A397557 A sequence constructed by greedily sampling the zeta distribution for parameter value 3 to minimize discrepancy.

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

Offset

1,4

Comments

The probability mass function for this zeta distribution is given by p(i) = 1/(zeta(3)*i^3).

A general family of these distributions is given by p(i) = 1/(zeta(s)*i^s) where s > 1. This sequence is the s=3 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 geometric mean approaches exp(-zeta'(3)/zeta(3)) = 1.1791840098089555... in the limit. In general, if the sequence was formed by every k^s occurrences, it would approach e^(-zeta'(s)/zeta(s)).

Unlike A381617, this sequence has a finite arithmetic mean, equal to zeta(2)/zeta(3) = 1.368432777... and in general for power s it equals zeta(s-1)/zeta(s).

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) the number of occurrences of k among the first n-1 terms; the expected count among n terms is n*p(k). We pick the k maximizing n*p(k) - c(k).
| n | n*p(1) - c(1) | n*p(2) - c(2) | n*p(3) - c(3) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     0.831     |       -       |       -       |   1    |
| 2 |     0.663     |     0.207     |       -       |   1    |
| 3 |     0.495     |     0.311     |       -       |   1    |
| 4 |     0.327     |     0.415     |       -       |   2    |
| 5 |     1.159     |    -0.480     |     0.154     |   1    |
| 6 |     0.991     |    -0.376     |     0.184     |   1    |

Mathematica

probCountDiff[j_, k_, count_] := k/(Zeta[3] j^3) - 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]
A397557 = samplePDF[87]

Prog

(Python)
from sympy import zeta
def prob_count_diff(j, k, count):
    return k/(zeta(3)*j**3) - count
def sample_zeta_distribution(num_coeffs):
    coeffs, unreached_val, counts = [], 1, {}
    for k in range(1, num_coeffs+1):
        diffs = [prob_count_diff(i, k, counts.get(i, 0)) for i in range(1, unreached_val+1)]
        most_probable = diffs.index(max(diffs)) + 1
        unreached_val += most_probable == unreached_val
        coeffs.append(most_probable)
        counts[most_probable] = counts.get(most_probable, 0) + 1
    return coeffs
A397557 = sample_zeta_distribution(87)

Crossrefs

Cf. A381522, A381617.

Sequence in context: A184097 A345932 A205399 * A135303 A036065 A082907

Adjacent sequences: A397552 A397555 A397556 * A397558 A397559 A397560

Keyword

nonn,new

Author

Jwalin Bhatt, Jun 30 2026

Status

approved