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a(1)=1; then for n even, a(n)=(sum of previous terms times n) plus 1, for n odd, a(n)=(sum of previous terms times n) minus 1.
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%I #15 Sep 28 2018 23:35:18

%S 1,3,11,61,379,2731,22301,203897,2064455,22938391,277554529,

%T 3633441109,51170962283,771500662115,12399117783989,211611610180081,

%U 3822234708877711,72847296804492847,1460993008134550985

%N a(1)=1; then for n even, a(n)=(sum of previous terms times n) plus 1, for n odd, a(n)=(sum of previous terms times n) minus 1.

%C Conjecture: There are infinitely many primes in this sequence.

%C The sequence would have been a little nicer if the even terms had a minus one and the odd a plus one, so the first term would not have to be an exception.

%C Except for the first two terms, it appears that a(n) are the first differences of A002467. - _Carl Najafi_, Sep 27 2018

%H Muniru A Asiru, <a href="/A095237/b095237.txt">Table of n, a(n) for n = 1..100</a>

%F a(n) = (n+1)! - floor(((n+1)!+1)/e) - n! + floor((n!+1)/e), n > 1. - _Gary Detlefs_, Nov 07 2010

%p Digits:=100: a:=n->factorial(n+1)-floor((factorial(n+1)+1)/exp(1))-factorial(n)+floor((factorial(n)+1)/exp(1)): 1,seq(a(n),n=2..20); # _Muniru A Asiru_, Sep 28 2018

%o (PARI) a=vector(100) s=1 for(i=2,100,if(Mod(i,2)==0,a[i]=s*i+1,a[i]=s*i-1);s+=a[i])

%Y Cf. A095236.

%K nonn

%O 1,2

%A _Amarnath Murthy_, Jun 13 2004

%E Edited by _Johan Claes_, Jun 16 2004