A278449 a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where c=3 and b(1) is chosen such that the sequence neither explodes nor goes to 1.

1, 3, 6, 9, 13, 16, 20, 24, 29, 33, 37, 42, 47, 51, 56, 61, 66, 71, 76, 81, 86, 92, 97, 102, 108, 113, 118, 124, 129, 135, 141, 146, 152, 158, 163, 169, 175, 181, 187, 193, 199, 205, 210, 216, 222, 229, 235, 241, 247, 253, 259, 265, 271, 278, 284, 290, 296, 303, 309, 315, 322, 328, 334, 341, 347, 354, 360, 367, 373, 379

Offset

1,2

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 (due to rounding to the nearest integer a(n-1)a(n+1)/a(n)^2 is not always less than 1).

In this case b(1) = 1.0828736095... A278809. 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.

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 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."

b(n) = n*log_3((n+1)*log_3((n+2)*log_3(...))) ~ n*log_3(n). - Andrey Zabolotskiy, Dec 01 2016

Links

Rok Cestnik, Table of n, a(n) for n = 1..1000

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

Example

a(2) = round(3^1.08...) = round(3.28...) = 3.
a(3) = round(3^(3.28.../2)) = round(6.07...) = 6.
a(4) = round(3^(6.07.../3)) = round(9.26...) = 9.

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[];
      }];
    ];
  ];
bnlist = {N[b1]};
bn = b1;
For[i = 1, i < n, ++i,
  bn = N[c^(bn/i), acc];
  If[bn > th, Break[]];
  bnlist = Append[bnlist, N[bn]];
  ];
anlist = Map[Round[#] &, bnlist]

Crossrefs

For decimal expansion of b(1) see A278809.

For different values of c see A278448, A278450, A278451, A278452.

For b(1)=0 see A278453.

Sequence in context: A289037 A060605 A325228 * A006590 A061781 A123753

Adjacent sequences: A278446 A278447 A278448 * A278450 A278451 A278452

Keyword

nonn

Author

Rok Cestnik, Nov 22 2016

Status

approved