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 A247908 Least number k such that e - 2*k/u(2*k) < 1/n^n, where u is defined as in Comments. 10

%I #6 Sep 27 2014 19:02:25

%S 1,2,3,3,4,4,5,6,6,7,8,8,9,9,10,11,11,12,12,13,14,14,15,16,16,17,17,

%T 18,19,19,20,20,21,22,22,23,23,24,25,25,26,26,27,28,28,29,29,30,30,31,

%U 32,32,33,33,34,35,35,36,36,37,38,38,39,39,40,41,41,42

%N Least number k such that e - 2*k/u(2*k) < 1/n^n, where u is defined as in Comments.

%C The sequence u is define recursively by u(n) = u(n-1) + u(n-2)/(n-2), with u(1) = 0 and u(2) = 1. Let d(n) = a(n+1) - a(n). It appears that d(n) is in {0,1} for n >= 1, that d(n+1) - d(n) is in {2,3}, and that similar bounds hold for higher differences.

%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.

%H Clark Kimberling, <a href="/A247908/b247908.txt">Table of n, a(n) for n = 1..1000</a>

%e Approximations for the first few terms of e - 2*n/u(2*n) and 1/n^n are shown here:

%e n ... e-2*n/u(2*n) .... 1/n^n

%e 1 ... 0.71828 ........ 1

%e 2 ... 0.0516152 ....... 0.25

%e 3 ... 0.0013007 ....... 0.037037

%e 4 ... 0.0000184967 .... 0.00390625

%e a(2) = 2 because e - 4/u(4) < 1/2^2 < e - 2/u(2).

%t \$RecursionLimit = 1000; \$MaxExtraPrecision = 1000;

%t z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);

%t f[n_] := f[n] = Select[Range[z], E - 2 #/u[2 #] < 1/n^n &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247908 *)

%t w = Differences[u]

%t Flatten[Position[w, 0]] (* A247909 *)

%t Flatten[Position[w, 1]] (* A247910 *)

%Y Cf. A247909, A247910, A247911, A247914.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 27 2014

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Last modified November 30 23:40 EST 2023. Contains 367464 sequences. (Running on oeis4.)