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A178823
a(1) = 1, a(n+1) = least k >= a(n) such that the sum of the number of letters in the English name of all values in the sequence through a(n), excluding spaces and hyphens (A005589), is prime.
0
1, 4, 5, 11, 12, 13, 24, 73, 1103, 1115, 1117, 1117, 1117, 1117, 1117, 1140, 1144, 1201, 1217, 1217, 1323, 1326, 1340, 1344, 1374, 1413, 1413, 1413, 1413, 1424, 1441, 1441, 1480, 1484
OFFSET
1,2
EXAMPLE
a(1) = 1 by definition.
a(2) = 4 because "one" plus "four" has 3 + 4 = 7 letters, with 7 prime.
a(3) = 5 because "one" plus "four" plus "five" gives 3 + 4 + 4 = 11, a prime.
a(4) = 11 because "one" plus "four" plus "five" plus "eleven" gives 3 + 4 + 4 + 6 = 17 is prime.
a(5) = 12 because "one" plus "four" plus "five" plus "eleven" plus "twelve" gives 3 + 4 + 4 + 6 + 6 = 23 is prime.
a(6) = 13 because "one" plus "four" plus "five" plus "eleven" plus "twelve" plus "thirteen" gives 3 + 4 + 4 + 6 + 6 + 8 = 31 is prime.
a(7) = 24 because "one" plus "four" plus "five" plus "eleven" plus "twelve" plus "thirteen" plus "twentyfour" gives 3 + 4 + 4 + 6 + 6 + 8 + 10 = 41 is prime.
a(8) = 73 because "one" plus "four" plus "five" plus "eleven" plus "twelve" plus "thirteen" plus "twentyfour" plus "seventythree" gives 3 + 4 + 4 + 6 + 6 + 8 + 10 + 12 = 53 is prime.
a(9) = 1103 because "one" plus "four" plus "five" plus "eleven" plus "twelve" plus "thirteen" plus "twentyfour" plus "seventythree" plus "one thousand one hundred three" gives 3 + 4 + 4 + 6 + 6 + 8 + 10 + 12 + 26 = 79 is prime.
a(10) = 1115 "one" plus "four" plus "five" plus "eleven" plus "twelve" plus "thirteen" plus "twentyfour" plus "seventythree" plus "one thousand one hundred three" plus "one thousand one hundred fifteen" gives 3 + 4 + 4 + 6 + 6 + 8 + 10 + 12 + 26 + 28 = 107 is prime.
CROSSREFS
KEYWORD
nonn,easy,word
AUTHOR
Jonathan Vos Post, Dec 26 2010
EXTENSIONS
a(11)-a(34) from Nathaniel Johnston, Jan 04 2011
STATUS
approved