OFFSET
1,1
COMMENTS
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).
In this case b(1) = 2.8718808270... A278808. 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.
The value of b(1) is found through trial and error. 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.
b(n) = n*log_2((n+1)*log_2((n+2)*log_2(...))) ~ n*log_2(n). - Andrey Zabolotskiy, Dec 01 2016
LINKS
Rok Cestnik, Table of n, a(n) for n = 1..1000
Rok Cestnik, Plot of the dependence of b(1) on c
EXAMPLE
a(2) = round(2^2.87...) = round(7.32...) = 7.
a(3) = round(2^(7.32.../2)) = round(12.64...) = 13.
a(4) = round(2^(12.64.../3)) = round(18.55...) = 19.
MATHEMATICA
c = 2;
n = 100;
acc = Round[n*1.2];
th = 1000000;
b1 = 0;
For[p = 0, p < acc, ++p,
For[d = 0, d < 9, ++d,
b1 = b1 + 1/10^p;
bn = b1;
For[i = 1, i < Round[n*1.2], ++i,
bn = N[c^(bn/i), acc];
If[bn > th, Break[]];
];
If[bn > th, {
b1 = b1 - 1/10^p;
Break[];
}];
];
];
bnlist = {N[b1]};
bn = b1;
For[i = 1, i < n, ++i,
bn = N[c^(bn/i), acc];
If[bn > th, Break[]];
bnlist = Append[bnlist, N[bn]];
];
anlist = Map[Round[#] &, bnlist]
CROSSREFS
KEYWORD
nonn
AUTHOR
Rok Cestnik, Nov 22 2016
STATUS
approved