Jwalin Bhatt - cp's OEIS frontend

A386904 A sequence constructed by greedily sampling the Borel distribution for parameter value 1/2 to minimize discrepancy.

1, 2, 1, 1, 3, 1, 4, 1, 2, 1, 1, 5, 1, 2, 1, 1, 3, 1, 1, 2, 1, 6, 1, 1, 2, 1, 1, 3, 1, 2, 1, 4, 1, 1, 2, 1, 1, 7, 1, 2, 1, 3, 1, 1, 8, 1, 2, 1, 1, 5, 1, 2, 1, 1, 3, 1, 1, 2, 1, 4, 1, 1, 2, 1, 3, 1, 1, 2, 1, 1, 9, 1, 2, 1, 1, 4, 1, 3, 1, 2, 1, 1, 6, 1, 2, 1, 1, 1, 3, 1, 2, 1, 5, 1, 1, 2, 1, 1, 4, 1, 2, 1, 3, 1, 1, 2, 1, 1, 10

Offset: 1

Jwalin Bhatt, Aug 07 2025

The geometric mean approaches A386009 in the limit.

The Borel distribution with parameter value 1/2 has PDF p(i) = (i/2)^(i-1) / (exp(i/2)*i!).

			Let p(k) denote the probability of k and c(k) denote the count of occurrences of k so far, then the expected occurrences of k at n-th step are 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 |     0.606     |       -       |       -       |   1    |
| 2 |     0.213     |     0.367     |       -       |   2    |
| 3 |     0.819     |    -0.448     |     0.251     |   1    |
| 4 |     0.426     |    -0.264     |     0.334     |   1    |
| 5 |     0.032     |    -0.080     |     0.418     |   3    |

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

Cf. A386009, A386016.

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]
A386904=samplePDF[120]

A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], to minimize discrepancy.

Original entry on oeis.org

0, 1, 0, 2, 3, 0, 1, 4, 0, 2, 5, 1, 0, 3, 0, 1, 6, 2, 4, 0, 1, 0, 3, 2, 7, 0, 5, 1, 0, 2, 4, 1, 3, 0, 0, 1, 6, 2, 0, 3, 5, 1, 4, 8, 0, 2, 0, 1, 0, 3, 2, 1, 0, 4, 7, 0, 5, 1, 6, 2, 3, 0, 1, 0, 2, 4, 0, 1, 3, 0, 2, 5, 1, 0, 0, 3, 4, 1, 6, 2, 0, 1, 0, 7, 2, 3, 0

Offset: 1

Jwalin Bhatt, Aug 10 2025

The pdf of the sequence is the probability to see i as the remainder when you divide 1,2,3... with a random number from [1,10].
The respective probabilities are:
  p(0) = 7381/25200 = 0.292896...
  p(1) = 4861/25200 = 0.192896...
  p(2) = 3601/25200 = 0.142896...
  p(3) = 2761/25200 = 0.109563...
  p(4) = 2131/25200 = 0.084563...
  p(5) = 1627/25200 = 0.064563...
  p(6) = 1207/25200 = 0.047896...
  p(7) = 121/3600   = 0.033611...
  p(8) = 19/900     = 0.021111...
  p(9) = 1/100      = 0.01
The sequence repeats after 25200 terms. If the random number is picked from [1,n] the period of the sequence is given by n*lcm{1,2,...,n} (A081528).
The arithmetic mean of this sequence is 2.25.
For a sequence with length n, the n-th root of the product of nonzero terms is (2**(5333/12600))*(3**(559/3150))*(5**(1627/25200))*(7**(121/3600)) which is 1.9302557640832...

			Let p(k) denote the probability of k and c(k) denote the number of occurrences of k among the first n-1 terms.
We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
| n | (c(0)+1)/p(0) | (c(1)+1)/p(1) | (c(2)+1)/p(2) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     3.414     |       -       |       -       |   0    |
| 2 |     6.828     |     5.181     |       -       |   1    |
| 3 |     6.828     |    10.362     |     6.998     |   0    |
| 4 |    10.242     |    10.362     |     6.998     |   2    |

Jwalin Bhatt, Table of n, a(n) for n = 1..25200

Cf. A081528.

samplePDF[numCoeffs_] := Module[{coeffs, counts},
  coeffs = {}; counts = ConstantArray[0, 10];
  Do[
    minTime = Infinity;
    Do[
      time = 10*(counts[[i]] + 1)/(HarmonicNumber[10] - HarmonicNumber[i - 1]);
      If[time < minTime,minIndex = i;minTime = time],{i, 1, 10}];
    counts[[minIndex]] += 1;
    coeffs = Append[coeffs, minIndex - 1],
    {numCoeffs}
   ];
  coeffs
 ]
A386950 = samplePDF[120]

from sympy import harmonic
def sample_pdf(num_coeffs):
  coeffs, counts = [], [0]*10
  for _ in range(num_coeffs):
    min_time = float('inf')
    for i, count in enumerate(counts):
      time = 10*(count+1) / (harmonic(10)-harmonic(i))
      if time < min_time:
        min_index, min_time = i, time
    counts[min_index] += 1
    coeffs.append(min_index)
  return coeffs
A386950 = sample_pdf(120)

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]

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

Original entry on oeis.org

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, 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, 1, 5, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 4, 1, 1, 7

Offset: 1

Jwalin Bhatt, Jul 14 2025

The geometric mean approaches A386009 in the limit.

