A385873 A sequence constructed by greedily sampling the Poisson distribution for parameter value 1 so as to minimize discrepancy.

1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 4, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 5, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 1, 4

Offset

1,2

Comments

The geometric mean approaches A385686 = exp((Sum_{k>=2} log(k)/k!)/(e-1)) in the limit.

The Poisson distribution used here is p(i) = 1/((e-1)*i!).

Links

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

Wikipedia, Poisson 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) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     0.581     |     0.290     |     0.096     |   1    |
| 2 |     0.163     |     0.581     |     0.193     |   2    |
| 3 |     0.745     |    -0.127     |     0.290     |   1    |
| 4 |     0.327     |     0.163     |     0.387     |   3    |
| 5 |     0.909     |     0.454     |    -0.515     |   1    |

Mathematica

probCountDiff[j_, k_, count_]:=N[k/((E-1)*Factorial[j])]-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]
A385873=samplePDF[120]

Prog

(Python)
from mpmath import iv
def prob_count_diff(j, k, count):
    return k/((iv.e-1)*iv.factorial(j)) - count
def sample_poisson_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
A385873 = sample_poisson_distribution(120)  # Jwalin Bhatt, Dec 16 2025

Crossrefs

Cf. A383238, A385685, A385686.

Sequence in context: A265578 A279288 A078734 * A028293 A092782 A119647

Adjacent sequences: A385870 A385871 A385872 * A385874 A385875 A385876

Keyword

nonn

Author

Jwalin Bhatt, Jul 11 2025

Status

approved