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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
 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, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 41, 41, 42 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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. REFERENCES Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19. LINKS Clark Kimberling, Table of n, a(n) for n = 1..1000 EXAMPLE Approximations for the first few terms of e - 2*n/u(2*n) and 1/n^n are shown here: n ... e-2*n/u(2*n) .... 1/n^n 1 ... 0.71828 ........ 1 2 ... 0.0516152 ....... 0.25 3 ... 0.0013007 ....... 0.037037 4 ... 0.0000184967 .... 0.00390625 a(2) = 2 because e - 4/u(4) < 1/2^2 < e - 2/u(2). MATHEMATICA \$RecursionLimit = 1000; \$MaxExtraPrecision = 1000; z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2); f[n_] := f[n] = Select[Range[z], E - 2 #/u[2 #] < 1/n^n &, 1]; u = Flatten[Table[f[n], {n, 1, z}]] (* A247908 *) w = Differences[u] Flatten[Position[w, 0]] (* A247909 *) Flatten[Position[w, 1]] (* A247910 *) CROSSREFS Cf. A247909, A247910, A247911, A247914. Sequence in context: A079952 A055930 A090638 * A057363 A073869 A060143 Adjacent sequences: A247905 A247906 A247907 * A247909 A247910 A247911 KEYWORD nonn,easy AUTHOR Clark Kimberling, Sep 27 2014 STATUS approved

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Last modified February 7 12:58 EST 2023. Contains 360123 sequences. (Running on oeis4.)