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a(n) = round(sqrt(Fibonacci(n))).
1

%I #22 Jun 19 2024 19:50:53

%S 0,1,1,1,2,2,3,4,5,6,7,9,12,15,19,25,31,40,51,65,82,105,133,169,215,

%T 274,348,443,564,717,912,1160,1476,1877,2388,3038,3864,4915,6252,7953,

%U 10116,12868,16368,20821,26484,33688,42852,54509,69336,88197,112189,142706

%N a(n) = round(sqrt(Fibonacci(n))).

%F Also the rounded values of the sequence of real numbers g(n) defined by g(0) = 0, g(1) = 1, then g(n) = (g(n-1)^2 + g(n-2)^2)^(1/2). - Corrected by _Thomas Ordowski_, Jan 05 2013

%F This is a special case of the following conjectured relationship: Given a starting sequence of g(0)=0, g(1)=1, ..., g(m)=1 (i.e., 0 then m-1 1's) and then a recurrence relation g(n) = b_1*g(n-1)^c_1 + b_2*g(n-2)^c_2 + ... + b_m*g(n-m)^c_m where the b_i's and c_i's are real numbers > 0 then (b_1*(g(n-1)^c_1)^x + b_2*(g(n-2)^c_2)^x + ... + b_m*(g(n-m)^c_m)^x)^(1/x) = g(n)^(1/x) where x is a real number > 0. - _Gerald McGarvey_, Dec 12 2004

%t Table[ Round[ Sqrt[ Fibonacci[ n]]], {n, 0, 50}] (* _Robert G. Wilson v_ Dec 10 2004 *)

%o (Python)

%o from gmpy2 import isqrt, fib

%o def A100665(n): return int((m:=isqrt(k:=fib(n)))+(k-m*(m+1)>=1)) # _Chai Wah Wu_, Jun 19 2024

%Y Cf. A000045.

%K nonn

%O 0,5

%A _Gerald McGarvey_, Dec 04 2004

%E More terms from _Robert G. Wilson v_, Dec 10 2004