A181756 a(1) = 1, a(2) = 10. 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, 10, 11, 1001, 110001, 1001000001, 1100010000000001, 10010000010000000000000001, 110001000000000100000000000000000000000001, 10010000010000000000000001000000000000000000000000000000000000000001

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) = 110/10 = 11

... a(4) = 11011/11 = 1001

... a(5) = 110111001/1001 = 110001.

The above calculations are in base 10, but we get exactly the same results when working in an arbitrary base b. For example, with starting values a(1) = 1, a(2) = b (= 10 in base b), we get

... a(3) = (b^2+b)/b = b+1 which equals 11 in base b,

... a(4) = (b^4+b^3+b+1)/(b+1) = b^3+1 = 1001 in base b,

and so on.

For similarly defined sequences see A181754, A181755 and A181864 through A181870.

Crossrefs

Cf. A000045, A181754, A181755, A181864, A181865, A181866, A181867, A181868, A181869, A181870

Programs

Maple

#A181756
M:=10:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=10:
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) = 10, 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) = 100^F(n)*a(n)+1,

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

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