 3 letters. Sentence length: 7,113,621
The 10,000,000th word is 'thousand' and has 8 letters. Sentence length: 70,995,756
```

## Go

This is a naive non-optimized implementation that stores each word of the sentence so far. It uses the `sayOrdinal` and `say` functions from the Spelling of ordinal numbers task (omitted from this listing).

```package main

import (
"fmt"
"strings"
"unicode"
)

func main() {
f := NewFourIsSeq()
fmt.Print("The lengths of the first 201 words are:")
for i := 1; i <= 201; i++ {
if i%25 == 1 {
fmt.Printf("\n%3d: ", i)
}
_, n := f.WordLen(i)
fmt.Printf(" %2d", n)
}
fmt.Println()
fmt.Println("Length of sentence so far:", f.TotalLength())
/* For debugging:
log.Println("sentence:", strings.Join(f.words, " "))
for i, w := range f.words {
log.Printf("%3d: %2d %q\n", i, countLetters(w), w)
}
log.Println(f.WordLen(2202))
log.Println("len(f.words):", len(f.words))
log.Println("sentence:", strings.Join(f.words, " "))
*/
for i := 1000; i <= 1e7; i *= 10 {
w, n := f.WordLen(i)
fmt.Printf("Word %8d is %q, with %d letters.", i, w, n)
fmt.Println("  Length of sentence so far:", f.TotalLength())
}
}

type FourIsSeq struct {
i     int      // index of last word processed
words []string // strings.Join(words," ") gives the sentence so far
}

func NewFourIsSeq() *FourIsSeq {
return &FourIsSeq{
//words: strings.Fields("Four is the number of letters in the first word of this sentence,"),
words: []string{
"Four", "is", "the", "number",
"of", "letters", "in", "the",
"first", "word", "of", "this", "sentence,",
},
}
}

// WordLen returns the w'th word and its length (only counting letters).
func (f *FourIsSeq) WordLen(w int) (string, int) {
for len(f.words) < w {
f.i++
n := countLetters(f.words[f.i])
ns := say(int64(n))
os := sayOrdinal(int64(f.i+1)) + ","
// append something like: "two in the second,"
f.words = append(f.words, strings.Fields(ns)...)
f.words = append(f.words, "in", "the")
f.words = append(f.words, strings.Fields(os)...)
}
word := f.words[w-1]
return word, countLetters(word)
}

// TotalLength returns the total number of characters (including blanks,
// commas, and punctuation) of the sentence so far constructed.
func (f FourIsSeq) TotalLength() int {
cnt := 0
for _, w := range f.words {
cnt += len(w) + 1
}
return cnt - 1
}

func countLetters(s string) int {
cnt := 0
for _, r := range s {
if unicode.IsLetter(r) {
cnt++
}
}
return cnt
}

// ...
// the contents of
// https://rosettacode.org/wiki/Spelling_of_ordinal_numbers#Go
// omitted from this listing
// ...
```
Output:
```The lengths of the first 201 words are:
1:   4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26:   3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51:   2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76:   3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101:  11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126:   4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151:   2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176:   3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201:  11
Length of sentence so far: 1203
Word     1000 is "in", with 2 letters.  Length of sentence so far: 6279
Word    10000 is "in", with 2 letters.  Length of sentence so far: 64140
Word   100000 is "one", with 3 letters.  Length of sentence so far: 659474
Word  1000000 is "the", with 3 letters.  Length of sentence so far: 7113621
Word 10000000 is "thousand", with 8 letters.  Length of sentence so far: 70995756
```

Uses the solution Spelling_of_ordinal_numbers#Haskell and tying-the-knot technique to create the infinite sentence.

```import Data.Char

sentence = start ++ foldMap add (zip [2..] \$ tail \$ words sentence)
where
start = "Four is the number of letters in the first word of this sentence, "
add (i, w) = unwords [spellInteger (alphaLength w), "in the", spellOrdinal i ++ ", "]

alphaLength w = fromIntegral \$ length \$ filter isAlpha w

main = mapM_ (putStrLn . say) [1000,10000,100000,1000000]
where
ws = words sentence
say n =
let (a, w:_) = splitAt (n-1) ws
in "The " ++ spellOrdinal n ++ " word is \"" ++ w ++ "\" which has " ++
spellInteger (alphaLength  w) ++ " letters. The sentence length is " ++
show (length \$ unwords a) ++ " chars."
```
```λ> take 200 sentence
"Four is the number of letters in the first word of this sentence, two in the second, three in the third, six in the fourth, two in the fifth, seven in the sixth, two in the seventh, three in the eight"

λ> main
The one thousandth word is "in" which has two letters. The sentence length is 6349 chars.
The ten thousandth word is "in" which has two letters. The sentence length is 66051 chars.
The one hundred thousandth word is "one" which has three letters. The sentence length is 683690 chars.
The one millionth word is "the" which has three letters. The sentence length is 7349567 chars.```

## Java

Take care with the requirements. As noted, numberToString(23) = twenty-three. Therefore numberToString(723423) = seven hundred twenty-three thousand four hundred twenty-three (note the two "-").
The discussion was helpful by providing the first 2202 words of the sentence.

```import java.util.HashMap;
import java.util.Map;

public class FourIsTheNumberOfLetters {

public static void main(String[] args) {
String [] words = neverEndingSentence(201);
System.out.printf("Display the first 201 numbers in the sequence:%n%3d: ", 1);
for ( int i = 0 ; i < words.length ; i++ ) {
System.out.printf("%2d ", numberOfLetters(words[i]));
if ( (i+1) % 25 == 0 ) {
System.out.printf("%n%3d: ", i+2);
}
}
System.out.printf("%nTotal number of characters in the sentence is %d%n", characterCount(words));
for ( int i = 3 ; i <= 7 ; i++ ) {
int index = (int) Math.pow(10, i);
words = neverEndingSentence(index);
String last = words[words.length-1].replace(",", "");
System.out.printf("Number of letters of the %s word is %d. The word is \"%s\".  The sentence length is %,d characters.%n", toOrdinal(index), numberOfLetters(last), last, characterCount(words));
}
}

@SuppressWarnings("unused")
private static void displaySentence(String[] words, int lineLength) {
int currentLength = 0;
for ( String word : words ) {
if ( word.length() + currentLength > lineLength ) {
String first = word.substring(0, lineLength-currentLength);
String second = word.substring(lineLength-currentLength);
System.out.println(first);
System.out.print(second);
currentLength = second.length();
}
else {
System.out.print(word);
currentLength += word.length();
}
if ( currentLength == lineLength ) {
System.out.println();
currentLength = 0;
}
System.out.print(" ");
currentLength++;
if ( currentLength == lineLength ) {
System.out.println();
currentLength = 0;
}
}
System.out.println();
}

private static int numberOfLetters(String word) {
return word.replace(",","").replace("-","").length();
}

private static long characterCount(String[] words) {
int characterCount = 0;
for ( int i = 0 ; i < words.length ; i++ ) {
characterCount += words[i].length() + 1;
}
//  Extra space counted in last loop iteration
characterCount--;
return characterCount;
}

private static String[] startSentence = new String[] {"Four", "is", "the", "number", "of", "letters", "in", "the", "first", "word", "of", "this", "sentence,"};

private static String[] neverEndingSentence(int wordCount) {
String[] words = new String[wordCount];
int index;
for ( index = 0 ; index < startSentence.length && index < wordCount ; index++ ) {
words[index] = startSentence[index];
}
int sentencePosition = 1;
while ( index < wordCount ) {
//  X in the Y
//  X
sentencePosition++;
String word = words[sentencePosition-1];
for ( String wordLoop : numToString(numberOfLetters(word)).split(" ") ) {
words[index] = wordLoop;
index++;
if ( index == wordCount ) {
break;
}
}
// in
words[index] = "in";
index++;
if ( index == wordCount ) {
break;
}
//  the
words[index] = "the";
index++;
if ( index == wordCount ) {
break;
}
//  Y
for ( String wordLoop : (toOrdinal(sentencePosition) + ",").split(" ") ) {
words[index] = wordLoop;
index++;
if ( index == wordCount ) {
break;
}
}
}
return words;
}

private static final String[] nums = new String[] {
"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine",
"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"
};

private static final String[] tens = new String[] {"zero", "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};

private static final String numToString(long n) {
return numToStringHelper(n);
}

private static final String numToStringHelper(long n) {
if ( n < 0 ) {
return "negative " + numToStringHelper(-n);
}
int index = (int) n;
if ( n <= 19 ) {
return nums[index];
}
if ( n <= 99 ) {
return tens[index/10] + (n % 10 > 0 ? "-" + numToStringHelper(n % 10) : "");
}
String label = null;
long factor = 0;
if ( n <= 999 ) {
label = "hundred";
factor = 100;
}
else if ( n <= 999999) {
label = "thousand";
factor = 1000;
}
else if ( n <= 999999999) {
label = "million";
factor = 1000000;
}
else if ( n <= 999999999999L) {
label = "billion";
factor = 1000000000;
}
else if ( n <= 999999999999999L) {
label = "trillion";
factor = 1000000000000L;
}
else if ( n <= 999999999999999999L) {
factor = 1000000000000000L;
}
else {
label = "quintillion";
factor = 1000000000000000000L;
}
return numToStringHelper(n / factor) + " " + label + (n % factor > 0 ? " " + numToStringHelper(n % factor ) : "");
}

private static Map<String,String> ordinalMap = new HashMap<>();
static {
ordinalMap.put("one", "first");
ordinalMap.put("two", "second");
ordinalMap.put("three", "third");
ordinalMap.put("five", "fifth");
ordinalMap.put("eight", "eighth");
ordinalMap.put("nine", "ninth");
ordinalMap.put("twelve", "twelfth");
}

private static String toOrdinal(long n) {
String spelling = numToString(n);
String[] split = spelling.split(" ");
String last = split[split.length - 1];
String replace = "";
if ( last.contains("-") ) {
String[] lastSplit = last.split("-");
String lastWithDash = lastSplit[1];
String lastReplace = "";
if ( ordinalMap.containsKey(lastWithDash) ) {
lastReplace = ordinalMap.get(lastWithDash);
}
else if ( lastWithDash.endsWith("y") ) {
lastReplace = lastWithDash.substring(0, lastWithDash.length() - 1) + "ieth";
}
else {
lastReplace = lastWithDash + "th";
}
replace = lastSplit[0] + "-" + lastReplace;
}
else {
if ( ordinalMap.containsKey(last) ) {
replace = ordinalMap.get(last);
}
else if ( last.endsWith("y") ) {
replace = last.substring(0, last.length() - 1) + "ieth";
}
else {
replace = last + "th";
}
}
split[split.length - 1] = replace;
return String.join(" ", split);
}

}
```
Output:
```Display the first 201 numbers in the sequence:
1:  4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26:  3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51:  2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76:  3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101: 11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126:  4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151:  2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176:  3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201: 11
Total number of characters in the sentence is 1203
Number of letters of the one thousandth word is 2. The word is "in".  The sentence length is 6,249 characters.
Number of letters of the ten thousandth word is 2. The word is "in".  The sentence length is 64,097 characters.
Number of letters of the one hundred thousandth word is 3. The word is "one".  The sentence length is 659,455 characters.
Number of letters of the one millionth word is 3. The word is "the".  The sentence length is 7,113,560 characters.
Number of letters of the ten millionth word is 8. The word is "thousand".  The sentence length is 70,995,729 characters.
```

