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

Author

Rok Cestnik, Nov 28 2016

Keywords

• nonn
• cons
• nice

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

Examples

			5.75819593911037494197402886500932909247424264705531...
		

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

Crossrefs

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

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

Programs

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]