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%I #31 Mar 02 2019 04:21:48
%S 5,7,5,8,1,9,5,9,3,9,1,1,0,3,7,4,9,4,1,9,7,4,0,2,8,8,6,5,0,0,9,3,2,9,
%T 0,9,2,4,7,4,2,4,2,6,4,7,0,5,5,3,1,5,4,1,5,1,4,1,2,5,9,9,0,6,1,9,7,1,
%U 0,7,5,9,8,9,1,5,8,7,2,3,0,8,3,3,3,7,8,7,0,6,9,5,8,7,9,1,1,5,7,2,0,0,5,6,2,9,5,0,5,6,3,2,1,1,0,5,7,1,4,7,1,3,5,9,5,0,6,0,7,7
%N Decimal expansion of c in the sequence b(n+1) = c^(b(n)/n) A278453, where b(1)=0 and c is chosen such that the sequence neither explodes nor goes to 1.
%C There exists a unique value of c 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.
%C If c were chosen smaller the sequence b(n) would approach 1, if it were chosen greater it 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 c is found through trial and error. Suppose one starts with c = 5, the sequence b(n) would continue b(2) = 1, b(3) = 2.23..., b(4) = 3.31..., b(5) = 3.80..., b(6) = 3.39..., b(7) = 2.48..., b(8) = 1.77... and from there one can see that such a sequence is tending to 1. One continues by trying a larger value, say c = 6, which gives rise to b(2) = 1, b(3) = 2.44, b(4) = 4.31..., b(5) = 6.92..., b(6) = 11.94..., b(7) = 35.38... and from there one can see that such a sequence is escalating too fast. Therefore, one now knows that the true value of c is between 5 and 6.
%C c satisfies 2*log_c(3*log_c(4*log_c(...))) = 1. - _Andrey Zabolotskiy_, Dec 02 2016
%C No closed form expression is known. Probably transcendental but this is unproved. - _Robert G. Wilson v_, Dec 02 2016
%H Robert G. Wilson v, <a href="/A278813/b278813.txt">Table of n, a(n) for n = 1..2500</a> (first 1000 from Rok Cestnik)
%H Rok Cestnik, <a href="/A278813/a278813.pdf">Plot of the dependence of b(1) on c</a>
%e 5.75819593911037494197402886500932909247424264705531...
%t b1 = 0;
%t n = 100;
%t acc = Round[n*1.2];
%t th = 1000000;
%t c = 0;
%t For[p = 0, p < acc, ++p, For[d = 0, d < 9, ++d, c = c + 1/10^p;
%t bn = b1;
%t For[i = 1, i < Round[n*1.2], ++i, bn = N[c^(bn/i), acc];
%t If[bn > th, Break[]];];
%t If[bn > th, {c = c - 1/10^p;
%t Break[];
%t }];
%t ];
%t ];
%t N[c, n]
%Y For sequence round(b(n)) see A278453.
%Y For different values of b(1) see A278808, A278809, A278810, A278811, A278812.
%K nonn,cons,nice
%O 1,1
%A _Rok Cestnik_, Nov 28 2016
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