## Julia

The functions num2text and numtext2ordinal are from the "Spelling of ordinal numbers" and "Number names" tasks, updated for Julia 1.0 and to remove the "and" words.
```using DataStructures # for deque

const seed = "Four is the number of letters in the first word of this sentence, "
const (word2, word3) = ("in", "the")

lettercount(w) = length(w) -  length(collect(eachmatch(r"-", w)))
splits(txt) = [x.match for x in eachmatch(r"[\w\-]+", txt)]
todq(sentence) = (d = Deque{String}(); map(x->push!(d, x), splits(sentence)[2:end]); d)

struct CountLetters
seedsentence::String
words::Deque{String}
commasafter::Vector{Int}
CountLetters(s) = new(s, todq(s), [13])
CountLetters() = CountLetters(seed)
end

function Base.iterate(iter::CountLetters, state = (1, 5, ""))
if length(iter.words) < 1
return nothing
end
returnword = popfirst!(iter.words)
nextwordindex = state[1] + 1
wordlen = lettercount(returnword)
wordvec = vcat(num2text(wordlen), word2, word3, splits(numtext2ordinal(num2text(nextwordindex))))
map(x -> push!(iter.words, x), wordvec)
push!(iter.commasafter, length(iter.words))
added = length(returnword) + (nextwordindex in iter.commasafter ? 2 : 1)
(wordlen, (nextwordindex, state[2] + added, returnword))
end

Base.eltype(iter::CountLetters) = Int

function firstN(n = 201)
countlet = CountLetters()
print("It is interesting how identical lengths align with 20 columns.\n   1:   4")
iter_result = iterate(countlet)
itercount = 2
while iter_result != nothing
(wlen, state) = iter_result
if itercount % 20 == 0
elseif itercount >= n
break
end
iter_result = iterate(countlet, state)
itercount += 1
end
println()
end

function sumwords(iterations)
countlet = CountLetters()
iter_result = iterate(countlet)
itercount = 2
while iter_result != nothing
(wlen, state) = iter_result
if itercount == iterations
return state
end
iter_result = iterate(countlet, state)
itercount += 1
end
throw("Iteration failed on \"Four is the number\" task.")
end

firstN()

for n in [2202, 1000, 10000, 100000, 1000000, 10000000]
(itercount, totalletters, lastword) = sumwords(n)
println("\$n words -> \$itercount iterations, \$totalletters letters total, ",
"last word \"\$lastword\" with \$(length(lastword)) letters.")
end
```
Output:
```
It is interesting how identical lengths align with 20 columns.

1:   4   2   3   6   2   7   2   3   5   4   2   4   8   3   2   3   6   5   2   3
21:   5   3   2   3   6   3   2   3   5   5   2   3   5   3   2   3   7   5   2   3
41:   6   4   2   3   5   4   2   3   5   3   2   3   8   4   2   3   7   5   2   3
61:  10   5   2   3  10   3   2   3   9   5   2   3   9   3   2   3  11   4   2   3
81:  10   3   2   3  10   5   2   3   9   4   2   3  11   5   2   3  12   3   2   3
101:  11   5   2   3  12   3   2   3  11   5   2   3  11   3   2   3  13   5   2   3
121:  12   4   2   3  11   4   2   3   9   3   2   3  11   5   2   3  12   4   2   3
141:  11   5   2   3  12   3   2   3  11   5   2   3  11   5   2   3  13   4   2   3
161:  12   3   2   3  11   5   2   3   8   3   2   3  10   4   2   3  11   3   2   3
181:  10   5   2   3  11   4   2   3  10   4   2   3  10   3   2   3  12   5   2   3
201:  11

2202 words -> 2202 iterations, 14035 letters total, last word "ninety-ninth" with 12 letters.
1000 words -> 1000 iterations, 6290 letters total, last word "in" with 2 letters.
10000 words -> 10000 iterations, 64320 letters total, last word "in" with 2 letters.
100000 words -> 100000 iterations, 661369 letters total, last word "one" with 3 letters.
1000000 words -> 1000000 iterations, 7127541 letters total, last word "the" with 3 letters.
10000000 words -> 10000000 iterations, 71103026 letters total, last word "thousand" with 8 letters.

