A396405 - OEIS

%I #16 Jun 09 2026 00:05:27

%S 1,1,1,1,2,1,1,1,1,2,1,3,1,1,1,2,1,1,1,1,2,4,1,1,3,1,1,2,1,1,1,1,2,1,

%T 1,5,1,3,1,2,1,1,1,1,2,1,4,1,1,1,2,1,3,1,6,1,1,2,1,1,1,1,2,1,1,3,1,1,

%U 2,1,1,4,1,7,1,2,5,1,1,1,3,1,2,1,1,1,1,2

%N A sequence constructed by greedily sampling the probability distribution given by 1/log_2(i+1)^2 - 1/log_2(i+2)^2 to minimize ratio discrepancy.

%C The probability mass function used here is a simple example of a heavy log tail distribution. It can be seen as a discrete analog of the log-Cauchy distribution. As a consequence of the heavy tail, the arithmetic mean of this sequence diverges to infinity but it has a finite geometric mean.

%C A general family of these distributions is given by 1/log_2(i+1)^p - 1/log_2(i+2)^p where p>0.

%C Properties change sharply at p=1, the arithmetic mean remains infinite for all p but the geometric mean becomes finite for p>1 and infinite for p<=1.

%C | mean       | 0 < p <= 1 | p > 1  |

%C |------------|------------|--------|

%C | arithmetic | oo         | oo     |

%C | geometric  | oo         | finite |

%C The Fisher information is always finite and given by Sum_{i=1..inf} ((ln(log_2(i+2))/(log_2(i+2))^p - ln(log_2(i+1))/(log_2(i+1))^p)^2) / (1/(log_2(i+1))^p - 1/(log_2(i+2))^p).

%C This sequence is generated by the D'Hondt (or Jefferson) apportionment method for the given distribution.

%C The geometric mean equals Product_{i>=1} (1+1/i)^(1/log_2(i+2)^2) = 2.097184811... and for general power p in the family above it equals Product_{i>=1} (1+1/i)^(1/log_2(i+2)^p). - _Jwalin Bhatt_, Jun 03 2026

%H Jwalin Bhatt, <a href="/A396405/b396405.txt">Table of n, a(n) for n = 1..10000</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/D%27Hondt_method">D'Hondt method</a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Log-Cauchy_distribution">Log-Cauchy distribution</a>.

%e Let p(k) denote the probability of k and c(k) denote the count of occurrences of k so far.

%e We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.

%e Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).

%e | n | (c(1)+1)/p(1) | (c(2)+1)/p(2) | (c(3)+1)/p(3) | choice |

%e |---|---------------|---------------|---------------|--------|

%e | 1 |     1.661     |       -       |       -       |   1    |

%e | 2 |     3.322     |     6.753     |       -       |   1    |

%e | 3 |     4.983     |     6.753     |       -       |   1    |

%e | 4 |     6.645     |     6.753     |       -       |   1    |

%e | 5 |     8.306     |     6.753     |       -       |   2    |

%e | 6 |     8.306     |    13.506     |    15.499     |   1    |

%t pdf[i_] := 1/Log2[i+1]^2 - 1/Log2[i+2]^2

%t samplePDF[numCoeffs_] := Module[

%t   {coeffs = {}, counts = {0}, minTime, minIndex, time},

%t Do[

%t     minTime = Infinity;

%t     Do[

%t       time = (counts[[i]] + 1)/pdf[i];

%t       If[time < minTime, minIndex = i; minTime = time],

%t       {i, 1, Length[counts]}

%t     ];

%t     If[minIndex == Length[counts], AppendTo[counts, 0]];

%t     counts[[minIndex]] += 1;

%t     AppendTo[coeffs, minIndex],

%t     {numCoeffs}

%t   ];

%t   coeffs

%t ]

%t A396405 = samplePDF[120]

%o (Python)

%o from mpmath import iv

%o def pdf(k):

%o     return 1/iv.log(k+1,2)**2 - 1/iv.log(k+2,2)**2

%o def sample_log_tail_distribution(num_coeffs):

%o   coeffs, counts = [], [0]

%o   for _ in range(num_coeffs):

%o     min_time = iv.inf

%o     for i, count in enumerate(counts, start=1):

%o       time = (count+1) / pdf(i)

%o       if time < min_time:

%o         min_index, min_time = i, time

%o     if min_index == len(counts):

%o       counts.append(0)

%o     counts[min_index-1] += 1

%o     coeffs.append(min_index)

%o   return coeffs

%o A396405 = sample_log_tail_distribution(120)

%Y Cf. A393100, A394370, A394401.

%K nonn

%O 1,5

%A _Jwalin Bhatt_, May 24 2026