A383238 -id:A383238 - cp's OEIS frontend

A382961 A sequence constructed by greedily sampling the logarithmic distribution for parameter value 1/2 so as to minimize discrepancy.

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

Offset: 1

Jwalin Bhatt, Apr 10 2025

The geometric mean approaches A381898 = exp(-PolyLog'(1,1/2)/log(2)) in the limit.

The logarithmic distribution PDF is p(i) = 1/(log(2)*(2^i)*i).

			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 |     0.721     |       -       |       -       |   1    |
| 2 |     0.442     |     0.360     |       -       |   1    |
| 3 |     0.164     |     0.541     |       -       |   2    |
| 4 |     0.885     |    -0.278     |     0.240     |   1    |
| 5 |     0.606     |    -0.098     |     0.300     |   1    |
| 6 |     0.328     |     0.082     |     0.360     |   3    |

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

Cf. A381617, A381898, A381900, A383238.

probCountDiff[j_, k_, count_] := k/(Log[2]*(2^j)*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]
A382961 = samplePDF[120]

A241773 A sequence constructed by greedily sampling the Gauss-Kuzmin distribution log_2[(i+1)^2/(i^2+2*i)] to minimize discrepancy.

Original entry on oeis.org

1, 2, 3, 1, 4, 1, 5, 2, 1, 6, 1, 7, 1, 2, 3, 1, 8, 1, 9, 2, 1, 4, 1, 10, 1, 3, 2, 1, 11, 1, 2, 5, 1, 12, 1, 3, 1, 2, 4, 1, 13, 1, 2, 6, 1, 14, 1, 3, 1, 2, 15, 1, 16, 1, 2, 4, 1, 3, 1, 5, 7, 1, 2, 1, 17, 1, 2, 3, 1, 18, 1, 8, 2, 1, 4, 1, 6, 1, 2, 3, 1, 5, 1, 19

Offset: 1

Jonathan Deane, Apr 28 2014

Let the sequence be A = {a(i)}, i = 1, 2, 3,... and define p(i) =
log_2[(i+1)^2/(i^2+2*i)]. Additionally, define u(j, k) = k*p(j) - N(j, k), where N(j, k) is the number of occurrences of j in {a(i)}, i = 1,..., k-1. Refer to the first argument of u as the "index" of u. Then A is defined by a(1) = 1 and, for i > 1, a(i) = m, where m is the index of the maximal element of the set {u(j, i)}, j = 1, 2, 3,... That there is a single maximal element for all i is guaranteed by the fact that p(i) - p(j) is irrational for all i not equal to j.
Interpreting sequence A as the partial coefficients of the continued fraction expansion of a real number C, say, then C = 1.44224780173510148... which is, by construction, normal (in the continued fraction sense).
The geometric mean of the sequence equals Khintchine's constant K=2.685452001 = A002210 since the frequency of the integers agrees with the Gauss-Kuzmin distribution. - Jwalin Bhatt, Feb 11 2025

			Example from _Jwalin Bhatt_, Jun 10 2025: (Start)
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(n) 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 |     0.415     |     0.169     |     0.093     |   1    |
| 2 |    -0.169     |     0.339     |     0.186     |   2    |
| 3 |     0.245     |    -0.490     |     0.279     |   3    |
| 4 |     0.660     |    -0.320     |    -0.627     |   1    |
(End)

Jonathan Deane, Table of n, a(n) for n = 1..10000
Jonathan Deane, An integer sequence whose members obey a given p.d.f.

Cf. A002210, A084580, A089618, A381617, A382961, A383238, A383899.

pdf := i -> -log[2](1 - 1/(i+1)^2);
gen_seq := proc(n)
local i, j, N, A, u, mm, ndig;
ndig := 40; N := 'N';
for i from 1 to n do    N[i] := 0;      end do;
A := 'A'; A[1] := 1; N[1] := 1;
for i from 2 to n do
        u := 'u';
        for j from 1 to n do
                u[j] := i*pdf(j) - N[j];
        end do;
        mm := max_maxind(evalf(convert(u, list), ndig));
        if mm[3] then
                A[i] := mm[1];
                N[mm[1]] := N[mm[1]] + 1;
        else
                return();
        end if;
end do;
return(convert(A, list));
end:
max_maxind := proc(inl)
local uniq, mxind, mx, i;
uniq := `true`;
if nops(inl) = 1 then return([1, inl[1], uniq]); end if;
mxind := 1; mx := inl[1];
for i from 2 to nops(inl) do
        if inl[i] > mx then
                mxind := i;
                mx := inl[i];
                uniq := `true`;
        elif inl[i] = mx then
                uniq := `false`;
        end if;
end do;
return([mxind, mx, uniq]);
end:
gen_seq(100);

probCountDiff[j_, k_, count_] := k*-Log[2, 1 - (1/((j + 1)^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]
A241773 = samplePDF[120]  (* Jwalin Bhatt, Jun 04 2025 *)

from fractions import Fraction
def prob_count_diff(j, k, count):
    return  - ((1 - Fraction(1,(j+1)*(j+1)))**k) * (2**count)
def sample_pdf(n):
    coeffs, unreached_val, counts = [], 1, {}
    for k in range(1, n+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
A241773 = sample_pdf(120)  # Jwalin Bhatt, Feb 09 2025

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

Original entry on oeis.org

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

Jwalin Bhatt, Jul 11 2025

nonn

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!).

			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    |

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

Cf. A383238, A385685, A385686.

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]

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A382961 A sequence constructed by greedily sampling the logarithmic distribution for parameter value 1/2 so as to minimize discrepancy.

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A241773 A sequence constructed by greedily sampling the Gauss-Kuzmin distribution log_2[(i+1)^2/(i^2+2*i)] to minimize discrepancy.

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

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Programs

Mathematica