```

## Kotlin

This pulls in (slightly adjusted) code from related tasks to convert numbers to text or ordinals.

```// version 1.1.4-3

val names = mapOf(
1 to "one",
2 to "two",
3 to "three",
4 to "four",
5 to "five",
6 to "six",
7 to "seven",
8 to "eight",
9 to "nine",
10 to "ten",
11 to "eleven",
12 to "twelve",
13 to "thirteen",
14 to "fourteen",
15 to "fifteen",
16 to "sixteen",
17 to "seventeen",
18 to "eighteen",
19 to "nineteen",
20 to "twenty",
30 to "thirty",
40 to "forty",
50 to "fifty",
60 to "sixty",
70 to "seventy",
80 to "eighty",
90 to "ninety"
)

val bigNames = mapOf(
1_000L to "thousand",
1_000_000L to "million",
1_000_000_000L to "billion",
1_000_000_000_000L to "trillion",
1_000_000_000_000_000_000L to "quintillion"
)

val irregOrdinals = mapOf(
"one" to "first",
"two" to "second",
"three" to "third",
"five" to "fifth",
"eight" to "eighth",
"nine" to "ninth",
"twelve" to "twelfth"
)

fun String.toOrdinal(): String {
if (this == "zero") return "zeroth"  // or alternatively 'zeroeth'
val splits = this.split(' ', '-')
val last = splits[splits.lastIndex]
return if (irregOrdinals.containsKey(last)) this.dropLast(last.length) + irregOrdinals[last]!!
else if (last.endsWith("y")) this.dropLast(1) + "ieth"
else this + "th"
}

fun numToText(n: Long, uk: Boolean = false): String {
if (n == 0L) return "zero"
val neg = n < 0L
val maxNeg = n == Long.MIN_VALUE
var nn = if (maxNeg) -(n + 1) else if (neg) -n else n
val digits3 = IntArray(7)
for (i in 0..6) {  // split number into groups of 3 digits from the right
digits3[i] = (nn % 1000).toInt()
nn /= 1000
}

fun threeDigitsToText(number: Int) : String {
val sb = StringBuilder()
if (number == 0) return ""
val hundreds = number / 100
val remainder = number % 100
if (hundreds > 0) {
sb.append(names[hundreds], " hundred")
if (remainder > 0) sb.append(if (uk) " and " else " ")
}
if (remainder > 0) {
val tens = remainder / 10
val units = remainder % 10
if (tens > 1) {
sb.append(names[tens * 10])
if (units > 0) sb.append("-", names[units])
}
else sb.append(names[remainder])
}
return sb.toString()
}

val strings = Array(7) { threeDigitsToText(digits3[it]) }
var text = strings[0]
var andNeeded = uk && digits3[0] in 1..99
var big = 1000L
for (i in 1..6) {
if (digits3[i] > 0) {
var text2 = strings[i] + " " + bigNames[big]
if (text.isNotEmpty()) {
text2 += if (andNeeded) " and " else " "  // no commas inserted in this version
andNeeded = false
}
else andNeeded = uk && digits3[i] in 1..99
text = text2 + text
}
big *= 1000
}
if (maxNeg) text = text.dropLast(5) + "eight"
if (neg) text = "minus " + text
return text
}

val opening = "Four is the number of letters in the first word of this sentence,".split(' ')

val String.adjustedLength get() = this.replace(",", "").replace("-", "").length  // no ',' or '-'

fun getWords(n: Int): List<String> {
val words = mutableListOf<String>()
if (n > opening.size) {
var k = 2
while (true) {
val len = words[k - 1].adjustedLength
val text = numToText(len.toLong())
val splits = text.split(' ')
val text2 = numToText(k.toLong()).toOrdinal() + ","  // add trailing comma
val splits2 = text2.split(' ')
if (words.size >= n) break
k++
}
}
return words
}

fun getLengths(n: Int): Pair<List<Int>, Int> {
val words = getWords(n)
val lengths = words.take(n).map { it.adjustedLength }
val sentenceLength = words.sumBy { it.length } + words.size - 1  // includes hyphens, commas & spaces
return Pair(lengths, sentenceLength)
}

fun getLastWord(n: Int): Triple<String, Int, Int> {
val words = getWords(n)
val nthWord = words[n - 1]
val sentenceLength = words.sumBy { it.length } + words.size - 1  // includes hyphens, commas & spaces
return Triple(nthWord, nthWordLength, sentenceLength)
}

fun main(args: Array<String>) {
var n = 201
println("The lengths of the first \$n words are:\n")
val (list, sentenceLength) = getLengths(n)
for (i in 0 until n) {
if (i % 25 == 0) {
if (i > 0) println()
print("\${"%3d".format(i + 1)}: ")
}
print("%3d".format(list[i]))
}
println("\n\nLength of sentence = \$sentenceLength\n")
n = 1_000
do {
var (word, wLen, sLen) = getLastWord(n)
if (word.endsWith(",")) word = word.dropLast(1)  // strip off any trailing comma
println("The length of word \$n [\$word] is \$wLen")
println("Length of sentence = \$sLen\n")
n *= 10
}
while (n <= 10_000_000)
}
```
Output:
```The lengths of the first 201 words are:

1:   4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26:   3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51:   2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76:   3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101:  11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126:   4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151:   2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176:   3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201:  11

Length of sentence = 1203

The length of word 1000 [in] is 2
Length of sentence = 6279

The length of word 10000 [in] is 2
Length of sentence = 64140

The length of word 100000 [one] is 3
Length of sentence = 659474

The length of word 1000000 [the] is 3
Length of sentence = 7113621

The length of word 10000000 [thousand] is 8
Length of sentence = 70995756
```

## Mathematica/Wolfram Language

```(*==Number names==*)

(*Mathematica has a built-in function for getting the name of an integer. It's a semantically rich function (dealing with many languages and grammatical variants), and consequently it would slow down our algorithm significantly. So, I've used the built-in function to seed/memoize special-purpose functions. Furthermore, the problem is suited to using a representation of the sentence that is an array of strings rather than a single monolithic string, and so these integer name functions will return arrays of strings.*)

(*We'll define the function for small integers to use the built-in function and to trigger memoization on the first invocation. After that, the basic strategy is to chunk up an integer into groups of three digits, apply the name to what those digits usually represent and then add the 'scaling' term (e.g. 'thousand', 'million'). Just for laziness, I'll skip trying to handle the 'thousand and zero' case and just fall back to the built-in function--it shouldn't be called often enough to matter. Since this problem won't need number names exceeding the 'million' scale, I won't optimize beyond that.*)
IntNameWords[n_]:=(IntNameWords[n]=StringSplit[IntegerName[n,"Words"]])/;n<1000;
IntNameWords[n_]:=StringSplit[IntegerName[n,"Words"]]/;Divisible[n,1000];
IntNameWords[n_]:=Flatten[Riffle[IntNameWords/@QuotientRemainder[n,1000],"thousand"]]/;n<1000000;
IntNameWords[n_]:=Flatten[Riffle[IntNameWords/@QuotientRemainder[n,1000000],"million"]]/;n<1000000000;
IntNameWords[n_]:=StringSplit[IntegerName[n,"Words"]];
(*I'm using Scan to trigger the memoization.*)
Scan[IntNameWords,Range[999]];

(*The strategy is similar for ordinals. Note that I'm tacking on a comma to the ordinals. This makes this function quite specialized to this specific problem.*)
OrdNameWords[n_]:=(OrdNameWords[n]=StringSplit[IntegerName[n,"Ordinal"]<>","])/;n<1000;
OrdNameWords[n_]:=StringSplit[IntegerName[n,"Ordinal"]<>","]/;Divisible[n,1000];
OrdNameWords[n_]:=StringSplit[IntegerName[n,"Ordinal"]<>","];
(*Triggering memoization again.*)
Scan[OrdNameWords,Range[999]];

(*==Helper/driver functions==*)

(*This could be generalized, but for this problem, the '-' and ',' are the only non-letter characters we need to worry about.*)
LetterCount[str_]:=StringLength[StringDelete[str,"-"|","]];

(*The seed/initial part of the sentence.*)
SentenceHeadWords=StringSplit["Four is the number of letters in the first word of this sentence,"];

(*Output formatters*)
DisplayWordLengthSequence[wordSeq_]:=StringRiffle[{"First "<>StringRiffle[IntNameWords[Length@wordSeq]]<>" numbers in sequence:",LetterCount/@wordSeq},"\n"];
DisplayCharacterCount[wordSeq_]:=StringRiffle[{"String length of sentence with "<>StringRiffle[IntNameWords[Length@wordSeq]]<>" words:",SentenceCharacterCount[wordSeq]}];
DisplayWordInfo[wordSeq_,wordIdx_]:=StringForm["The `` word is '``' consisting of `` letters.",StringRiffle[OrdNameWords[Length@wordSeq]],wordSeq[[wordIdx]],StringRiffle[IntNameWords[StringLength@wordSeq[[wordIdx]]]]];

(*There is a space between each 'word', so we can just add 1 less than the number of 'words' to get total characters in the full string representation of the sentence (if we were to create it). I could also subract another 1 for the trailing comma, but the requirements weren't precise in this regard.*)
SentenceCharacterCount[chunks:{__String}]:=Total[StringLength[chunks]]+Length[chunks]-1;

(*==Implementation A==*)

(*A simple functional implementation that continues to extend the 'sentence' one fragment at a time until the number of words exceeds the requested number. This implementation takes several seconds to complete the 100,000 word case.*)
ExtendCharChunks[{fragCt_,wordCt_,chunks_}]:=
With[
{nextFrag=Flatten[{IntNameWords[LetterCount[chunks[[1+fragCt]]]],"in","the",OrdNameWords[1+fragCt]}]},
{1+fragCt,wordCt+Length[nextFrag],Flatten[{chunks,nextFrag}]}
];
SentenceChunksFun[chunkCt_]:=Take[Last[NestWhile[ExtendCharChunks,ExtendCharChunks[{0,0,{}}],#[[2]]<chunkCt&]],chunkCt];

(*==Implementation B==*)

(*This implementation uses a pre-allocated array, an iterative strategy, and inlining of the fragment construction. It performs much better than the previous implementation but still takes about 20 seconds for the 10 million word case. One could try compiling the function for greater performance.*)
SentenceChunksArray[targetCount_]:=
Block[
{
chunks=ConstantArray["",targetCount],
wordIdx=0,
fragmentIdx=0
},
++fragmentIdx;
While[
(*Since each new fragment is longer than one word, it is likely that we will try to insert more words into the array than it has been allocated to hold. This generates and error message, but does not otherwise interfere with processing (the insertion simply fails). I could include more checks, but it didn't seem necessary for this task.*)
wordIdx<targetCount,
Scan[(chunks[[++wordIdx]]=#)&,{Splice[IntNameWords[LetterCount[chunks[[++fragmentIdx]]]]],"in","the",Splice[OrdNameWords[fragmentIdx]]}]
];
chunks
];

(*==Output==*)

StringRiffle[
{
DisplayWordLengthSequence[SentenceChunksArray[201]],
DisplayCharacterCount[SentenceChunksArray[201]],
DisplayWordInfo[SentenceChunksArray[1000],1000],
DisplayWordInfo[SentenceChunksArray[10000],10000],
DisplayWordInfo[SentenceChunksArray[100000],100000],
DisplayWordInfo[SentenceChunksArray[1000000],1000000],
DisplayWordInfo[SentenceChunksArray[10000000],10000000]
},
"\n\n"
]
```
Output:
```First two hundred one numbers in sequence:
{4, 2, 3, 6, 2, 7, 2, 3, 5, 4, 2, 4, 8, 3, 2, 3, 6, 5, 2, 3, 5, 3, 2, 3, 6, 3, 2, 3, 5, 5, 2, 3, 5, 3, 2, 3, 7, 5, 2, 3, 6, 4, 2, 3, 5, 4, 2, 3, 5, 3, 2, 3, 8, 4, 2, 3, 7, 5, 2, 3, 10, 5, 2, 3, 10, 3, 2, 3, 9, 5, 2, 3, 9, 3, 2, 3, 11, 4, 2, 3, 10, 3, 2, 3, 10, 5, 2, 3, 9, 4, 2, 3, 11, 5, 2, 3, 12, 3, 2, 3, 11, 5, 2, 3, 12, 3, 2, 3, 11, 5, 2, 3, 11, 3, 2, 3, 13, 5, 2, 3, 12, 4, 2, 3, 11, 4, 2, 3, 9, 3, 2, 3, 11, 5, 2, 3, 12, 4, 2, 3, 11, 5, 2, 3, 12, 3, 2, 3, 11, 5, 2, 3, 11, 5, 2, 3, 13, 4, 2, 3, 12, 3, 2, 3, 11, 5, 2, 3, 8, 3, 2, 3, 10, 4, 2, 3, 11, 3, 2, 3, 10, 5, 2, 3, 11, 4, 2, 3, 10, 4, 2, 3, 10, 3, 2, 3, 12, 5, 2, 3, 11}

String length of sentence with two hundred one words: 1203

The one thousandth, word is 'in' consisting of two letters.

The ten thousandth, word is 'in' consisting of two letters.

The one hundred thousandth, word is 'one' consisting of three letters.

The one millionth, word is 'the' consisting of three letters.

The ten millionth, word is 'thousand' consisting of eight letters.```

## Nim

