A181754 a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).

1, 2, 6, 21, 601, 21001, 60100001, 2100100000001, 601000010000000000001, 2100100000001000000000000000000001, 6010000100000000000010000000000000000000000000000000001

Offset: 1

Author

Peter Bala, Nov 09 2010

Keywords

• easy
• nonn
• base

Comments

The calculations for the first few values of the sequence are

... a(3) = 12/2 = 6

... a(4) = 126/6 = 21

... a(5) = 12621/21 = 601

... a(6) = 12621601/601 = 21001.

Similar sequences may be formed by

1) starting with different initial values. See A181755 and A181756.

2) concatenating the k-th powers of the first n-1 terms of the sequence before dividing by a(n-1). See A181864, A181865 and A181866.

3) concatenating the k-th powers of the first n-1 terms of the sequence in reverse order before dividing by a(n-1). See A181867, A181868, A181869 and A181870.

Crossrefs

Cf. A000045, A181755, A181756, A181864, A181865, A181866, A181867, A181868, A181869, A181870

Programs

Maple

#A181754
M:=11:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1],string):
s[2]:=cat(s[1],convert(a[2],string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1],convert(a[n],string));
end do:
seq(a[n],n = 1..M);

Formula

DEFINITION

a(1) = 1, a(2) = 2, and for n >= 3

(1)... a(n) = concatenate(a(1),a(2),...,a(n-1))/a(n-1).

RECURRENCE RELATION

For n >= 2

(2)... a(n+2) = 10^F(n)*a(n)+1,

where F(n) = A000045(n) are the Fibonacci numbers.

For n >= 2, a(n) has F(n-1) digits.