The Borel distribution has PDF p(i) = (i/2)^(i-1) / (exp(i/2)*i!).

			Let p(k) denote the probability of k and c(k) denote the count of occurrences of k so far.
We take the ratio of the actual occurrences c(k)+1 to the probability and pick the one with the lowest value.
Since p(k) is monotonic decreasing, we only need to compute c(k) once we see c(k-1).
| n | (c(1)+1)/p(1) | (c(2)+1)/p(2) | (c(3)+1)/p(3) | choice |
|---|---------------|---------------|---------------|--------|
| 1 |     1.648     |       -       |       -       |   1    |
| 2 |     3.297     |     5.436     |       -       |   1    |
| 3 |     4.946     |     5.436     |       -       |   1    |
| 4 |     6.594     |     5.436     |       -       |   2    |
| 5 |     6.594     |    10.873     |    11.951     |   1    |

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

Cf. A386009, A386904.

pdf[i_] := ((i/2)^(i - 1))/((E^(i/2)) * Factorial[i])
samplePDF[numCoeffs_] := Module[
  {coeffs = {}, counts = {0}, minTime, minIndex, time},
Do[
    minTime = Infinity;
    Do[
      time = (counts[[i]] + 1)/pdf[i];
      If[time < minTime, minIndex = i; minTime = time],
      {i, 1, Length[counts]}
    ];
    If[minIndex == Length[counts], AppendTo[counts, 0]];
    counts[[minIndex]] += 1;
    AppendTo[coeffs, minIndex],
    {numCoeffs}
  ];
  coeffs
]
A386016 = samplePDF[120]

A386009 Decimal expansion of Product_{k>=2} k^( ((k/2)^(k-1)) / (exp(k/2)*k!) ).

Original entry on oeis.org

1, 5, 4, 9, 2, 5, 1, 0, 4, 3, 4, 2, 5, 6, 3, 4, 0, 4, 8, 4, 0, 0, 5, 4, 4, 0, 2, 8, 1, 5, 8, 9, 4, 3, 5, 5, 5, 3, 0, 8, 8, 3, 5, 2, 6, 7, 0, 1, 5, 8, 3, 6, 8, 5, 4, 7, 2, 3, 3, 4, 6, 6, 4, 9, 5, 4, 7, 2, 2, 4, 6, 5, 2, 1, 8, 3, 3, 4, 7, 1, 6, 1, 9, 4, 2

Offset: 1

Jwalin Bhatt, Jul 14 2025

The geometric mean of the Borel distribution with parameter value 1/2 (A386016) approaches this constant. In general, for parameter value p it approaches Product_{k>=2} k^(((p*k)^(k-1))/((e^(p*k))*k!)).

			1.549251043425634048400544028158943555...

Cf. A386016.

Exp[NSum[((k/2)^(k-1) * Log[k])/(E^(k/2) * Factorial[k]), {k, 2, Infinity}, WorkingPrecision -> 120, NSumTerms -> 1000]]

Equals exp( Sum_{k>=2} log(k) * (((k/2)^(k-1)) / (exp(k/2)*k!)) ).

A385686 Decimal expansion of exp((Sum_{k>=2} log(k)/k!)/(e-1)).

Original entry on oeis.org

1, 4, 2, 1, 0, 3, 7, 9, 5, 9, 7, 3, 1, 9, 6, 0, 7, 1, 5, 3, 3, 7, 8, 1, 4, 4, 8, 9, 0, 5, 9, 2, 8, 5, 6, 9, 5, 3, 9, 8, 2, 5, 7, 1, 7, 4, 2, 9, 3, 2, 0, 0, 7, 8, 6, 8, 1, 0, 2, 8, 0, 5, 1, 8, 1, 5, 8, 2, 2, 1, 6, 1, 7, 5, 8, 0, 8, 3, 0, 7, 1, 7, 9, 7, 5

Offset: 1

Jwalin Bhatt, Jul 06 2025

The geometric mean of the Poisson distribution with parameter value 1 (A385685) approaches this constant.

			1.4210379597319607153378144890592856953982...