```import strutils, strformat, tables

####################################################################################################
# Cardinal and ordinal strings.

const

Small = ["zero",    "one",     "two",       "three",    "four",
"five",    "six",     "seven",     "eight",    "nine",
"ten",     "eleven",  "twelve",    "thirteen", "fourteen",
"fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]

Tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]

Illions = ["", " thousand", " million", " billion", " trillion", " quadrillion", " quintillion"]

IrregularOrdinals = {"one": "first", "two": "second", "three": "third", "five": "fifth",
"eight": "eighth", "nine": "ninth", "twelve": "twelfth"}.toTable()

#---------------------------------------------------------------------------------------------------

func spellCardinal(n: int64): string =
## Spell an integer as a cardinal.

var n = n

if n < 0:
result = "negative "
n = -n

if n < 20:
result &= Small[n]

elif n < 100:
result &= Tens[n div 10]
let m = n mod 10
if m != 0: result &= '-' & Small[m]

elif n < 1000:
result &= Small[n div 100] & " hundred"
let m = n mod 100
if m != 0: result &= ' ' & m.spellCardinal()

else:
# Work from right to left.
var sx = ""
var i = 0
while n > 0:
let m = n mod 1000
n = n div 1000
if m != 0:
var ix = m.spellCardinal() & Illions[i]
if sx.len > 0: ix &= " " & sx
sx = ix
inc i
result &= sx

#---------------------------------------------------------------------------------------------------

func spellOrdinal(n: int64): string =
## Spell an integer as an ordinal.

result = n.spellCardinal()
var parts = result.rsplit({' ', '-'}, maxsplit = 1)
let tail = parts[^1]
if tail in IrregularOrdinals:
result[^tail.len..^1] = IrregularOrdinals[tail]
elif tail.endsWith('y'):
result[^1..^1]= "ieth"
else:
result &= "th"

####################################################################################################
# Sentence building.

type Sentence = seq[string]

#---------------------------------------------------------------------------------------------------

iterator words(sentence: var Sentence): tuple[idx: int; word: string] =
## Yield the successive words of the sentence with their index.

yield (0, "Four")
var idx = 1
var last = 0
while true:
yield (idx, sentence[idx])
inc idx
if idx == sentence.len:
inc last
# For the position, we need to split the ordinal as it may contain spaces.

#---------------------------------------------------------------------------------------------------

iterator letterCounts(sentence: var Sentence): tuple[idx: int; word: string; count: int] =
## Secondary iterator used to yield the number of letters in addition to the index and the word.

for i, word in sentence.words():
yield (i, word, word.count(Letters))

####################################################################################################
# Drivers.

# Constant to initialize the sentence.
const Init = "Four is the number of letters in the first word of this sentence,".splitWhitespace()

#---------------------------------------------------------------------------------------------------

proc displayLetterCounts(pos: Positive) =
## Display the number of letters of the word at position "pos".

var sentence = Init
echo fmt"Number of letters in first {pos} words in the sequence:"
var valcount = 0   # Number of values displayed in the current line.
var length = 0

for i, word, letterCount in sentence.letterCounts():
if i == pos:
# Terminated.
dec length  # Adjust space count.
echo ""
break

if valcount == 0: stdout.write fmt"{i+1:>3}:"
stdout.write fmt"{letterCount:>3}"
inc valcount
inc length, word.len + 1  # +1 for space.

if valcount == 12:
# Terminate line.
echo ""
valcount = 0

echo fmt"Length of sentence: {length}"

#---------------------------------------------------------------------------------------------------

proc displayWord(pos: Positive) =
## Display the word at position "pos".

var sentence = Init
let idx = pos - 1
var length = 0
for i, word in sentence.words():
length += word.len + 1
if i == idx:
dec length    # Adjust space count.
let w = word.strip(leading = false, chars = {','})  # Remove trailing ',' if needed.
echo fmt"Word {pos} is ""{w}"" with {w.count(Letters)} letters."
echo fmt"Length of sentence: {length}"
break

#———————————————————————————————————————————————————————————————————————————————————————————————————

displayLetterCounts(201)
for n in [1_000, 10_000, 100_000, 1_000_000, 10_000_000]:
echo ""
displayWord(n)
```
Output:
```Number of letters in first 201 words in the sequence:
1:  4  2  3  6  2  7  2  3  5  4  2  4
13:  8  3  2  3  6  5  2  3  5  3  2  3
25:  6  3  2  3  5  5  2  3  5  3  2  3
37:  7  5  2  3  6  4  2  3  5  4  2  3
49:  5  3  2  3  8  4  2  3  7  5  2  3
61: 10  5  2  3 10  3  2  3  9  5  2  3
73:  9  3  2  3 11  4  2  3 10  3  2  3
85: 10  5  2  3  9  4  2  3 11  5  2  3
97: 12  3  2  3 11  5  2  3 12  3  2  3
109: 11  5  2  3 11  3  2  3 13  5  2  3
121: 12  4  2  3 11  4  2  3  9  3  2  3
133: 11  5  2  3 12  4  2  3 11  5  2  3
145: 12  3  2  3 11  5  2  3 11  5  2  3
157: 13  4  2  3 12  3  2  3 11  5  2  3
169:  8  3  2  3 10  4  2  3 11  3  2  3
181: 10  5  2  3 11  4  2  3 10  4  2  3
193: 10  3  2  3 12  5  2  3 11
Length of sentence: 1203

Word 1000 is "in" with 2 letters.
Length of sentence: 6249

Word 10000 is "in" with 2 letters.
Length of sentence: 64097

Word 100000 is "one" with 3 letters.
Length of sentence: 659455

Word 1000000 is "the" with 3 letters.
Length of sentence: 7113560

Word 10000000 is "thousand" with 8 letters.
Length of sentence: 70995729```

## Perl

Uses `Lingua::EN::Numbers` module to generate number names. State variable in extend_to routine keeps track of last word tallied.

