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).

2, 1, 12, 101, 10012, 10000101, 1000000010012, 100000000000010000101, 1000000000000000000001000000010012, 1000000000000000000000000000000000100000000000010000101

Offset: 1

Author

Peter Bala, Nov 28 2010

Keywords

• nonn
• easy
• base

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.

Crossrefs

A000045, A181754, A181755, A181756, A181864, A181865, A181866, A181868, A181869, A181870

Programs

Maple

#A181867
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);

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.