A278813 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.

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, 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, 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

Offset

1,1

Comments

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.

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.

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 satisfies 2*log_c(3*log_c(4*log_c(...))) = 1. - Andrey Zabolotskiy, Dec 02 2016

No closed form expression is known. Probably transcendental but this is unproved. - Robert G. Wilson v, Dec 02 2016

Links

Robert G. Wilson v, Table of n, a(n) for n = 1..2500 (first 1000 from Rok Cestnik)

Rok Cestnik, Plot of the dependence of b(1) on c

Example

5.75819593911037494197402886500932909247424264705531...

Mathematica

b1 = 0;
n = 100;
acc = Round[n*1.2];
th = 1000000;
c = 0;
For[p = 0, p < acc, ++p, For[d = 0, d < 9, ++d, c = c + 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, {c = c - 1/10^p;
      Break[];
      }];
    ];
  ];
N[c, n]

Crossrefs

For sequence round(b(n)) see A278453.

For different values of b(1) see A278808, A278809, A278810, A278811, A278812.

Sequence in context: A247872 A306400 A090987 * A217167 A348052 A195498

Adjacent sequences: A278810 A278811 A278812 * A278814 A278815 A278816

Keyword

nonn,cons,nice

Author

Rok Cestnik, Nov 28 2016

Status

approved