Translation of: Raku
```use feature 'state';
use Lingua::EN::Numbers qw(num2en num2en_ordinal);

my @sentence = split / /, 'Four is the number of letters in the first word of this sentence, ';

sub extend_to {
my(\$last) = @_;
state \$index = 1;
until (\$#sentence > \$last) {
push @sentence, split ' ', num2en(alpha(\$sentence[\$index])) . ' in the ' . no_c(num2en_ordinal(1+\$index)) . ',';
\$index++;
}
}

sub alpha { my(\$s) = @_; \$s =~ s/\W//gi; length \$s }
sub no_c  { my(\$s) = @_; \$s =~ s/\ and|,//g;   return \$s }
sub count { length(join ' ', @sentence[0..-1+\$_[0]]) . " characters in the sentence, up to and including this word.\n" }

print "First 201 word lengths in the sequence:\n";
extend_to(201);
for (0..200) {
printf "%3d", alpha(\$sentence[\$_]);
print "\n" unless (\$_+1) % 32;
}
print "\n" . count(201) . "\n";

for (1e3, 1e4, 1e5, 1e6, 1e7) {
extend_to(\$_);
print
ucfirst(num2en_ordinal(\$_)) .  " word, '\$sentence[\$_-1]' has " . alpha(\$sentence[\$_-1]) .  " characters. \n" .
count(\$_) . "\n";
}
```
Output:
```First 201 word lengths in the sequence:
4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6  3  2  3  5  5  2  3
5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3  2  3  8  4  2  3  7  5  2  3 10  5  2  3
10  3  2  3  9  5  2  3  9  3  2  3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3
12  3  2  3 11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11  4  2  3
9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5  2  3 11  5  2  3 13  4  2  3
12  3  2  3 11  5  2  3  8  3  2  3 10  4  2  3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3
10  3  2  3 12  5  2  3 11
1203 characters in the sentence, up to and including this word.

One thousandth word, 'in' has 2 characters.
6249 characters in the sentence, up to and including this word.

Ten thousandth word, 'in' has 2 characters.
64097 characters in the sentence, up to and including this word.

One hundred thousandth word, 'one' has 3 characters.
659455 characters in the sentence, up to and including this word.

One millionth word, 'the' has 3 characters.
7113560 characters in the sentence, up to and including this word.

Ten millionth word, 'thousand' has 8 characters.
70995729 characters in the sentence, up to and including this word.```

## Phix

Note that my version of Number_names includes "and" (and ","), that others do not, hence the kill_and()/grr below and the minor mismatch of sentence lengths.

```with javascript_semantics
include demo\rosetta\number_names.exw -- see note

-- as per Spelling_of_ordinal_numbers#Phix:
constant {irregs,ordinals} = columnize({{"one","first"},
{"two","second"},
{"three","third"},
{"five","fifth"},
{"eight","eighth"},
{"nine","ninth"},
{"twelve","twelfth"}})

function ordinl(string s)
integer i
for i=length(s) to 1 by -1 do
integer ch = s[i]
if ch=' ' or ch='-' then exit end if
end for
integer k = find(s[i+1..\$],irregs)
if k then
s = s[1..i]&ordinals[k]
elsif s[\$]='y' then
s[\$..\$] = "ieth"
else
s &= "th"
end if
return s
end function
--/copy of Spelling_of_ordinal_numers#Phix

function count_letters(string s)
integer res = 0
for i=1 to length(s) do
integer ch = s[i]
if (ch>='A' and ch<='Z')
or (ch>='a' and ch<='z') then
res += 1
end if
end for
return res
end function

sequence words = split("Four is the number of letters in the first word of this sentence,")
integer fi = 1

function kill_and(sequence s)
--grr...
for i=length(s) to 1 by -1 do
if s[i] = "and" then
s[i..i] = {}
end if
end for
return s
end function

function word_len(integer w)
-- Returns the w'th word and its length (only counting letters).
while length(words)<w do
fi += 1
integer n = count_letters(words[fi])
sequence ns = kill_and(split(spell(n)))
sequence os = kill_and(split(ordinl(spell(fi)) & ","))
-- append eg {"two","in","the","second,"}
words &= ns&{"in","the"}&os
end while
string word = words[w]
return {word, count_letters(word)}
end function

function total_length()
-- Returns the total number of characters (including blanks,
-- commas, and punctuation) of the sentence so far constructed.
integer res = 0
for i=1 to length(words) do
res += length(words[i])+1
end for
return res
end function

procedure main()
printf(1,"The lengths of the first 201 words are:\n")
for i=1 to 201 do
if mod(i,25)==1 then
printf(1,"\n%3d: ", i)
end if
printf(1," %2d", word_len(i)[2])
end for
printf(1,"\nLength of sentence so far:%d\n", total_length())
for p=3 to iff(platform()=JS?5:7) do
integer i = power(10,p)
{string w, integer n} = word_len(i)
printf(1,"Word %8d is \"%s\", with %d letters.", {i, w, n})
printf(1,"  Length of sentence so far:%d\n", total_length())
end for
end procedure
main()
```

Note you will have to comment out main()..EOF in the number_names.exw file for pwa/p2js to accept it/not complain about multiple main() or #ilASM (which I may yet remove). Also as indicated 100K words is simply too much for your typical browser, let alone a million, so we limit the last part to 10K under pwa/p2js.

Output:
```The lengths of the first 201 words are:

1:   4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26:   3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51:   2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76:   3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101:  11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126:   4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151:   2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176:   3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201:  11
Length of sentence so far:1204
Word     1000 is "in", with 2 letters.  Length of sentence so far:6280
Word    10000 is "in", with 2 letters.  Length of sentence so far:64692
Word   100000 is "one", with 3 letters.  Length of sentence so far:671578
Word  1000000 is "the", with 3 letters.  Length of sentence so far:7235383
Word 10000000 is "thousand,", with 8 letters.  Length of sentence so far:72079160
```

## Python

```# Python implementation of Rosetta Code Task
# http://rosettacode.org/wiki/Four_is_the_number_of_letters_in_the_...
# Uses inflect
# https://pypi.org/project/inflect/

import inflect

def count_letters(word):
"""
count letters ignore , or -, or space
"""
count = 0
for letter in word:
if letter != ',' and letter !='-' and letter !=' ':
count += 1

return count

def split_with_spaces(sentence):
"""
Takes string with partial sentence and returns
list of words with spaces included.

