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A117030 a(1) = a(2) = 1; a(n) = a(n-1)*a(n-2) - a(n-3) - a(n-4) - ... - a(1) for n>2. 1

%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(n-1)*a(n-2) - a(n-3) - a(n-4) - ... - 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*n2-sum; 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*b-sum(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

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Last modified December 15 17:03 EST 2019. Contains 330000 sequences. (Running on oeis4.)