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a(n) = a(n-1) * a(n-2) - 1.
4

%I #46 Sep 08 2022 08:45:01

%S 2,2,3,5,14,69,965,66584,64253559,4278258972455,274893365303916717344,

%T 1176065006579831683931038116759519,

%U 323292467474902888912311701915651931811127805144397535

%N a(n) = a(n-1) * a(n-2) - 1.

%H Vincenzo Librandi, <a href="/A055937/b055937.txt">Table of n, a(n) for n = 1..19</a>

%F a(n) ~ c^(phi^n), where c = 1.26679081808631741720378389711... and phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - _Vaclav Kotesovec_, Jun 15 2019

%t RecurrenceTable[{a[1]==a[2]==2, a[n]==a[n-1]*a[n-2] -1}, a, {n, 15}] (* _Vincenzo Librandi_, Sep 30 2015 *)

%t nxt[{a_,b_}]:={b,a*b-1}; NestList[nxt,{2,2},15][[All,1]] (* _Harvey P. Dale_, Dec 05 2020 *)

%o (Magma) I:=[2,2]; [n le 2 select I[n] else Self(n-1)*Self(n-2)-1: n in [1..15]]; // _Vincenzo Librandi_, Sep 30 2015

%o (PARI) a(n) = if(n<3, 2, a(n-1)*a(n-2)-1);

%o vector(15, n, a(n)) \\ _Altug Alkan_, Sep 30 2015

%o (Sage)

%o @cached_function

%o def a(n):

%o if n == 1 or n == 2:

%o return 2

%o else:

%o return a(n - 1) * a(n - 2) - 1

%o [a(n) for n in range(1, 16)] # _G. C. Greubel_, Jun 07 2019

%Y A152446 uses largest prime < a(n-1)*a(n-2) instead of subtracting 1. - _Joshua D. Olson_, Sep 29 2015

%Y Cf. A007660, A121258.

%K easy,nonn

%O 1,1

%A _Robert G. Wilson v_, Jul 05 2000