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a(n+1) = a(n) + (a(n))^3 with a(0)=1.
1

%I #23 Sep 08 2022 08:45:07

%S 1,2,10,1010,1030302010,1093688489093570254240903010,

%T 1308220420050293140207308568343275558610275754246390852904259630756188515781804010

%N a(n+1) = a(n) + (a(n))^3 with a(0)=1.

%C a(6) contains 82 digits. - _Harvey P. Dale_, Feb 19 2011 [corrected by _Altug Alkan_, Dec 22 2015]

%C For n > 2, the last three digits are always "010". - _Altug Alkan_, Dec 22 2015

%C a(7) has 244 digits, a(8) has 731. - _Robert Israel_, Dec 23 2015

%H Robert Israel, <a href="/A074333/b074333.txt">Table of n, a(n) for n = 0..8</a>

%p a[0]:= 1:

%p for n from 1 to 8 do a[n]:= a[n-1] + a[n-1]^3 od:

%p seq(a[i],i=0..8); # _Robert Israel_, Dec 23 2015

%t NestList[#+#^3&,1,8] (* _Harvey P. Dale_, Feb 19 2011 *)

%o (PARI) a(n) = if(n==0, 1, a(n-1)+(a(n-1))^3); \\ _Altug Alkan_, Dec 22 2015

%o (Magma) [n le 1 select 1 else Self(n-1)+Self(n-1)^3: n in [1..10]]; // _Vincenzo Librandi_, Dec 23 2015

%K nonn,easy

%O 0,2

%A Emrehan Halici (emrehan(AT)halici.com.tr), Sep 22 2002

%E a(5) from _Harvey P. Dale_, Feb 19 2011

%E a(6) from _Robert Israel_, Dec 23 2015