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A181867
a(1) = 2, a(2) = 1. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence in reverse order and then dividing the resulting number by a(n-1).
9
2, 1, 12, 101, 10012, 10000101, 1000000010012, 100000000000010000101, 1000000000000000000001000000010012, 1000000000000000000000000000000000100000000000010000101
OFFSET
1,1
COMMENTS
Compare with A181754. Here we concatenate the terms of the sequence in reverse order before dividing by a(n-1).
The calculations for the first few values of the sequence are
... a(3) = 12/1 = 12
... a(4) = 1212/12 = 101
... a(5) = 1011212/101 = 10012
... a(6) = 100121011212/10012 = 10000101.
For similarly defined sequences see A181864 through A181870.
FORMULA
DEFINITION
a(1) = 2, a(2) = 1, and for n >= 3
(1)... a(n) = concatenate(a(n-1),a(n-2),...,a(1))/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = a(n) + 10^(F(n)-1),
where F(n) = A000045(n) are the Fibonacci numbers.
a(n) has F(n) digits.
MAPLE
M:=10:
a:=array(1..M):s:=array(1..M):
a[1]:=2: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,base
AUTHOR
Peter Bala, Nov 28 2010
STATUS
approved