Leading space is attached to first word.
Later spaces attached to prior word.
"""
sentence_list = []
curr_word = ""
for c in sentence:
if c == " " and curr_word != "":
# append space to end of non-empty words
# assumed no more than 1 consecutive space.
sentence_list.append(curr_word+" ")
curr_word = ""
else:
curr_word += c

# add trailing word that does not end with a space

if len(curr_word) > 0:
sentence_list.append(curr_word)

return sentence_list

def my_num_to_words(p, my_number):
"""
Front end to inflect's number_to_words

Get's rid of ands and commas in large numbers.
"""

number_string_list = p.number_to_words(my_number, wantlist=True, andword='')

number_string = number_string_list[0]

for i in range(1,len(number_string_list)):
number_string += " " + number_string_list[i]

return number_string

def build_sentence(p, max_words):
"""

Builds at most max_words of the task following the pattern:

Four is the number of letters in the first word of this sentence, two in the second,
three in the third, six in the fourth, two in the fifth, seven in the sixth,

"""

# start with first part of sentence up first comma as a list

sentence_list = split_with_spaces("Four is the number of letters in the first word of this sentence,")

num_words = 13

# which word number we are doing next
# two/second is first one in loop

word_number = 2

# loop until sentance is at least as long as needs be

while num_words < max_words:
# Build something like
# ,two in the second

# get second or whatever we are on

ordinal_string = my_num_to_words(p, p.ordinal(word_number))

# get two or whatever the length is of the word_number word

word_number_string = my_num_to_words(p, count_letters(sentence_list[word_number - 1]))

new_string = " "+word_number_string+" in the "+ordinal_string+","

new_list = split_with_spaces(new_string)

sentence_list += new_list

num_words += len(new_list)

# increment word number

word_number += 1

return sentence_list, num_words

def word_and_counts(word_num):
"""

Print's lines like this:

Word     1000 is "in", with 2 letters.  Length of sentence so far: 6279

"""

sentence_list, num_words = build_sentence(p, word_num)

word_str = sentence_list[word_num - 1].strip(' ,')

num_letters = len(word_str)

num_characters = 0

for word in sentence_list:
num_characters += len(word)

print('Word {0:8d} is "{1}", with {2} letters.  Length of the sentence so far: {3}  '.format(word_num,word_str,num_letters,num_characters))

p = inflect.engine()

sentence_list, num_words = build_sentence(p, 201)

print(" ")
print("The lengths of the first 201 words are:")
print(" ")

print('{0:3d}:  '.format(1),end='')

total_characters = 0

for word_index in range(201):

word_length = count_letters(sentence_list[word_index])

total_characters += len(sentence_list[word_index])

print('{0:2d}'.format(word_length),end='')
if (word_index+1) % 20 == 0:
# newline every 20
print(" ")
print('{0:3d}:  '.format(word_index + 2),end='')
else:
print(" ",end='')

print(" ")
print(" ")
print("Length of the sentence so far: "+str(total_characters))
print(" ")

"""

Expected output this part:

Word     1000 is "in", with 2 letters.  Length of the sentence so far: 6279
Word    10000 is "in", with 2 letters.  Length of the sentence so far: 64140
Word   100000 is "one", with 3 letters.  Length of the sentence so far: 659474
Word  1000000 is "the", with 3 letters.  Length of the sentence so far: 7113621
Word 10000000 is "thousand", with 8 letters.  Length of the sentence so far: 70995756

"""

word_and_counts(1000)
word_and_counts(10000)
word_and_counts(100000)
word_and_counts(1000000)
word_and_counts(10000000)
```

Output:

```The lengths of the first 201 words are:

1:   4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3
21:   5  3  2  3  6  3  2  3  5  5  2  3  5  3  2  3  7  5  2  3
41:   6  4  2  3  5  4  2  3  5  3  2  3  8  4  2  3  7  5  2  3
61:  10  5  2  3 10  3  2  3  9  5  2  3  9  3  2  3 11  4  2  3
81:  10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101:  11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3
121:  12  4  2  3 11  4  2  3  9  3  2  3 11  5  2  3 12  4  2  3
141:  11  5  2  3 12  3  2  3 11  5  2  3 11  5  2  3 13  4  2  3
161:  12  3  2  3 11  5  2  3  8  3  2  3 10  4  2  3 11  3  2  3
181:  10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201:  11

Length of the sentence so far: 1203

Word     1000 is "in", with 2 letters.  Length of the sentence so far: 6279
Word    10000 is "in", with 2 letters.  Length of the sentence so far: 64140
Word   100000 is "one", with 3 letters.  Length of the sentence so far: 659474
Word  1000000 is "the", with 3 letters.  Length of the sentence so far: 7113621
Word 10000000 is "thousand", with 8 letters.  Length of the sentence so far: 70995756
```

## Raku

(formerly Perl 6)

Works with: Rakudo version 2017.09

Uses the Lingua::EN::Numbers module to generate both cardinal and ordinal numbers. This module places commas in number words between 3-orders-of-magnitude clusters. E.G. `12345678.&ordinal` becomes: twelve million, three hundred forty-five thousand, six hundred seventy-eighth. Uses a custom 'no-commas' routine to filter them out for accurate character counts. Generates the 'sentence' lazily so only the words needed are ever calculated and reified.

```use Lingua::EN::Numbers;
no-commas(True);

my \$index = 1;
my @sentence = flat 'Four is the number of letters in the first word of this sentence, '.words,
{ @sentence[\$index++].&alpha.&cardinal, 'in', 'the', |(\$index.&ordinal ~ ',').words } ... * ;

sub alpha ( \$str ) { \$str.subst(/\W/, '', :g).chars }
sub count ( \$index ) { @sentence[^\$index].join(' ').chars ~ " characters in the sentence, up to and including this word.\n" }

say 'First 201 word lengths in the sequence:';
put ' ', map { @sentence[\$_].&alpha.fmt("%2d") ~ (((1+\$_) %% 25) ?? "\n" !! '') }, ^201;
say 201.&count;

for 1e3, 1e4, 1e5, 1e6, 1e7 {
say "{.&ordinal.tc} word, '{@sentence[\$_ - 1]}', has {@sentence[\$_ - 1].&alpha} characters. ", .&count
}
```
Output:
```First 201 word lengths in the sequence:
4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
11
1203 characters in the sentence, up to and including this word.

One thousandth word, 'in', has 2 characters. 6249 characters in the sentence, up to and including this word.

Ten thousandth word, 'in', has 2 characters. 64097 characters in the sentence, up to and including this word.

One hundred thousandth word, 'one', has 3 characters. 659455 characters in the sentence, up to and including this word.

One millionth word, 'the', has 3 characters. 7113560 characters in the sentence, up to and including this word.

Ten millionth word, 'thousand', has 8 characters. 70995729 characters in the sentence, up to and including this word.```

## REXX

```/*REXX pgm finds/shows the number of letters in the  Nth  word in a constructed sentence*/
@= 'Four is the number of letters in the first word of this sentence,'             /*···*/
/* [↑]   the start of a long sentence. */
parse arg N M                                    /*obtain optional argument from the CL.*/
if N='' | N="," then N= 201                      /*Not specified?  Then use the default.*/
if M='' | M="," then M=1000 10000 100000 1000000 /* "      "         "   "   "     "    */
@abcU= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'              /*define the uppercase Latin alphabet. */
!.=.;     #.=.;      q=1;       w=length(N)      /* [↓]  define some helpful low values.*/
call tell N
if N<0  then say y     ' is the length of word '         a          "  ["word(@, a)"]"
say                                              /* [↑]  N negative?  Just show 1 number*/
say 'length of sentence= '   length(@)           /*display the length of the @ sentence.*/

if M\==''  then do k=1  for words(M)  while M\=0 /*maybe handle counts  (if specified). */
x=word(M, k)                     /*obtain the  Kth  word of the M list. */
call tell  -x                    /*invoke subroutine (with negative arg)*/
say
say y     ' is the length of word '      x       "  ["word(@, x)"]"
say 'length of sentence= '  length(@)    /*display length of @ sentence.*/
end   /*k*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
wordLen: arg ?;         return length(?) - length( space( translate(?, , @abcU), 0) )
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell: parse arg z,,\$;   idx=1;    a=abs(z);     group=25     /*show 25 numbers per line.*/
/*Q is the last number spelt by \$SPELL#*/
do j=1  for a                            /*traipse through all the numbers to N.*/
do 2                                   /*perform loop twice  (well ··· maybe).*/
y=wordLen( word(@, j) )                /*get the  Jth  word from the sentence.*/
if y\==0  then leave                   /*Is the word spelt?   Then we're done.*/
q=q + 1                                /*bump the on─going (moving) # counter.*/
if #.q==.  then #.q=\$spell#(q 'Q ORD') /*need to spell A as an ordinal number?*/
_=wordLen( word(@, q) )           /*use the length of the ordinal number.*/
if !._==.  then !._=\$spell#(_ 'Q')     /*Not spelled?   Then go and spell it. */
@=@  !._   'in the'    #.q","          /*append words to never─ending sentence*/
end   /*2*/                            /* [↑]   Q ≡ Quiet      ORD ≡ ORDinal  */

\$=\$ || right(y, 3)                       /* [↓]  append a justified # to a line.*/
if j//group==0 & z>0  then do; say right(idx, w)'►'\$;   idx=idx+group;   \$=;   end
end   /*j*/                              /* [↑]  show line if there's enough #s.*/

if \$\=='' & z>0 then say right(idx, w)'►'\$ /*display if there are residual numbers*/
return
```

