A296397 - OEIS

%I #22 Mar 08 2021 09:20:00

%S 1,1,2,3,7,23,173,4231,772535,3430031573,2767984611331999,

%T 9880508763685677890784167,28372546978138838124644984908123272195533,

%U 290052121708444744262218759616469916140851065875997330620050069911

%N a(n) = a(n-1) * a(n-2) + a(n-3) * Product_{k=0..n-4} a(k)^2, with a(0) = a(1) = 1, a(2) = 2.

%C The recurrence for b(n) is similar to Fibonacci except for the reciprocal.

%C An infinite coprime sequence defined by recursion. - _Michael Somos_, Dec 14 2017

%H Michael Penn, <a href="https://youtu.be/T_lHqiW81sw?t=377">2 viewer suggested recursive sequences.</a>, YouTube video, 2021.

%H Larry Powell, <a href="http://math.stackexchange.com/questions/2561924">Any insight in the half reciprocal Fibonacci sequence?</a> Math StackExchange, Dec 11 2017.

%F a(n) = a(n-1) * a(n-2) + a(n-1) * a(n-3) * a(n-4) - a(n-2) * a(n-3)^2 * a(n-4) for all n>=4.

%F a(n) = numerator of b(n) where b(0) = b(1) = 1, b(n) = b(n-1) + 1/b(n-2).

%t a[ n_] := Which[ n < 1, Boole[n == 0], n < 4, n, True, a[n - 1] a[n - 2] + a[n - 3] Product[ a[k], {k,0, n - 4}]^2];

%t Numerator@ RecurrenceTable[{a[n] == a[n - 1] + 1/a[n - 2], a[0] == a[1] == 1}, a, {n, 0, 13}] (* _Robert G. Wilson v_, Dec 11 2017 *)

%o (PARI) {a(n) = if( n<1, n==0, n<4, n, a(n-1) * a(n-2) + a(n-3) * prod(k=0, n-4, a(k))^2)};

%K nonn

%O 0,3

%A _Michael Somos_, Dec 11 2017