Hello Math-Fun, a(n) is the letter-length of a(n+1), written in English: S = 3, 6, 11, 23, 323, ... Could someone check and extend S (strictly monotonically increasing)? Always use the smallest necessary integer to extend S. Best, É.
More "constructively", let a(0)=3, and then define a(n+1) as the smallest number whose name written out in English has a(n) letters. I suspect it's easy to prove your parenthesized comment "strictly monotonically increasing". The entries are clearly correct up to a(3); I haven't checked to be sure that THREE HUNDRED TWENTY THREE is the smallest number whose name has 23 letters, but it seems pretty clear; the longest number names below 100 have 12 letters like SEVENTY THREE, and ONE HUNDRED or TWO HUNDRED don't supply enough letters to make it to 23. The next entry is the first hard one. The wordiest three-digit numbers have 24 letters, and 373 is the smallest exemplar. Thus, the smallest of the wordiest 30-digit numbers is 373,373,373,373,373,373,373,373,373 ("three hundred seventy three octillion ..."), and it has only 321 letters. So it is clear that we must break into the nonillions to find the smallest 323-letter number a(5). It's also clear that a(5) < 2e30, since ONE NONILLION, with 12 letters, is more than long enough to put us over the top. So a(5) = 1e30 + the smallest number with a 311-letter name. (I have a faint worry that I have committed a logical error to get here. Other eagle-eyed math-funsters should examine my reasoning carefully.) The smallest exemplar of the wordiest 27-digit numbers is 373,373,373,373,373,373,373,373,373, and it has 288 letters. This is 23 letters shy of 311; 9 of these letters are provided by the word OCTILLION, so it seems to me that a(5) = 1e30 + 1e27K + 373,373,373,373,373,373,373,373,373, where K is the smallest 14-letter number. This is ONE HUNDRED FOUR, but now at least one of my logical errors is exposed: ONE HUNDRED THREE is both smaller and longer, and perhaps there is a smaller 27-digit number with only 287 letters. In fact there is: it's the one that starts with 333. So: my present contender for a(5) is 1,103,333,373,373,373,373,373,373,373,373,373. One would need an extended naming system like Knuth's myllions or the one from The Book Of Numbers to make a stab at a(6), but this number would clearly consist of a prefix that might be as long as six digits slapped onto a long chain of 373's. On Wed, Jan 14, 2009 at 9:30 AM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Hello Math-Fun,
a(n) is the letter-length of a(n+1), written in English:
S = 3, 6, 11, 23, 323, ...
Could someone check and extend S (strictly monotonically increasing)? Always use the smallest necessary integer to extend S.
Best, É.
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This is iterating on A080777 ----- Original Message ----- From: "Allan Wechsler" <acwacw@gmail.com> To: "math-fun" <math-fun@mailman.xmission.com> Sent: Wednesday, January 14, 2009 2:10 PM Subject: Re: [math-fun] English sequence More "constructively", let a(0)=3, and then define a(n+1) as the smallest number whose name written out in English has a(n) letters. I suspect it's easy to prove your parenthesized comment "strictly monotonically increasing". The entries are clearly correct up to a(3); I haven't checked to be sure that THREE HUNDRED TWENTY THREE is the smallest number whose name has 23 letters, but it seems pretty clear; the longest number names below 100 have 12 letters like SEVENTY THREE, and ONE HUNDRED or TWO HUNDRED don't supply enough letters to make it to 23. The next entry is the first hard one. The wordiest three-digit numbers have 24 letters, and 373 is the smallest exemplar. Thus, the smallest of the wordiest 30-digit numbers is 373,373,373,373,373,373,373,373,373 ("three hundred seventy three octillion ..."), and it has only 321 letters. So it is clear that we must break into the nonillions to find the smallest 323-letter number a(5). It's also clear that a(5) < 2e30, since ONE NONILLION, with 12 letters, is more than long enough to put us over the top. So a(5) = 1e30 + the smallest number with a 311-letter name. (I have a faint worry that I have committed a logical error to get here. Other eagle-eyed math-funsters should examine my reasoning carefully.) The smallest exemplar of the wordiest 27-digit numbers is 373,373,373,373,373,373,373,373,373, and it has 288 letters. This is 23 letters shy of 311; 9 of these letters are provided by the word OCTILLION, so it seems to me that a(5) = 1e30 + 1e27K + 373,373,373,373,373,373,373,373,373, where K is the smallest 14-letter number. This is ONE HUNDRED FOUR, but now at least one of my logical errors is exposed: ONE HUNDRED THREE is both smaller and longer, and perhaps there is a smaller 27-digit number with only 287 letters. In fact there is: it's the one that starts with 333. So: my present contender for a(5) is 1,103,333,373,373,373,373,373,373,373,373,373. One would need an extended naming system like Knuth's myllions or the one from The Book Of Numbers to make a stab at a(6), but this number would clearly consist of a prefix that might be as long as six digits slapped onto a long chain of 373's. On Wed, Jan 14, 2009 at 9:30 AM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Hello Math-Fun,
a(n) is the letter-length of a(n+1), written in English:
S = 3, 6, 11, 23, 323, ...
Could someone check and extend S (strictly monotonically increasing)? Always use the smallest necessary integer to extend S.
Best, Ã.
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This type of question reminds me of my son's first experience learning to count: he got caught in an infinite loop! ONE TWO THREE FOUR FIVE SIX SEVEN EIGHT NINE TEN ELEVEN EIGHT NINE TEN ELEVEN EIGHT NINE etc. On Jan 14, 2009, at 9:30 AM, Eric Angelini wrote:
Hello Math-Fun,
a(n) is the letter-length of a(n+1), written in English:
S = 3, 6, 11, 23, 323, ...
Could someone check and extend S (strictly monotonically increasing)? Always use the smallest necessary integer to extend S.
Best, É.
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On Thursday 15 January 2009, Veit Elser wrote:
This type of question reminds me of my son's first experience learning to count: he got caught in an infinite loop! ... EIGHT NINE TEN ELEVEN EIGHT NINE ...
My daughter recently went 1,2,3,...,13,14,16,16,16,16,16,... . I assume some 4-bit counter overflowed and scrobbled a linked-list pointer, or something. -- g
participants (5)
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Allan Wechsler -
David Wilson -
Eric Angelini -
Gareth McCaughan -
Veit Elser