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A321676
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a(1)=1; for n > 1, a(n) is a(n-1) plus the number of vowels in the name of a(n-1) in US English.
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0
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1, 3, 5, 7, 9, 11, 14, 18, 22, 24, 27, 30, 31, 34, 37, 40, 41, 44, 47, 50, 51, 54, 57, 60, 61, 64, 67, 70, 72, 75, 79, 83, 87, 91, 95, 99, 103, 109, 115, 122, 128, 135, 142, 148, 155, 162, 168, 175, 183, 191, 199, 207, 212, 217, 224, 230, 234, 240, 244, 250
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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This sequence takes into account the numbers written as words; for example, "fifty-seven" contains three vowels, so 3 is added to 57 to create the next term. The word "and" is not included in US English (cf. A158352), so 115 is written as "one hundred fifteen". This sequence is puzzling as it shares its first 6 terms with the odd numbers before jumping to 14, then 18. When only given the first 8 terms it can be very difficult to spot the rule.
Assumes "y" in e.g. "fifty" is not a vowel. - Chai Wah Wu, Dec 17 2018
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LINKS
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Table of n, a(n) for n=1..60.
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FORMULA
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a(n) = A139282(n) for n >= 2. - Chai Wah Wu, Dec 17 2018
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MATHEMATICA
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vowelCount[n_] := StringCount[IntegerName[n, "Words"], {"a", "e", "i", "o", "u"}]; f[n_] := n + vowelCount[n]; NestList[f, 1, 100] (* Amiram Eldar, Dec 10 2018 *)
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PROG
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(Python 3.3)
from itertools import accumulate, repeat
from num2words import num2words
A321676_list = list(accumulate(repeat(1, 100), lambda x, _ : x+sum(num2words(x).count(d) for d in 'aeiou'))) # Chai Wah Wu, Dec 17 2018
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CROSSREFS
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Cf. A005589, A037196, A139282, A158352.
Sequence in context: A318919 A279539 A174059 * A268292 A240992 A165704
Adjacent sequences: A321673 A321674 A321675 * A321677 A321678 A321679
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KEYWORD
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nonn,word,easy
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AUTHOR
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Andrew Toothill, Dec 02 2018
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EXTENSIONS
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More terms from Amiram Eldar, Dec 10 2018
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STATUS
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approved
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