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A278449 a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=3 and b(1) is chosen such that the sequence neither explodes nor goes to 1. 7

%I #21 Dec 02 2016 00:15:33

%S 1,3,6,9,13,16,20,24,29,33,37,42,47,51,56,61,66,71,76,81,86,92,97,102,

%T 108,113,118,124,129,135,141,146,152,158,163,169,175,181,187,193,199,

%U 205,210,216,222,229,235,241,247,253,259,265,271,278,284,290,296,303,309,315,322,328,334,341,347,354,360,367,373,379

%N a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=3 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) = 1.0828736095... A278809. 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=3 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_3((n+1)*log_3((n+2)*log_3(...))) ~ n*log_3(n). - _Andrey Zabolotskiy_, Dec 01 2016

%H Rok Cestnik, <a href="/A278449/b278449.txt">Table of n, a(n) for n = 1..1000</a>

%H Rok Cestnik, <a href="/A278449/a278449.pdf">Plot of the dependence of b(1) on c</a>

%e a(2) = round(3^1.08...) = round(3.28...) = 3.

%e a(3) = round(3^(3.28.../2)) = round(6.07...) = 6.

%e a(4) = round(3^(6.07.../3)) = round(9.26...) = 9.

%t c = 3;

%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 A278809.

%Y For different values of c see A278448, A278450, A278451, A278452.

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

%K nonn

%O 1,2

%A _Rok Cestnik_, Nov 22 2016

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