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%I #54 Feb 27 2019 17:34:00
%S 0,1,1,2,3,5,8,13,21,36,58,96,159,262,431,712,1172,1934,3189,5256,
%T 8667,14289,23559,38841,64039,105583,174076,287003,473188,780155,
%U 1286258,2120681,3496412,5764609,9504233,15669832,25835185,42595018,70227313,115785266
%N Number of squares in the interval [e^(n-1), e^n).
%C The lower endpoint e^(n-1) is included; the upper endpoint is not included. The terms a(0) to a(8) coincide with the Fibonacci numbers.
%H Alois P. Heinz, <a href="/A306486/b306486.txt">Table of n, a(n) for n = 0..4607</a> (first 501 terms from Alexei Kourbatov)
%F a(n) = ceiling(sqrt(exp(n))) - ceiling(sqrt(exp(n-1))).
%F From _Alois P. Heinz_, Feb 19 2019: (Start)
%F Lim_{n->oo} a(n+1)/a(n) = sqrt(e) = 1.64872127... = A019774.
%F a(n) = A005181(n+1) - A005181(n). (End)
%F a(n) = (1-1/sqrt(e))*e^(n/2)+O(1) ~ 0.39346934...*e^(n/2) ~ A290506*e^(n/2). - _Alexei Kourbatov_, Feb 20 2019
%e Between exp(2) and exp(3) there are two squares, namely, 9 and 16; therefore, a(3)=2.
%p a:= n-> (f-> f(n)-f(n-1))(i-> ceil(exp(i/2))):
%p seq(a(n), n=0..44); # _Alois P. Heinz_, Feb 18 2019
%o (PARI) a(n)=ceil(sqrt(exp(n)))-ceil(sqrt(exp(n-1)));
%o for(n=0,50,print1(a(n)", "))
%Y Cf. A000045, A000290, A001113, A005181, A019774, A290506, A306604.
%K nonn
%O 0,4
%A _Alexei Kourbatov_, Feb 18 2019
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