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A181868
a(1) = 10, a(2) = 1. For n >= 3, a(n) = concatenate a(n-1), a(n-2), ..., a(1) and then divide the resulting number by a(n-1).
10
10, 1, 110, 1001, 1000110, 10000001001, 100000000001000110, 10000000000000000010000001001, 10000000000000000000000000000100000000001000110, 1000000000000000000000000000000000000000000000010000000000000000010000001001
OFFSET
1,1
COMMENTS
Compare with A181756.
The calculations for the first few values of the sequence are
... a(3) = 110/1 = 110
... a(4) = 110110/110 = 1001
... a(5) = 1001110110 /1001 = 1000110
... a(6) = 10001101001110110/1000110 = 10000001001.
The above calculations are in base 10, but we get exactly the same results when working in an arbitrary base b with initial values a(1) = 10 (= b in base b), a(2) = 1.
For similarly defined sequences see A181754 through A181756 and A181864 through A181870.
FORMULA
a(n+2) = a(n) + 10^(A000032(n+1)-1), n>=1.
a(n) has A000032(n-1) digits.
MAPLE
M:=10:
a:=array(1..M):s:=array(1..M):
a[1]:=10:a[2]:=1:
s[1]:=convert(a[1], string):
s[2]:=cat(convert(a[2], string), s[1]):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(convert(a[n], string), s[n-1]);
end do:
seq(a[n], n = 1..M);
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Nov 28 2010
STATUS
approved