A278453 a(n) = nearest integer to b(n) = c^(b(n-1)/(n-1)), where b(1)=0 and c is chosen such that the sequence neither explodes nor goes to 1.

0, 1, 2, 4, 6, 8, 10, 12, 15, 17, 19, 22, 25, 27, 30, 33, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 75, 78, 81, 84, 88, 91, 94, 98, 101, 104, 108, 111, 114, 118, 121, 125, 128, 132, 135, 139, 142, 146, 149, 153, 157, 160, 164, 167, 171, 175, 178, 182, 186, 189, 193, 197, 201, 204, 208, 212, 216

Offset

1,3

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

Its value: c = 5.7581959391... A278813. If c 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 c is found through trial and error. Suppose one starts with c = 5, the sequence 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.

b(n) = n*log_c((n+1)*log_c((n+2)*log_c(...))). At n=1 this gives the relation between c and b(1). It follows that b(n) ~ n*log_c(n). - Andrey Zabolotskiy, Nov 30 2016

Links

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

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

Formula

a(n) = round(n*log_c((n+1)*log_c((n+2)*log_c(...)))). - Andrey Zabolotskiy, Nov 30 2016

Example

a(2) = round(5.75...^0) = round(1) = 1.
a(3) = round(5.75...^(1/2)) = round(2.39...) = 2.
a(4) = round(5.75...^(2.39.../3)) = round(4.05...) = 4.

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[];
      }];
    ];
  ];
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 c see A278813.

For different values of b(1) see A278448, A278449, A278450, A278451, A278452.

Sequence in context: A292651 A113903 A248563 * A347009 A130261 A186384

Adjacent sequences: A278450 A278451 A278452 * A278454 A278455 A278456

Keyword

nonn

Author

Rok Cestnik, Nov 22 2016

Status

approved