A381617 A sequence constructed by greedily sampling the zeta distribution for parameter value 2 to minimize discrepancy.

1, 2, 1, 1, 3, 1, 4, 1, 1, 2, 1, 1, 5, 1, 2, 1, 1, 6, 1, 3, 1, 1, 2, 1, 7, 1, 1, 8, 1, 2, 1, 1, 4, 1, 3, 1, 2, 1, 1, 1, 9, 1, 2, 1, 1, 10, 1, 2, 1, 1, 3, 1, 5, 1, 1, 2, 1, 11, 1, 1, 4, 1, 2, 1, 1, 3, 1, 1, 2, 1, 12, 1, 1, 13, 1, 2, 1, 1, 3, 1, 6, 1, 2, 1, 1, 1, 4, 1, 2, 1, 1, 14

Offset

1,2

Comments

The zeta distribution probability mass function is p(i) = 1/(zeta(2)*i^2).

From Jwalin Bhatt, Jun 09 2026: (Start)

A general family of these distributions is given by p(i) = 1/(zeta(s)*i^s) where s > 1. This sequence is the s=2 case.

Properties change sharply at s=2, the geometric mean remains finite for all s but the arithmetic mean becomes finite for s > 2 and infinite for s <= 2.

| mean | 1 < s <= 2 | s > 2 |

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

| arithmetic | oo | finite |

| geometric | finite | finite |

The arithmetic mean of this sequence diverges to infinity but for general power s in the family above, it approaches zeta(s-1)/zeta(s) in the limit for s>2.

The geometric mean of this sequence approaches Product_{i>=2} i^(1/(zeta(2)*i^2)) = exp(-zeta'(2)/zeta(2)) = 1.768198078153... (A381456) in the limit and for general power s in the family above it approaches exp(-zeta'(s)/zeta(s)) in the limit. (End)

Links

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

Wikipedia, Zeta 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 |     0.607     |       -       |       -       |   1    |
| 2 |     0.215     |     0.303     |       -       |   2    |
| 3 |     0.823     |    -0.544     |     0.202     |   1    |
| 4 |     0.431     |    -0.392     |     0.270     |   1    |
| 5 |     0.039     |    -0.240     |     0.337     |   3    |

Mathematica

probCountDiff[j_, k_, count_] := k/(Zeta[2] 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]
A381617 = samplePDF[120]

Prog

(Python)
from mpmath import iv
zeta2 = iv.pi**2/6
def prob_count_diff(j, k, count):
    return  k/(zeta2*j*j) - count
def sample_zeta_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
A381617 = sample_zeta_distribution(120)  # Jwalin Bhatt, Dec 18 2025

Crossrefs

Cf. A241773, A381522.

Sequence in context: A089141 A336084 A375128 * A396506 A386904 A245717

Adjacent sequences: A381614 A381615 A381616 * A381618 A381619 A381620

Keyword

nonn,changed

Author

Jwalin Bhatt, Mar 10 2025

Status

approved