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%I #20 Apr 13 2020 12:05:37
%S 1,1,2,2,3,4,5,7,9,12,15,19,24,31,39,50,63,80,101,128,161,204,257,324,
%T 409,515,650,819,1032,1301,1639,2066,2603,3280,4133,5207,6561,8266,
%U 10415,13122,16533,20831,26245,33067,41662,52491,66135,83325
%N a(n) = ceiling(sqrt(2*a(n-1)*a(n-2))), a(1) = a(2) = 1.
%C a(n)/a(n-1) ~ cube root of 2.
%C a(n)/a(n-3) ~ 2.
%e a(12) = ceiling(sqrt(2*a(11)*a(10))) = ceiling(sqrt(2*15*12)) = ceiling(sqrt(360)) = 19.
%t a[n_] := a[n] = If[n<3, 1, Ceiling[Sqrt[2 a[n-1] a[n-2]]]]; Array[a, 50] (* _Giovanni Resta_, Nov 26 2019 *)
%t RecurrenceTable[{a[1]==a[2]==1,a[n]==Ceiling[Sqrt[2a[n-1]a[n-2]]]},a,{n,50}] (* _Harvey P. Dale_, Apr 13 2020 *)
%o (Python)
%o import math
%o r = []
%o r.append(1)
%o r.append(1)
%o i = 2
%o while i < 1001:
%o r.append(math.ceil(math.sqrt(2*r[i-1]*r[i-2])))
%o i += 1
%o print(r)
%Y Cf. A017981.
%K nonn
%O 1,3
%A _Oren Meisner_, Aug 14 2018