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

1,2

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.

Links

Table of n, a(n) for n=1..10.

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.

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);

Crossrefs

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

Sequence in context: A041917 A318089 A109280 * A287874 A064841 A286200

Adjacent sequences: A181753 A181754 A181755 * A181757 A181758 A181759

Keyword

easy,nonn,base

Author

Peter Bala, Nov 09 2010

Status

approved