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A000149 a(n) = floor(e^n).

%I A000149 M1751 N0695 #65 Aug 20 2025 16:19:02
%S A000149 1,2,7,20,54,148,403,1096,2980,8103,22026,59874,162754,442413,1202604,
%T A000149 3269017,8886110,24154952,65659969,178482300,485165195,1318815734,
%U A000149 3584912846,9744803446,26489122129,72004899337,195729609428,532048240601,1446257064291
%N A000149 a(n) = floor(e^n).
%C A000149 A000079(n) <= a(n) <= A000244(n); for n > 0: A064780(n) = a(n+1) - a(n). - _Reinhard Zumkeller_, Mar 17 2015
%C A000149 Satisfies Benford's law [Whyman et al., 2016]. - _N. J. A. Sloane_, Feb 12 2017
%D A000149 Federal Works Agency, Work Projects Administration for the City of NY, Tables of the Exponential Function. National Bureau of Standards, Washington, DC, 1939.
%D A000149 A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 230.
%D A000149 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A000149 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A000149 T. D. Noe, <a href="/A000149/b000149.txt">Table of n, a(n) for n=0..300</a>
%H A000149 R. K. Guy and N. J. A. Sloane, <a href="/A005180/a005180.pdf">Correspondence</a>, 1988.
%H A000149 Johan Kok, <a href="https://arxiv.org/abs/2507.16500">Integer sequences with conjectured relation with certain graph parameters of the family of linear Jaco graphs</a>, arXiv:2507.16500 [math.CO], 2025. See p. 9.
%H A000149 G. Whyman, N. Ohtori, E. Shulzinger and Ed. Bormashenko, <a href="https://doi.org/10.1016/j.physa.2016.06.054">Revisiting the Benford law: When the Benford-like distribution of leading digits in sets of numerical data is expectable?</a>, Physica A: Statistical Mechanics and its Applications, 461 (2016), 595-601.
%H A000149 <a href="/index/Be#Benford">Index entries for sequences related to Benford's law</a>
%F A000149 a(n)^(1/n) converges to e because |1-a(n)/e^n|=|e^n-a(n)|/e^n < e^(-n) and so a(n)^(1/n)=(e^n*(1+o(1)))^(1/n)=e*(1+o(1)). - _Hieronymus Fischer_, Jan 22 2006
%t A000149 a[n_]:=Floor[E^n]; (* _Vladimir Joseph Stephan Orlovsky_, Dec 12 2008 *)
%t A000149 Floor[E^Range[0,30]] (* _Harvey P. Dale_, Apr 01 2012 *)
%o A000149 (PARI) a(n) = floor(exp(n)); \\ _Arkadiusz Wesolowski_, Nov 26 2011
%o A000149 (PARI) apply( A000149(n)=exp(n)\1, [0..30]) \\ An error message will say so if default(realprecision) must be increased, for large n. - _M. F. Hasler_, May 27 2018
%o A000149 (Haskell)
%o A000149 a000149 = floor . (exp 1 ^)
%o A000149 a000149_list = let e = exp 1 in map floor $ iterate (* e) 1
%o A000149 -- _Reinhard Zumkeller_, Mar 17 2015
%o A000149 (Python)
%o A000149 from sympy import floor, E
%o A000149 def a(n):  return floor(E**n)
%o A000149 print([a(n) for n in range(29)]) # _Michael S. Branicky_, Jul 20 2021
%Y A000149 Bisection: A116472.
%Y A000149 Cf. A001113, A003619, A000079, A000244, A064780 (first differences, apart from initial term).
%Y A000149 Cf. A000227 (round e^n), A001671 (ceiling e^n).
%K A000149 nonn,easy
%O A000149 0,2
%A A000149 _N. J. A. Sloane_