The   \$SPELL#.REX   routine can be found here   ───►   \$SPELL#.REX.

output:
```  1►  4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26►  3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51►  2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76►  3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101► 11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126►  4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151►  2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176►  3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201► 11

length of sentence=  1203

2  is the length of word  1000   [in]
length of sentence=  6279

2  is the length of word  10000   [in]
length of sentence=  64140

3  is the length of word  100000   [one]
length of sentence=  659474

3  is the length of word  1000000   [the]
length of sentence=  7113621
```

## Rust

Translation of: C
```struct NumberNames {
cardinal: &'static str,
ordinal: &'static str,
}

impl NumberNames {
fn get_name(&self, ordinal: bool) -> &'static str {
if ordinal {
return self.ordinal;
}
self.cardinal
}
}

const SMALL_NAMES: [NumberNames; 20] = [
NumberNames {
cardinal: "zero",
ordinal: "zeroth",
},
NumberNames {
cardinal: "one",
ordinal: "first",
},
NumberNames {
cardinal: "two",
ordinal: "second",
},
NumberNames {
cardinal: "three",
ordinal: "third",
},
NumberNames {
cardinal: "four",
ordinal: "fourth",
},
NumberNames {
cardinal: "five",
ordinal: "fifth",
},
NumberNames {
cardinal: "six",
ordinal: "sixth",
},
NumberNames {
cardinal: "seven",
ordinal: "seventh",
},
NumberNames {
cardinal: "eight",
ordinal: "eighth",
},
NumberNames {
cardinal: "nine",
ordinal: "ninth",
},
NumberNames {
cardinal: "ten",
ordinal: "tenth",
},
NumberNames {
cardinal: "eleven",
ordinal: "eleventh",
},
NumberNames {
cardinal: "twelve",
ordinal: "twelfth",
},
NumberNames {
cardinal: "thirteen",
ordinal: "thirteenth",
},
NumberNames {
cardinal: "fourteen",
ordinal: "fourteenth",
},
NumberNames {
cardinal: "fifteen",
ordinal: "fifteenth",
},
NumberNames {
cardinal: "sixteen",
ordinal: "sixteenth",
},
NumberNames {
cardinal: "seventeen",
ordinal: "seventeenth",
},
NumberNames {
cardinal: "eighteen",
ordinal: "eighteenth",
},
NumberNames {
cardinal: "nineteen",
ordinal: "nineteenth",
},
];

const TENS: [NumberNames; 8] = [
NumberNames {
cardinal: "twenty",
ordinal: "twentieth",
},
NumberNames {
cardinal: "thirty",
ordinal: "thirtieth",
},
NumberNames {
cardinal: "forty",
ordinal: "fortieth",
},
NumberNames {
cardinal: "fifty",
ordinal: "fiftieth",
},
NumberNames {
cardinal: "sixty",
ordinal: "sixtieth",
},
NumberNames {
cardinal: "seventy",
ordinal: "seventieth",
},
NumberNames {
cardinal: "eighty",
ordinal: "eightieth",
},
NumberNames {
cardinal: "ninety",
ordinal: "ninetieth",
},
];

struct NamedNumber {
cardinal: &'static str,
ordinal: &'static str,
number: usize,
}

impl NamedNumber {
fn get_name(&self, ordinal: bool) -> &'static str {
if ordinal {
return self.ordinal;
}
self.cardinal
}
}

const N: usize = 7;
const NAMED_NUMBERS: [NamedNumber; N] = [
NamedNumber {
cardinal: "hundred",
ordinal: "hundredth",
number: 100,
},
NamedNumber {
cardinal: "thousand",
ordinal: "thousandth",
number: 1000,
},
NamedNumber {
cardinal: "million",
ordinal: "millionth",
number: 1000000,
},
NamedNumber {
cardinal: "billion",
ordinal: "billionth",
number: 1000000000,
},
NamedNumber {
cardinal: "trillion",
ordinal: "trillionth",
number: 1000000000000,
},
NamedNumber {
number: 1000000000000000,
},
NamedNumber {
cardinal: "quintillion",
ordinal: "quintillionth",
number: 1000000000000000000,
},
];

fn big_name(n: usize) -> &'static NamedNumber {
for i in 1..N {
if n < NAMED_NUMBERS[i].number {
return &NAMED_NUMBERS[i - 1];
}
}
&NAMED_NUMBERS[N - 1]
}

fn count_letters(s: &str) -> usize {
let mut count = 0;
for c in s.chars() {
if c.is_alphabetic() {
count += 1;
}
}
count
}

struct WordList {
words: Vec<(usize, usize)>,
string: String,
}

impl WordList {
fn new() -> WordList {
WordList {
words: Vec::new(),
string: String::new(),
}
}
fn append(&mut self, s: &str) {
let offset = self.string.len();
self.string.push_str(s);
self.words.push((offset, offset + s.len()));
}
fn extend(&mut self, s: &str) {
let len = self.words.len();
let mut w = &mut self.words[len - 1];
w.1 += s.len();
self.string.push_str(s);
}
fn len(&self) -> usize {
self.words.len()
}
fn sentence_length(&self) -> usize {
let n = self.words.len();
if n == 0 {
return 0;
}
self.string.len() + n - 1
}
fn get_word(&self, index: usize) -> &str {
let w = &self.words[index];
&self.string[w.0..w.1]
}
}

fn append_number_name(words: &mut WordList, n: usize, ordinal: bool) -> usize {
let mut count = 0;
if n < 20 {
words.append(SMALL_NAMES[n].get_name(ordinal));
count += 1;
} else if n < 100 {
if n % 10 == 0 {
words.append(TENS[n / 10 - 2].get_name(ordinal));
} else {
words.append(TENS[n / 10 - 2].get_name(false));
words.extend("-");
words.extend(SMALL_NAMES[n % 10].get_name(ordinal));
}
count += 1;
} else {
let big = big_name(n);
count += append_number_name(words, n / big.number, false);
if n % big.number == 0 {
words.append(big.get_name(ordinal));
count += 1;
} else {
words.append(big.get_name(false));
count += 1;
count += append_number_name(words, n % big.number, ordinal);
}
}
count
}

fn sentence(count: usize) -> WordList {
let mut result = WordList::new();
const WORDS: &'static [&'static str] = &[
"Four",
"is",
"the",
"number",
"of",
"letters",
"in",
"the",
"first",
"word",
"of",
"this",
"sentence,",
];
for s in WORDS {
result.append(s);
}
let mut n = result.len();
let mut i = 1;
while count > n {
let count = count_letters(result.get_word(i));
n += append_number_name(&mut result, count, false);
result.append("in");
result.append("the");
n += 2;
n += append_number_name(&mut result, i + 1, true);
result.extend(",");
i += 1;
}
result
}

fn main() {
let mut n = 201;
let s = sentence(n);
println!("Number of letters in first {} words in the sequence:", n);
for i in 0..n {
if i != 0 {
if i % 25 == 0 {
println!();
} else {
print!(" ");
}
}
print!("{:2}", count_letters(s.get_word(i)));
}
println!();
println!("Sentence length: {}", s.sentence_length());
n = 1000;
while n <= 10000000 {
let s = sentence(n);
let word = s.get_word(n - 1);
print!(
"The {}th word is '{}' and has {} letters. ",
n,
word,
count_letters(word)
);
println!("Sentence length: {}", s.sentence_length());
n *= 10;
}
}
```
Output:
```Number of letters in first 201 words in the sequence:
4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
11
Sentence length: 1203
The 1000th word is 'in' and has 2 letters. Sentence length: 6279
The 10000th word is 'in' and has 2 letters. Sentence length: 64140
The 100000th word is 'one' and has 3 letters. Sentence length: 659474
The 1000000th word is 'the' and has 3 letters. Sentence length: 7113621
The 10000000th word is 'thousand' and has 8 letters. Sentence length: 70995756
```

