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%I #8 Aug 08 2014 15:15:29
%S 1,5,3,51,301,51001,30100001,5100100000001,301000010000000000001,
%T 5100100000001000000000000000000001,
%U 3010000100000000000010000000000000000000000000000000001
%N 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).
%C The calculations for the first few values of the sequence are
%C ... a(3) = 15/5 = 3
%C ... a(4) = 153/3 = 51
%C ... a(5) = 15351/51 = 301
%C ... a(6) = 15351301/301 = 51001.
%C For similarly defined sequences see A181754, A184756 and A181864 through A181870.
%F DEFINITION
%F a(1) = 1, a(2) = 5, and for n >= 3
%F (1)... a(n) = concatenate(a(1),a(2),...,a(n-1))/a(n-1).
%F RECURRENCE RELATION
%F For n >= 2
%F (2)... a(n+2) = 10^F(n)*a(n)+1,
%F where F(n) = A000045(n) are the Fibonacci numbers.
%F For n >= 2, a(n) has F(n-1) digits.
%p #A181755
%p M:=11:
%p a:=array(1..M):s:=array(1..M):
%p a[1]:=1:a[2]:=5:
%p s[1]:=convert(a[1],string):
%p s[2]:=cat(s[1],convert(a[2],string)):
%p for n from 3 to M do
%p a[n] := parse(s[n-1])/a[n-1];
%p s[n]:= cat(s[n-1],convert(a[n],string));
%p end do:
%p seq(a[n],n = 1..M);
%t nxt[lst_]:=Module[{nt=FromDigits[Flatten[IntegerDigits/@lst]]/Last[ lst]},Flatten[{lst,nt}]]; Nest[nxt[#]&,{1,5},10] (* _Harvey P. Dale_, Aug 08 2014 *)
%Y Cf. A000045, A181754, A181756, A181864, A181865, A181866, A181867, A181868, A181869, A181870
%K easy,nonn,base
%O 1,2
%A _Peter Bala_, Nov 09 2010
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