A394401 A sequence constructed by greedily sampling the probability distribution given by 1/log_2(i+1) - 1/log_2(i+2) to minimize discrepancy.

1, 2, 3, 1, 4, 1, 5, 6, 1, 2, 7, 1, 8, 9, 1, 10, 1, 2, 3, 1, 11, 12, 1, 13, 2, 1, 4, 1, 14, 15, 1, 3, 2, 1, 16, 1, 17, 5, 1, 2, 18, 1, 19, 1, 20, 3, 1, 2, 21, 1, 4, 1, 6, 22, 1, 2, 23, 1, 24, 3, 1, 25, 1, 2, 7, 1, 26, 27, 1, 2, 1, 5, 4, 1, 3, 28, 1, 2, 1, 29

Offset

1,2

Comments

The probability mass function used here is a simple example of a heavy log tail distribution. It can be seen as a discrete analog of the log-Cauchy distribution. As a consequence of the heavy tail, both the arithmetic mean and the geometric mean of this sequence diverge to infinity.

A general family of these distributions is given by 1/log_2(i+1)^p - 1/log_2(i+2)^p where p>0.

Properties change sharply at p=1, the mean remains infinite for all p but the geometric mean becomes finite for p>1 and infinite for p<=1.

| mean | 0 < p <= 1 | p > 1 |

|------------|------------|--------|

| arithmetic | oo | oo |

| geometric | oo | finite | - Jwalin Bhatt, Jun 09 2026

The Fisher information is always finite and given by Sum_{i=1..inf} ((ln(log_2(i+2))/(log_2(i+2))^p - ln(log_2(i+1))/(log_2(i+1))^p)^2) / (1/(log_2(i+1))^p - 1/(log_2(i+2))^p).

Links

Jwalin Bhatt, Table of n, a(n) for n = 1..10000

Wikipedia, Log-Cauchy distribution

Example

Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms; then the expected number of occurrences of k among n random terms is given by n*p(k).
We subtract the actual occurrences c(k) from the expected occurrences and pick the one with the highest value.
| n | n*p(1) - c(1) | n*p(2) - c(2) | n*p(3) - c(3) | n*p(4) - c(4) | choice |
|---|---------------|---------------|---------------|---------------|--------|
| 1 |     0.369     |       -       |       -       |       -       |   1    |
| 2 |    -0.261     |     0.261     |       -       |       -       |   2    |
| 3 |     0.107     |    -0.607     |     0.207     |       -       |   3    |
| 4 |     0.476     |    -0.476     |    -0.722     |     0.175     |   1    |
| 5 |    -0.154     |    -0.345     |    -0.653     |     0.219     |   4    |

Mathematica

probCountDiff[j_, k_, count_] := k*(1/Log2[j+1] - 1/Log2[j+2]) - Lookup[count, j, 0]
samplePDF[n_] := Module[{coeffs, unreachedVal, counts, k, probCountDiffs, mostProbable},
  coeffs = ConstantArray[0, n]; unreachedVal = 1; counts = <||>;
  Do[probCountDiffs = Table[probCountDiff[i, k, counts], {i, 1, unreachedVal}];
    mostProbable = First@FirstPosition[probCountDiffs, Max[probCountDiffs]];
    If[mostProbable == unreachedVal, unreachedVal++]; coeffs[[k]] = mostProbable;
    counts[mostProbable] = Lookup[counts, mostProbable, 0] + 1; , {k, 1, n}]; coeffs]
A394401 = samplePDF[120]

Prog

(Python)
from mpmath import iv
def prob_count_diff(j, k, count):
    return  k*(1/iv.log(j+1, 2) - 1/iv.log(j+2, 2)) - count
def sample_log_tail_distribution(num_coeffs):
    coeffs, unreached_val, counts = [], 1, {}
    for k in range(1, num_coeffs+1):
        prob_count_diffs = [prob_count_diff(i, k, counts.get(i, 0)) for i in range(1, unreached_val+1)]
        most_probable = prob_count_diffs.index(max(prob_count_diffs)) + 1
        unreached_val += most_probable == unreached_val
        coeffs.append(most_probable)
        counts[most_probable] = counts.get(most_probable, 0) + 1
    return coeffs
A394401 = sample_log_tail_distribution(120)

Crossrefs

Cf. A386904, A393100, A393793, A396506.

Sequence in context: A205790 A279820 A235791 * A035426 A065516 A375511

Adjacent sequences: A394398 A394399 A394400 * A394402 A394403 A394404

Keyword

nonn

Author

Jwalin Bhatt, Mar 19 2026

Status

approved