## Wren

Translation of: Kotlin
Library: Wren-fmt
```import "/fmt" for Fmt

var names = {
1: "one",
2: "two",
3: "three",
4: "four",
5: "five",
6: "six",
7: "seven",
8: "eight",
9: "nine",
10: "ten",
11: "eleven",
12: "twelve",
13: "thirteen",
14: "fourteen",
15: "fifteen",
16: "sixteen",
17: "seventeen",
18: "eighteen",
19: "nineteen",
20: "twenty",
30: "thirty",
40: "forty",
50: "fifty",
60: "sixty",
70: "seventy",
80: "eighty",
90: "ninety"
}

var bigNames = {
1e3 : "thousand",
1e6 : "million",
1e9 : "billion",
1e12: "trillion",
}

var irregOrdinals = {
"one"   : "first",
"two"   : "second",
"three" : "third",
"five"  : "fifth",
"eight" : "eighth",
"nine"  : "ninth",
"twelve": "twelfth"
}

var strToOrd = Fn.new { |s|
if (s == "zero") return "zeroth" // or alternatively 'zeroeth'
var splits = s.replace("-", " ").split(" ")
var last = splits[-1]
return irregOrdinals.containsKey(last) ? s[0...-last.count] + irregOrdinals[last] :
last.endsWith("y") ? s[0...-1] + "ieth" : s + "th"
}

var numToText = Fn.new { |n, uk|
if (n == 0) return "zero"
var neg = n < 0
var nn = neg ? - n : n
var digits3 = List.filled(6, 0)
for (i in 0..5) {  // split number into groups of 3 digits from the right
digits3[i] = nn % 1000
nn = (nn / 1000).truncate
}

var threeDigitsToText = Fn.new { |number|
var sb = ""
if (number == 0) return ""
var hundreds = (number / 100).truncate
var remainder = number % 100
if (hundreds > 0) {
sb = sb + names[hundreds] + " hundred"
if (remainder > 0) sb = sb + (uk ? " and " : " ")
}
if (remainder > 0) {
var tens = (remainder / 10).truncate
var units = remainder % 10
if (tens > 1) {
sb = sb + names[tens * 10]
if (units > 0) sb = sb + "-" + names[units]
} else {
sb = sb + names[remainder]
}
}
return sb
}

var strings = List.filled(6, 0)
for (i in 0..5) strings[i] = threeDigitsToText.call(digits3[i])
var text = strings[0]
var andNeeded = uk && 1 <= digits3[0] && digits3[0] <= 99
var big = 1000
for (i in 1..5) {
if (digits3[i] > 0) {
var text2 = strings[i] + " " + bigNames[big]
if (!text.isEmpty) {
text2 = text2 + (andNeeded ? " and " : " ")  // no commas inserted in this version
andNeeded = false
} else {
andNeeded = uk && 1 <= digits3[i] && digits3[i] <= 99
}
text = text2 + text
}
big = big * 1000
}
if (neg) text = "minus " + text
return text
}

var opening = "Four is the number of letters in the first word of this sentence,".split(" ")

var adjustedLength = Fn.new { |s|  s.replace(",", "").replace("-", "").count } // no ',' or '-'

var getWords = Fn.new { |n|
var words = []
if (n > opening.count) {
var k = 2
while (true) {
var len = adjustedLength.call(words[k - 1])
var text = numToText.call(len, false)
var splits = text.split(" ")
var text2 = strToOrd.call(numToText.call(k, false)) + ","  // add trailing comma
var splits2 = text2.split(" ")
if (words.count >= n) break
k = k + 1
}
}
return words
}

var getLengths = Fn.new { |n|
var words = getWords.call(n)
var lengths = words.take(n).map { |w| adjustedLength.call(w) }.toList
// includes hyphens, commas & spaces
var sentenceLength = words.reduce(0) { |acc, w| acc + w.count } + words.count - 1
return [lengths, sentenceLength]
}

var getLastWord = Fn.new { |n|
var words = getWords.call(n)
var nthWord = words[n - 1]
// includes hyphens, commas & spaces
var sentenceLength = words.reduce(0) { |acc, w| acc + w.count } + words.count - 1
return [nthWord, nthWordLength, sentenceLength]
}

var n = 201
System.print("The lengths of the first %(n) words are:\n")
var res = getLengths.call(n)
var list = res[0]
var sentenceLength = res[1]
for (i in 0...n) {
if (i % 25 == 0) {
if (i > 0) System.print()
Fmt.write("\$3d: ", i + 1)
}
Fmt.write("\$3d", list[i])
}
Fmt.print("\n\nLength of sentence = \$,d\n", sentenceLength)

n = 1000
while (true) {
var res = getLastWord.call(n)
var word = res[0]
var wLen = res[1]
var sLen = res[2]
if (word.endsWith(",")) word = word[0...-1]  // strip off any trailing comma
Fmt.print("The length of word \$,d [\$s] is \$d", n, word, wLen)
Fmt.print("Length of sentence = \$,d\n", sLen)
n = n * 10
if (n > 1e7) break
}
```
Output:
```The lengths of the first 201 words are:

1:   4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26:   3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51:   2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76:   3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101:  11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126:   4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151:   2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176:   3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201:  11

Length of sentence = 1,203

The length of word 1,000 [in] is 2
Length of sentence = 6,279

The length of word 10,000 [in] is 2
Length of sentence = 64,140

The length of word 100,000 [one] is 3
Length of sentence = 659,474

The length of word 1,000,000 [the] is 3
Length of sentence = 7,113,621

The length of word 10,000,000 [thousand] is 8
Length of sentence = 70,995,756
```

## zkl

Uses the nth function from Spelling_of_ordinal_numbers#zkl

```   // Built the sentence in little chucks but only save the last one
// Save the word counts
fcn fourIsThe(text,numWords){
const rmc="-,";
seq:=(text - rmc).split().apply("len").copy();  // (4,2,3,6...)
szs:=Data(numWords + 100,Int).howza(0).extend(seq);	// bytes
cnt,lastWords := seq.len(),"";
total:=seed.len() - 1;  // don't count trailing space

foreach idx in ([1..]){
sz:=szs[idx];
a,b := nth(sz,False),nth(idx+1);  // "two","three hundred sixty-seventh"
lastWords="%s in the %s, ".fmt(a,b);
ws:=lastWords.counts(" ")[1];  // "five in the forty-ninth " --> 4
cnt+=ws; total+=lastWords.len();
lastWords.split().pump(szs.append,'-(rmc),"len");
if(cnt>=numWords){
if(cnt>numWords){
z,n:=lastWords.len(),z-2;
do(cnt - numWords){ n=lastWords.rfind(" ",n) - 1; }
lastWords=lastWords[0,n+1]; total-=(z - n);
}
break;
}
}
return(lastWords.strip(),szs,total);
}
fcn lastWord(sentence){ sentence[sentence.rfind(" ")+1,*] }```
```var seed="Four is the number of letters in the first word of this sentence, ";
sentence,szs,total := fourIsThe(seed,201);
print("  1:");
foreach n,x in ([1..201].zip(szs)){
print("%3d".fmt(x));
if(0 == n%25) print("\n%3d:".fmt(n+1));
}
println("\nLength of above sentence: ",total);```
Output:
```  1:  4  2  3  6  2  7  2  3  5  4  2  4  8  3  2  3  6  5  2  3  5  3  2  3  6
26:  3  2  3  5  5  2  3  5  3  2  3  7  5  2  3  6  4  2  3  5  4  2  3  5  3
51:  2  3  8  4  2  3  7  5  2  3 10  5  2  3 10  3  2  3  9  5  2  3  9  3  2
76:  3 11  4  2  3 10  3  2  3 10  5  2  3  9  4  2  3 11  5  2  3 12  3  2  3
101: 11  5  2  3 12  3  2  3 11  5  2  3 11  3  2  3 13  5  2  3 12  4  2  3 11
126:  4  2  3  9  3  2  3 11  5  2  3 12  4  2  3 11  5  2  3 12  3  2  3 11  5
151:  2  3 11  5  2  3 13  4  2  3 12  3  2  3 11  5  2  3  8  3  2  3 10  4  2
176:  3 11  3  2  3 10  5  2  3 11  4  2  3 10  4  2  3 10  3  2  3 12  5  2  3
201: 11
Length of above sentence: 1203
```
```n:=1000; do(5){
sentence,x,total := fourIsThe(seed,n);
word:=lastWord(sentence);
println("%,d words: \"%s\" [%d]. Length=%,d"
.fmt(n,word,word.len(),total));
n*=10;
}```
Output:
```1,000 words: "in" [2]. Length=6,247
10,000 words: "in" [2]. Length=64,095
100,000 words: "one" [3]. Length=659,453
1,000,000 words: "the" [3]. Length=7,140,558
10,000,000 words: "thousand" [8]. Length=71,250,727
```