%I
%S 1,1,1,0,2,3,3,10,28,279,7803,2177000,16987130758,
%T 36980983660158439,628200804994572838287982201,
%U 23231483704802676028750227275477328286998042,14594036764575342428539025427350979161630036659925283421091485142638200
%N a(1) = a(2) = 1; a(n) = a(n1)*a(n2)  a(n3)  a(n4)  ...  a(1) for n>2.
%C Form the product of the previous two terms and then subtract all other previous terms.
%C Additionally, with a(1)=1, a(2)=2, this gives: 1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9,10,10... cf. A008619.
%t f[s_] := Block[{}, Append[s, s[[ 1]]s[[ 2]]  Plus @@ Drop[s, 2]]]; Nest[f, {1, 1}, 15] (* _Robert G. Wilson v_, May 26 2006 *)
%o (C) #include <stdio.h> #include <inttypes.h> int main (void) { int64_t n1=1; int64_t n2=1; int i; int64_t sum=0,next; printf("%lld,%lld,",n1,n2); for (i=0;i<12;i++) { next=n1*n2sum; sum+=n1; n1=n2; n2=next; printf("%lld,",n2); } }
%o (PARI) {m=16;a=1;b=1;print1(a=1,",",b=1,",");v=[];for(n=3,m,print1(k=a*bsum(j=1,#v,v[j]),",");v=concat(v,a);a=b;b=k)} \\ _Klaus Brockhaus_
%Y Cf. A008619, A117157.
%K sign
%O 1,5
%A Gabriel Finch (salsaman(AT)xs4all.nl), Apr 16 2006
%E a(12) corrected; a(15) and a(16) from _Klaus Brockhaus_, Apr 17 2006
