A261147 English flagpole sequence of nonnegative integers (like A215693, but with a(1)=0).

0, 1, 3, 2, 4, 7, 5, 8, 13, 14, 9, 10, 12, 20, 21, 15, 22, 24, 23, 6, 11, 17, 16, 18, 25, 19, 27, 26, 28, 29, 38, 30, 31, 32, 33, 34, 35, 36, 37, 39, 41, 40, 42, 44, 43, 45, 47, 48, 49, 51, 46, 57, 53, 50, 55, 59, 52, 62, 58, 111, 65, 61, 54, 72, 63, 67, 56

Offset: 1

Author

Danny Rorabaugh, Nov 28 2015

Keywords

• nonn
• word

Comments

Concatenate the spelling of the terms in English (ignoring "and", punctuation, capitalization, and spaces); the i-th letter of this infinite word is in a(i). This is the lexicographically least such sequence of distinct nonnegative integers. A215693 is the lexicographically least such sequence of distinct positive integers.

In the first 100000 terms: the records for a(n)-n are set by a(1)=0, a(3)=3, a(6)=7, a(9)=13, a(14)=20, a(31)=38, a(60)=111, a(119)=211, a(18617)=18711; the records for n-a(n) are set by a(1)=0, a(4)=2, a(20)=6, a(335)=319.

Examples

			The sequence begins: 0,1,3,2,4,7,5,8,13....
The initial terms are spelled: ZERO;ONE;THREE;TWO;....
These letters are found in the spelling of the initial terms: Zero, onE, thRee, twO; fOur, seveN, fivE; eighT, tHirteen, ....
		

Links

• Danny Rorabaugh, Table of n, a(n) for n = 1..3000
• Wikipedia, English numerals
• Hans Havermann, List of a(n) for n = 1..100000

Crossrefs

Cf. A215693.

Programs

Sage

little = {0:"", 1:"one", 2:"two", 3:"three", 4:"four", 5:"five", 6:"six", 7:"seven", 8:"eight", 9:"nine", 10:"ten", 11:"eleven", 12:"twelve", 13:"thirteen", 14:"fourteen", 15:"fifteen", 16:"sixteen", 17:"seventeen", 18:"eighteen", 19:"nineteen"}
decade = {2:"twenty", 3:"thirty", 4:"forty", 5:"fifty", 6:"sixty", 7:"seventy", 8:"eighty", 9:"ninety"}
illion = {1:"thousand", 2:"million", 3:"billion", 4:"trillion"}
def nmb_wrd(n): # For positive integers n<10^15
    if n<20: return little[n]
    if n<100: return decade[floor(n/10)] + little[n%10]
    if n<1000: return little[floor(n/100)] + "hundred" + nmb_wrd(n%100)
    k = floor((len(str(n))-1)/3)
    return nmb_wrd(floor(n/10^(3*k))) + illion[k] + nmb_wrd(n%(10^(3*k)))
def English_fp(n, A = [0,1,3,2], i = 1):
    while len(A)A215693 is English_fp(113, [1,7,3])

Extensions

Corrected by Danny Rorabaugh, Nov 29 2015