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a(n) = (n-1)*a(n-1) - n*a(n-2), with a(1) = a(2) = -1.
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%I #21 Jun 05 2023 07:36:53

%S -1,-1,1,7,23,73,277,1355,8347,61573,523913,5024167,53479135,

%T 624890417,7946278813,109195935523,1612048228547,25439293045885,

%U 427278358483537,7609502950269503,143217213477235783,2840152418116022377

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

%H Mohammed Bouras, <a href="http://www.ssmrmh.ro/2022/08/15/a-new-sequence-of-prime-numbers/">A new sequence of prime numbers</a>, Romanian Math. Mag. (15 August 2022). See Table 1 p. 1.

%F a(n) = A051403(n-3) + n*A051403(n-4).

%F a(n)/(n^2 - n - 1) = 1/(2-3/(3-4/(4-5/(...(n-1)-n/(n-(n+1)))))), for n >= 2.

%t a[1]=a[2]=-1; a[n_]:=a[n]=(n-1)a[n-1]-n a[n-2]; Array[a,22] (* _Stefano Spezia_, Aug 23 2022 *)

%o (PARI) a(n) = if(n<=2, -1, 1/2*((n-1)*sum(k=0,n-3,k!) + n*(n-2)*sum(k=0,n-4,k!))) \\ _Jianing Song_, Oct 15 2022, following the formula above

%Y Cf. A051403, A356247.

%K sign,easy

%O 1,4

%A _Mohammed Bouras_, Aug 22 2022