OFFSET
0,2
COMMENTS
A rotation of the decimal digits to the left puts the leading digit in the rightmost position (123456 becomes 234561). The (even) integers that are cut in half by this operation form 10 families; a singleton {a(0)=0} and 9 infinite families, each consisting of any number of repetitions of an 18-digit pattern.
Each of those families corresponds to a single decadic solution of the equation x=2(10x+d) where d is a digit from 0 to 9 (namely x=-2d/19 which is a periodic decadic).
This sequence is strongly related to A146088:
A146088(8k+1) = a(9k+2)/2 = a(9k+1)
A146088(8k+2) = a(9k+3)/2
A146088(8k+3) = a(9k+4)/2 = a(9k+2)
A146088(8k+4) = a(9k+5)/2
A146088(8k+5) = a(9k+6)/2 = a(9k+3)
A146088(8k+6) = a(9k+7)/2
A146088(8k+7) = a(9k+8)/2 = a(9k+4)
A146088(8k+8) = a(9k+9)/2
Note that a(9k+1)/2 is not part of A146088, because rotating one place to the left a decimal expansion starting with 105263157894736842 yields a discarded zero leading digit which is not retrieved by rotating the digits to the right (right-rotation would double the pattern 052631578947368421 but not the number 52631578947368421).
Lists normally have offset 1, but there is a good reason to make an exception in this case. - N. J. A. Sloane, Dec 24 2012
LINKS
G. P. Michon, Numbers cut in half by rotating digits left (2006).
FORMULA
a(j)=j*105263157894736842 for j = 0 to 9
a(n+9)=10^18*a(n) + a(n) mod 10^18 for n>0
a(9k-9+j) = j*a(1)*(10^18k-1)/(10^18-1) for j=1..9 and k>0
e.g., a(1000) = a(9*112-9+1) = a(1)*(10^2016-1)/(10^18-1)
EXAMPLE
a(0) = 0 because rotating a lone zero gives 0 = 0/2.
a(1) = 105263157894736842 because its half equals 052631578947368421 (discarding leading 0).
a(3) = 315789473684210526 since a(3)/2 = 157894736842105263.
CROSSREFS
KEYWORD
base,nonn,easy
AUTHOR
Gerard P. Michon, Oct 28 2012
STATUS
approved