A278809 Decimal expansion of b(1) in the sequence b(n+1) = c^(b(n)/n) A278449, where c=3 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

1, 0, 8, 2, 8, 7, 3, 6, 0, 9, 5, 2, 0, 7, 3, 8, 6, 9, 4, 0, 8, 2, 8, 5, 0, 3, 1, 3, 4, 5, 3, 1, 0, 0, 8, 0, 2, 5, 7, 8, 6, 3, 4, 5, 4, 7, 8, 5, 3, 8, 5, 0, 6, 4, 3, 2, 8, 8, 4, 7, 8, 2, 1, 6, 8, 0, 6, 9, 2, 2, 7, 8, 8, 9, 5, 2, 9, 9, 5, 5, 7, 4, 7, 0, 6, 8, 1, 4, 4, 8, 7, 8, 6, 2, 3, 9, 2, 4, 4, 3, 1, 1, 5, 4, 5, 9, 9, 1, 8, 9, 2, 4, 3, 8, 8, 4, 0, 6, 3, 6, 2, 6, 1, 3, 5, 9, 3, 4, 0, 0

Offset: 1

Author

Rok Cestnik, Nov 28 2016

Keywords

• nonn
• cons

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.

If b(1) 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.

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 b(n) 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."

Examples

			1.08287360952073869408285031345310080257863454785385...
		

Links

• Rok Cestnik, Table of n, a(n) for n = 1..1000
• Rok Cestnik, Plot of the dependence of b(1) on c

Crossrefs

For sequence round(b(n)) see A278449.

For different values of c see A278808, A278810, A278811, A278812.

For b(1)=0 see A278813.

Programs

Mathematica

c = 3;
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[];
      }];
    ];
  ];
N[b1,n]
RealDigits[ Fold[ Log[3, #1*#2] &, 1, Reverse@ Range[2, 160]], 10, 111][[1]] (* Robert G. Wilson v, Dec 02 2016 *)

Formula

log_3(2*log_3(3*log_3(4*log_3(...)))). - Andrey Zabolotskiy, Dec 01 2016