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a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c = e = 2.71828... and b(1) is chosen such that the sequence neither explodes nor goes to 1.
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%I #21 Dec 03 2016 13:10:13

%S 1,4,7,11,15,19,23,28,33,37,42,48,53,58,64,69,75,80,86,92,97,103,109,

%T 115,121,127,133,139,146,152,158,165,171,177,184,190,197,203,210,216,

%U 223,230,236,243,250,256,263,270,277,284,290,297,304,311,318,325,332,339,346,353,360,367,375,382,389,396,403,410,418,425

%N a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c = e = 2.71828... and b(1) is chosen such that the sequence neither explodes nor goes to 1.

%C For the given c there exists a unique b(1) for which the sequence b(n) does not converge to 1 and at the same time always satisfies b(n-1)b(n+1)/b(n)^2 < 1 (due to rounding to the nearest integer a(n-1)a(n+1)/a(n)^2 is not always less than 1).

%C In this case b(1) = 1.3679012617... A278812. If b(1) were chosen smaller the sequence would approach 1, if it were chosen greater the sequence would at some point violate b(n-1)b(n+1)/b(n)^2 < 1 and from there on quickly escalate.

%C The value of b(1) is found through trial and error. Illustrative example for the case of c=2 (for c=e similar): "Suppose one starts with b(1) = 2, the sequence would continue b(2) = 4, b(3) = 4, b(4) = 2.51..., b(5) = 1.54... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say b(1) = 3, which gives rise to b(2) = 8, b(3) = 16, b(4) = 40.31... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of b(1) is between 2 and 3."

%C b(n) = n*log((n+1)*log((n+2)*log(...))) ~ n*log(n). - _Andrey Zabolotskiy_, Dec 01 2016

%H Rok Cestnik, <a href="/A278452/b278452.txt">Table of n, a(n) for n = 1..1000</a>

%H Rok Cestnik, <a href="/A278452/a278452.pdf">Plot of the dependence of b(1) on c</a>

%e a(2) = round(e^1.36...) = round(3.92...) = 4.

%e a(3) = round(e^(3.92.../2)) = round(7.12...) = 7.

%e a(4) = round(e^(7.12.../3)) = round(10.74...) = 11.

%t c = E;

%t n = 100;

%t acc = Round[n*1.2];

%t th = 1000000;

%t b1 = 0;

%t For[p = 0, p < acc, ++p,

%t For[d = 0, d < 9, ++d,

%t b1 = b1 + 1/10^p;

%t bn = b1;

%t For[i = 1, i < Round[n*1.2], ++i,

%t bn = N[c^(bn/i), acc];

%t If[bn > th, Break[]];

%t ];

%t If[bn > th, {

%t b1 = b1 - 1/10^p;

%t Break[];

%t }];

%t ];

%t ];

%t bnlist = {N[b1]};

%t bn = b1;

%t For[i = 1, i < n, ++i,

%t bn = N[c^(bn/i), acc];

%t If[bn > th, Break[]];

%t bnlist = Append[bnlist, N[bn]];

%t ];

%t anlist = Map[Round[#] &, bnlist]

%Y For decimal expansion of b(1) see A278812.

%Y For different values of c see A278448, A278449, A278450, A278451.

%Y For b(1)=0 see A278453.

%K nonn

%O 1,2

%A _Rok Cestnik_, Nov 22 2016