A181755 a(1) = 1, a(2) = 5. 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, 5, 3, 51, 301, 51001, 30100001, 5100100000001, 301000010000000000001, 5100100000001000000000000000000001, 3010000100000000000010000000000000000000000000000000001

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) = 15/5 = 3

... a(4) = 153/3 = 51

... a(5) = 15351/51 = 301

... a(6) = 15351301/301 = 51001.

For similarly defined sequences see A181754, A184756 and A181864 through A181870.

Crossrefs

Cf. A000045, A181754, A181756, A181864, A181865, A181866, A181867, A181868, A181869, A181870

Programs

Maple

#A181755
M:=11:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=5:
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);

Mathematica

nxt[lst_]:=Module[{nt=FromDigits[Flatten[IntegerDigits/@lst]]/Last[ lst]},Flatten[{lst,nt}]]; Nest[nxt[#]&,{1,5},10] (* Harvey P. Dale, Aug 08 2014 *)

Formula

DEFINITION

a(1) = 1, a(2) = 5, 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.