OFFSET
1,1
COMMENTS
This is defined to be the lexicographically earliest sequence with the following property:
If a(n) = 0, underline a(n+1); if a(n) = 1, underline a(n+2); if a(n) = 2, underline a(n+3); if a(n) = 3, underline a(n+4); if a(n) = 4, underline a(n+5); if a(n) = 5, underline a(n+6); if a(n) = 6, underline a(n+7); if a(n) = 7, underline a(n+8); if a(n) = 8, underline a(n+9); if a(n) = 9, underline a(n+10). Now, the subsequence of (once or more) underlined terms must be equal to the original sequence (copy #1), and the subsequence of non-underlined terms must also reproduce the original sequence (copy #2).
LINKS
Carole Dubois, Table of n, a(n) for n = 1..2007
EXAMPLE
The sequence starts (9,8,7,6,5,4,3,2,1,0,9,9,9,9,9,9,9,9,9,9,8,7,6,5,4,3,2,1,0,9,9,...)
Instead of underlining terms, we will put parentheses around the terms we want to emphasize:
a(1) = 9 produces parentheses around a(1 + 10 = 11):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,8,7,6,5,4,3,2,1,0,9,...
a(2) = 8 produces parentheses around a(2 + 9 = 11), which is now already done. Then,
a(3) = 7 produces parentheses around a(3 + 8 = 11), which is already done. Then,
a(4) = 6 produces parentheses around a(4 + 7 = 11) - already done. Then,
a(5) = 5 produces parentheses around a(5 + 6 = 11) - already done. Then,
a(6) = 4 produces parentheses around a(6 + 5 = 11) - already done. Then,
a(7) = 3 produces parentheses around a(7 + 4 = 11) - already done. Then,
a(8) = 2 produces parentheses around a(8 + 3 = 11) - already done. Then,
a(9) = 1 produces parentheses around a(9 + 2 = 11) - already done. Then,
a(10) = 0 produces parentheses around a(10 + 1 = 11) - already done. Now,
a(11) = 9 produces parentheses around a(11 + 10 = 21):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),7,6,5,4,3,2,1,0,9,...
a(12) = 9 produces parentheses around a(12 + 10 = 22):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),6,5,4,3,2,1,0,9,...
a(13) = 9 produces parentheses around a(13 + 10 = 23):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),5,4,3,2,1,0,9,...
a(14) = 9 produces parentheses around a(14 + 10 = 24):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),4,3,2,1,0,9,...
a(15) = 9 produces parentheses around a(15 + 10 = 25):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),(4),3,2,1,0,9,...
a(16) = 9 produces parentheses around a(16 + 10 = 26):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),(4),(3),2,1,0,9,...
a(17) = 9 produces parentheses around a(17 + 10 = 27):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),(4),(3),(2),1,0,9,...
a(18) = 9 produces parentheses around a(18 + 10 = 28):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),(4),(3),(2),(1),0,9,...
a(19) = 9 produces parentheses around a(19 + 10 = 29):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),(4),(3),(2),(1),(0),9,...
a(20) = 9 produces parentheses around a(20 + 10 = 30):
9,8,7,6,5,4,3,2,1,0,(9),9,9,9,9,9,9,9,9,9,(8),(7),(6),(5),(4),(3),(2),(1),(0),(9)...
Etc.
We see in this small example that the parenthesized terms reproduce the initial sequence:
(9),(8),(7),(6),(5),(4),(3),(2),(1),(0),(9)...
The same is true for the subsequence of non-parenthesized terms:
9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 9, 9, 9, 9, 9, 9, 9, 9, 9,...
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Eric Angelini and Carole Dubois, Apr 02 2019
STATUS
approved