cp's OEIS Frontend

A278451 a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=5 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

%I A278451 #21 Dec 02 2016 00:16:20
%S A278451 0,1,3,5,7,9,11,14,17,19,22,25,28,31,34,37,40,43,46,49,52,56,59,62,66,
%T A278451 69,73,76,80,83,87,90,94,98,101,105,109,112,116,120,123,127,131,135,
%U A278451 139,143,146,150,154,158,162,166,170,174,178,182,186,190,194,198,202,206,210,214,218,222,226,231,235,239
%N A278451 a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=5 and b(1) is chosen such that the sequence neither explodes nor goes to 1.
%C A278451 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 A278451 In this case b(1) = 0.1775819188... A278811. 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 A278451 The value of b(1) is found through trial and error. Illustrative example for the case of c=2 (for c=5 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 A278451 b(n) = n*log_5((n+1)*log_5((n+2)*log_5(...))) ~ n*log_5(n). - _Andrey Zabolotskiy_, Dec 01 2016
%H A278451 Rok Cestnik, <a href="/A278451/b278451.txt">Table of n, a(n) for n = 1..1000</a>
%H A278451 Rok Cestnik, <a href="/A278451/a278451.pdf">Plot of the dependence of b(1) on c</a>
%e A278451 a(2) = round(5^0.17...) = round(1.33...) = 1.
%e A278451 a(3) = round(5^(1.33.../2)) = round(2.91...) = 3.
%e A278451 a(4) = round(5^(2.91.../3)) = round(4.78...) = 5.
%t A278451 c = 5;
%t A278451 n = 100;
%t A278451 acc = Round[n*1.2];
%t A278451 th = 1000000;
%t A278451 b1 = 0;
%t A278451 For[p = 0, p < acc, ++p,
%t A278451   For[d = 0, d < 9, ++d,
%t A278451     b1 = b1 + 1/10^p;
%t A278451     bn = b1;
%t A278451     For[i = 1, i < Round[n*1.2], ++i,
%t A278451      bn = N[c^(bn/i), acc];
%t A278451      If[bn > th, Break[]];
%t A278451      ];
%t A278451     If[bn > th, {
%t A278451       b1 = b1 - 1/10^p;
%t A278451       Break[];
%t A278451       }];
%t A278451     ];
%t A278451   ];
%t A278451 bnlist = {N[b1]};
%t A278451 bn = b1;
%t A278451 For[i = 1, i < n, ++i,
%t A278451   bn = N[c^(bn/i), acc];
%t A278451   If[bn > th, Break[]];
%t A278451   bnlist = Append[bnlist, N[bn]];
%t A278451   ];
%t A278451 anlist = Map[Round[#] &, bnlist]
%Y A278451 For decimal expansion of b(1) see A278811.
%Y A278451 For different values of c see A278448, A278449, A278450, A278452.
%Y A278451 For b(1)=0 see A278453.
%K A278451 nonn
%O A278451 1,3
%A A278451 _Rok Cestnik_, Nov 22 2016