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 A247914 Least number k such that |(k+1)/u(k+1) - e| < 1/n^n, where u is defined as in Comments. 8

%I #4 Sep 27 2014 19:03:08

%S 1,3,4,5,6,7,9,10,11,12,14,15,16,17,19,20,21,22,23,25,26,27,28,30,31,

%T 32,33,34,36,37,38,39,40,42,43,44,45,46,48,49,50,51,52,54,55,56,57,58,

%U 59,61,62,63,64,65,67,68,69,70,71,73,74,75,76,77,78,80

%N Least number k such that |(k+1)/u(k+1) - e| < 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 {1,2} for n >= 1, that d(n+1) - d(n) is in {-1,0,1}, 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="/A247914/b247914.txt">Table of n, a(n) for n = 1..1000</a>

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

%e n ... |(n+1)/u(n+1)-e| .. 1/n^n

%e 1 ... 0.7182818285 ...... 1

%e 2 ... 0.28171817 ........ 0.25

%e 3 ... 0.051615161 ....... 0.037037

%e 4 ... 0.0089908988 ...... 0.00390625

%e 5 ... 0.0013006963 ...... 0.00032000

%e a(2) = 3 because |4/u(4) - e| < 1/2^2 < |3/u(3) - e|.

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

%t z = 500; 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], Abs[(# + 1)/u[# + 1] - E] < n^-n &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247914 *)

%t w = Differences[u]

%t f1 = Flatten[Position[w, 1]] (* A247915 *)

%t f2 = Flatten[Position[w, 2]] (* A247916 *)

%Y Cf. A247908, A247911, A247915, A247916.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 27 2014

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Last modified November 29 13:49 EST 2023. Contains 367445 sequences. (Running on oeis4.)