A386904 A sequence constructed by greedily sampling the Borel distribution for parameter value 1/2 to minimize discrepancy.

1, 2, 1, 1, 3, 1, 4, 1, 2, 1, 1, 5, 1, 2, 1, 1, 3, 1, 1, 2, 1, 6, 1, 1, 2, 1, 1, 3, 1, 2, 1, 4, 1, 1, 2, 1, 1, 7, 1, 2, 1, 3, 1, 1, 8, 1, 2, 1, 1, 5, 1, 2, 1, 1, 3, 1, 1, 2, 1, 4, 1, 1, 2, 1, 3, 1, 1, 2, 1, 1, 9, 1, 2, 1, 1, 4, 1, 3, 1, 2, 1, 1, 6, 1, 2, 1, 1, 1, 3, 1, 2, 1, 5, 1, 1, 2, 1, 1, 4, 1, 2, 1, 3, 1, 1, 2, 1, 1, 10

Offset

1,2

Comments

The geometric mean approaches A386009 in the limit.

The Borel distribution with parameter value 1/2 has PDF p(i) = (i/2)^(i-1) / (exp(i/2)*i!).

Links

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

Wikipedia, Borel distribution

Example

Let p(k) denote the probability of k and c(k) denote the count of occurrences of k so far, then the expected occurrences of k at n-th step are 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) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     0.606     |       -       |       -       |   1    |
| 2 |     0.213     |     0.367     |       -       |   2    |
| 3 |     0.819     |    -0.448     |     0.251     |   1    |
| 4 |     0.426     |    -0.264     |     0.334     |   1    |
| 5 |     0.032     |    -0.080     |     0.418     |   3    |

Mathematica

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]
A386904=samplePDF[120]

Prog

(Python)
from mpmath import iv
def prob_count_diff(j, k, count):
    return  k*(((iv.mpf(j)/2)**(j-1))/(iv.exp(iv.mpf(j)/2)*iv.factorial(j))) - count
def sample_borel_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
A386904 = sample_borel_distribution(120)  # Jwalin Bhatt, Dec 16 2025

Crossrefs

Cf. A386009, A386016.

Sequence in context: A375128 A381617 A396506 * A245717 A248008 A327981

Adjacent sequences: A386901 A386902 A386903 * A386905 A386906 A386907

Keyword

nonn

Author

Jwalin Bhatt, Aug 07 2025

Status

approved