A383238 A sequence constructed by greedily sampling the Poisson distribution for parameter value 1, 1/(e*(i-1)!) to minimize discrepancy selecting the smallest value in case of ties.

1, 2, 3, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 5, 1, 2, 3, 1, 2, 1, 2, 3, 4, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 5, 1, 2, 3, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 4, 1, 2, 3, 1, 2, 1, 2, 3, 1, 2, 6, 1, 2, 3, 1, 2

Offset

1,2

Comments

The geometric mean approaches A382095 = exp(Sum_{k>=2} log(k)/(k-1)!) in the limit.

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 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 |     0.367     |     0.367     |     0.183     |   1    |
| 2 |    -0.264     |     0.735     |     0.367     |   2    |
| 3 |     0.103     |     0.103     |     0.551     |   3    |
| 4 |     0.471     |     0.471     |    -0.264     |   1    |
| 5 |    -0.160     |     0.839     |    -0.080     |   2    |

Mathematica

probCountDiff[j_, k_, count_]:=N[k/(E*Factorial[j-1])]-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]
A383238=samplePDF[120]

Prog

(Python)
from mpmath import iv
def prob_count_diff(j, k, count):
    return  k/(iv.e*iv.factorial(j-1)) - 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
A383238 = sample_poisson_distribution(120)  # Jwalin Bhatt, Dec 16 2025

Crossrefs

Cf. A381617, A382093, A382095, A382961.

Sequence in context: A176506 A200596 A088422 * A100833 A097293 A296656

Adjacent sequences: A383235 A383236 A383237 * A383239 A383240 A383241

Keyword

nonn

Author

Jwalin Bhatt, Apr 20 2025

Status

approved