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A247778 Least k such that e - (1 + 1/k)^k < 1/n. 7

%I #7 Sep 25 2014 21:15:48

%S 1,2,4,5,6,8,9,10,12,13,15,16,17,19,20,21,23,24,25,27,28,29,31,32,34,

%T 35,36,38,39,40,42,43,44,46,47,49,50,51,53,54,55,57,58,59,61,62,63,65,

%U 66,68,69,70,72,73,74,76,77,78,80,81,82,84,85,87,88,89

%N Least k such that e - (1 + 1/k)^k < 1/n.

%C a(n+1) - a(n) is in {1,2} for n >= 1.

%H Clark Kimberling, <a href="/A247778/b247778.txt">Table of n, a(n) for n = 1..3000</a>

%e The values of e - (1 + 1/k)^k for k = 1..8 are approximately 0.718282, 0.468282, 0.347911, 0.276876, 0.229962, 0.196655, 0.171782, 0.152497, so that the first 6 terms of A247778 are 1,2,4,5,6,8.

%t z = 600; f[n_] := f[n] = Select[Range[z], E - (1 + 1/#)^# < 1/n &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (*A247778*)

%t d1 = Flatten[Position[Differences[u], 1]] (*A247779*)

%t d2 = Flatten[Position[Differences[u], 2]] (*A247780*)

%o (PARI) a(n) = {k = 1; while ((exp(1) - (1 + 1/k)^k) >= 1/n, k++); k;} \\ _Michel Marcus_, Sep 25 2014

%Y Cf. A247779, A247780.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 23 2014

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