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a(1)=0, a(1)=1, a(n+2)=abs(concatenate(a(n+1)a(n))-concatenate(a(n)a(n+1))).
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%I #14 Jan 01 2024 13:57:24

%S 0,1,9,72,243,47871,23523372,2434786275501,8244905115337247871,

%T 58101188398354233807319449027630,

%U 243478627550182449084906698122045988902204111779759,33753325643335988898828779215425644588407139004473126805509723691755094884662752129

%N a(1)=0, a(1)=1, a(n+2)=abs(concatenate(a(n+1)a(n))-concatenate(a(n)a(n+1))).

%C a(n)==0 mod 3 if n>2. Is a(n) always of the form 2^a*3^b*b(n) where b(n) is a squarefree number? As example : a(12)=3^12*11*192263*58877057*6250682413*588631991107100965223

%e a(2)=72 a(3)=243 then a(4)=abs(24372-72243)=47871

%t nxt[{a_,b_}]:=Module[{ida=IntegerDigits[a],idb=IntegerDigits[b]},{b,Abs[ FromDigits[ Join[ ida,idb]]-FromDigits[Join[idb,ida]]]}]; Transpose[ NestList[ nxt,{0,1},13]] [[1]] (* _Harvey P. Dale_, Sep 19 2014 *)

%K easy,nonn,base

%O 1,3

%A _Benoit Cloitre_, May 15 2002

%E One more term (a(12)) from _Harvey P. Dale_, Sep 19 2014