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%I #6 Oct 04 2023 22:37:35
%S 0,1,1,1,1,2,2,4,6,24,4738381338321616896,4738381338321616920,
%T 22452257707354557353808363243511480320
%N Adding, multiplying and exponentiating cycle of the previous two terms similar to A039941.
%F a(1)=0, a(2)=1, a(n): if n mod 3 is 0: a(n)=a(n-2) + a(n-1), if n mod 3 is 1: a(n)=a(n-2) * a(n-1), if n mod 3 is 2: a(n)=a(n-2)^a(n-1).
%e a(11) = a(9)^a(10)=6^24 because 11 mod 3 is 2.
%t nxt[{n_,a_,b_}]:={n+1,b,Which[Mod[n+1,3]==0,a+b,Mod[n+1,3]==1,a*b,True,a^b]}; NestList[ nxt,{2,0,1},12][[;;,2]] (* _Harvey P. Dale_, Oct 04 2023 *)
%Y Cf. A039941.
%K easy,nonn
%O 1,6
%A Anthony Peterson (civ2buf(AT)ricochet.com), Jul 09 2003
%E The next 2 terms are (6^24)^((6^24)*(6^24+24)) and (6^24)^((6^24) * (6^24 + 24)) + (6^24) * (6^24 + 24).
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