A278453 - OEIS

%I #20 Dec 01 2016 10:42:46

%S 0,1,2,4,6,8,10,12,15,17,19,22,25,27,30,33,35,38,41,44,47,50,53,56,59,

%T 62,65,68,71,75,78,81,84,88,91,94,98,101,104,108,111,114,118,121,125,

%U 128,132,135,139,142,146,149,153,157,160,164,167,171,175,178,182,186,189,193,197,201,204,208,212,216

%N a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), 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 (due to rounding to the nearest integer a(n-1)a(n+1)/a(n)^2 is not always less than 1).

%C Its value: c = 5.7581959391... A278813. If c 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 c is found through trial and error. Suppose one starts with c = 5, the sequence 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 b(n) = n*log_c((n+1)*log_c((n+2)*log_c(...))). At n=1 this gives the relation between c and b(1). It follows that b(n) ~ n*log_c(n). - _Andrey Zabolotskiy_, Nov 30 2016

%H Rok Cestnik, <a href="/A278453/b278453.txt">Table of n, a(n) for n = 1..1000</a>

%H Rok Cestnik, <a href="/A278453/a278453.pdf">Plot of the dependence of b(1) on c</a>

%F a(n) = round(n*log_c((n+1)*log_c((n+2)*log_c(...)))). - _Andrey Zabolotskiy_, Nov 30 2016

%e a(2) = round(5.75...^0) = round(1) = 1.

%e a(3) = round(5.75...^(1/2)) = round(2.39...) = 2.

%e a(4) = round(5.75...^(2.39.../3)) = round(4.05...) = 4.

%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,

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

%t     c = c + 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       c = c - 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 c see A278813.

%Y For different values of b(1) see A278448, A278449, A278450, A278451, A278452.

%K nonn

%O 1,3

%A _Rok Cestnik_, Nov 22 2016