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.

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

Author

Jonathan Vos Post, Dec 26 2010

Keywords

• nonn
• easy
• word

Examples

			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

Cf. A006944 (ordinals), A052360, A052362-A052363, A005589, A134629, A133418, A016037

Extensions

a(11)-a(34) from Nathaniel Johnston, Jan 04 2011