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A386016 A sequence constructed by greedily sampling the Borel distribution for parameter value 1/2 to minimize ratio discrepancy.

%I A386016 #17 Aug 22 2025 17:19:39
%S A386016 1,1,1,2,1,1,1,2,1,3,1,1,2,1,1,1,1,2,4,1,3,1,1,2,1,1,1,2,1,1,3,1,5,1,
%T A386016 2,1,1,1,2,4,1,1,3,1,2,1,1,1,2,1,1,1,1,6,3,2,1,1,1,2,1,4,1,1,2,1,3,1,
%U A386016 1,5,1,2,1,1,1,2,1,3,1,1,2,1,4,1,1,7
%N A386016 A sequence constructed by greedily sampling the Borel distribution for parameter value 1/2 to minimize ratio discrepancy.
%C A386016 The geometric mean approaches A386009 in the limit.
%C A386016 The Borel distribution has PDF p(i) = (i/2)^(i-1) / (exp(i/2)*i!).
%H A386016 Jwalin Bhatt, <a href="/A386016/b386016.txt">Table of n, a(n) for n = 1..10000</a>
%H A386016 Wikipedia, <a href="https://en.wikipedia.org/wiki/Borel_distribution">Borel distribution</a>
%e A386016 Let p(k) denote the probability of k and c(k) denote the count of occurrences of k so far.
%e A386016 We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
%e A386016 Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
%e A386016 | n | (c(1)+1)/p(1) | (c(2)+1)/p(2) | (c(3)+1)/p(3) | choice |
%e A386016 |---|---------------|---------------|---------------|--------|
%e A386016 | 1 |     1.648     |       -       |       -       |   1    |
%e A386016 | 2 |     3.297     |     5.436     |       -       |   1    |
%e A386016 | 3 |     4.946     |     5.436     |       -       |   1    |
%e A386016 | 4 |     6.594     |     5.436     |       -       |   2    |
%e A386016 | 5 |     6.594     |    10.873     |    11.951     |   1    |
%t A386016 pdf[i_] := ((i/2)^(i - 1))/((E^(i/2)) * Factorial[i])
%t A386016 samplePDF[numCoeffs_] := Module[
%t A386016   {coeffs = {}, counts = {0}, minTime, minIndex, time},
%t A386016 Do[
%t A386016     minTime = Infinity;
%t A386016     Do[
%t A386016       time = (counts[[i]] + 1)/pdf[i];
%t A386016       If[time < minTime, minIndex = i; minTime = time],
%t A386016       {i, 1, Length[counts]}
%t A386016     ];
%t A386016     If[minIndex == Length[counts], AppendTo[counts, 0]];
%t A386016     counts[[minIndex]] += 1;
%t A386016     AppendTo[coeffs, minIndex],
%t A386016     {numCoeffs}
%t A386016   ];
%t A386016   coeffs
%t A386016 ]
%t A386016 A386016 = samplePDF[120]
%Y A386016 Cf. A386009, A386904.
%K A386016 nonn,changed
%O A386016 1,4
%A A386016 _Jwalin Bhatt_, Jul 14 2025