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A306559
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A self-describing sequence mostly made of 1's and 0's emerging when written in English words (see the Comments section for an explanation).
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1
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1, 2, 3, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0
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OFFSET
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1,2
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COMMENTS
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We start with 1, 2, 3. From this point on, the sequence is extended by the rule that when written in English words, the lengths of runs of consonants between successive vowels is the sequence itself. The sequence is ONE, TWO, THREE, ZERO, ONE, ONE, ZERO, ONE, ZERO, ONE, ONE, ONE, ZERO, ONE, ONE, ONE, ZERO, ...
Those runs of consonants are better viewed with commas and spaces deleted: ONETWOTHREEZEROONEONEZEROONEZEROONEONEONEZEROONEONEONEZERO...
Note that there is no (thus ZERO) consonant between the two E's of THREE.
Jean-Marc Falcoz has computed 100000 terms and found no place where the sequence enters into a loop. It is conjectured that the sequence will grow infinitely without ever entering into such a loop.
The proportions of ONEs and ZEROs are conjectured to be 2/3 and 1/3.
_Alan C. Wechsler_ (Feb 23 2019) observes that if the first four terms are deleted, this appears to coincide with A039982. Indeed, the resulting pair of sequences agree for at least the first 28442 terms, so this is a very plausible conjecture. - N. J. A. Sloane, Feb 23 2019
After deleting the first three terms, this sequence is 0*W(0)*W^2(0)*W^3(0)*... where W is the morphism 0 -> 11, 1 -> 01. It can be shown that this is equivalent to the above conjecture (see link for proof). - Charlie Neder, Mar 04 2019
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LINKS
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EXAMPLE
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The #1 vowel (O) and #2 vowel (E) enclose indeed ONE consonant (N);
the #2 vowel (E) and #3 vowel (O) enclose indeed TWO consonants (TW);
the #3 vowel (O) and #4 vowel (E) enclose indeed THREE consonants (THR);
the #4 vowel (E) and #5 vowel (E) enclose indeed ZERO consonant;
the #5 vowel (E) and #6 vowel (E) enclose indeed ONE consonant (Z);
the #6 vowel (E) and #7 vowel (O) enclose indeed ONE consonant (R);
the #7 vowel (O) and #8 vowel (O) enclose indeed ZERO consonant;
etc.
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CROSSREFS
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KEYWORD
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base,nonn,word
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AUTHOR
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STATUS
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approved
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