Cf. A296301, A306243, A382095, A385685.

N[Exp [Sum[Log[i]/Factorial[i], {i, 2, Infinity}] / (E-1) ], 120]

PARI
```
prodinf(k=2, k^(1/k!))^(1/(exp(1)-1))
```

Equals exp((Sum_{k>=2} log(k)/k!)/(e-1)).
Equals (Product_{k>=2} k^(1/k!)) ^ (1/(e-1)).
From Vaclav Kotesovec, Jul 08 2025: (Start)
Equals exp(A306243/(exp(1) - 1)).
Equals A296301^(1/(exp(1) - 1)). (End)

A385685 Sequence where k is appended after every k! occurrences of 1, with multiple values following a 1 listed in order.

Original entry on oeis.org

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

Offset: 0

Jwalin Bhatt, Jul 06 2025

nonn

The frequencies of the terms follow the Poisson distribution with parameter value 1.

The geometric mean approaches A385686 in the limit. In general, for parameter value p it approaches Product_{k>=2} k^(((p^(k-1))*((e-1)^(-p)))/k!).

			Every 1 is followed by a 1 because 1! = 1,
after every (2!=2) ones we see a 2,
after every (3!=6) ones we see a 3 and so on.

Jwalin Bhatt, Table of n, a(n) for n = 0..9999
Wikipedia, Poisson distribution

Cf. A000142 (n!), A382093, A385686.

A385685[n_] := Module[{N1 = 0,NR= 1, result = {},i=1}, While[Length[result] < n,N1++;AppendTo[result, 1]; Do[If[Mod[N1, Factorial[k]] == 0,    AppendTo[result, k];f[k == NR + 1, NR++]],{k, 2, NR + 1}];If[Length[result] > n, result = Take[result, n]]];result];A385685[92] (* James C. McMahon, Jul 11 2025 *)

from itertools import islice
from math import factorial
def poisson_distribution_generator():
    num_ones, num_reached = 0, 1
    while num_ones := num_ones+1:
        yield 1
        for num in range(2, num_reached+2):
            if num_ones % factorial(num) == 0:
                yield num
                num_reached += num == num_reached+1
A385685 = list(islice(poisson_distribution_generator(), 120))

A385170 a(n) is the integer part of the reciprocal of the distance of x_n from its nearest integer, where x_n is the n-th extrema of gamma(x).

Original entry on oeis.org

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

Offset: 0

Jwalin Bhatt, Jun 20 2025

nonn

For n>=1, the nearest integer is -n and x_n drifts slowly towards it so that a(n) is a slowly increasing function of n.
a(n) = k occurs for the first time at n = A374856(k).
a(n) approximately equals floor(Pi/arctan(Pi/log(n))).
The corresponding gamma function value y_n = gamma(x_n) decreases in absolute value and A377506 captures its reciprocal (with a round).

			For n=10, x_10 = -9.702672... (A256687) and its nearest integer is -10 which is distance d = -9.702672... - (-10) = 0.297... away and a(n) = floor(1/d) = 3.

Jwalin Bhatt, Table of n, a(n) for n = 0..10000
Wikipedia, Gamma Extrema
Wikipedia, Digamma Roots

Cf. A256687, A374856, A377506.

a[n_] := Module[
{bounds, gammaExtrema, fractionalPart, intReci},
bounds = {-n - 1 + 1/(n + 3), -n - 0.5};
gammaExtrema = x /. FindRoot[PolyGamma[0, x + 1] == 0, {x, Sequence @@ bounds}];
gammaExtrema = -Abs[gammaExtrema];
fractionalPart = Mod[gammaExtrema, 1];
Floor[1 / fractionalPart]
]

from gmpy2 import mpq, get_context, digamma, sign, is_nan, RoundUp, RoundDown
def apply_on_interval(func, interval):
    ctx.round = RoundUp
    rounded_up = func(interval[0])
    ctx.round = RoundDown
    rounded_down = func(interval[1])
    return rounded_down, rounded_up
def digamma_sign_near_int(i, f):
    while True:
        d, u = apply_on_interval(lambda x: digamma(i + 1/x), [f, f])
        sign_d = sign(d)
        if not(is_nan(d)) and not(is_nan(u)) and (sign_d == sign(u)):
            return sign_d
        ctx.precision += 1
def generate_sequence(n):
    i, f, seq = -1, mpq('2'), [1]
    while len(seq) < n:
        if digamma_sign_near_int(i, f) != -1:
            f += 1
        seq.append(int(f)-1)
        i -= 1
    return seq
ctx = get_context()
ctx.precision = 2
A385170 = generate_sequence(101)

A383855 The n-th term of the sequence is k after every k*(k+1)/2 occurrences of 1, with multiple values following a 1 listed in order.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 4, 1, 1, 2, 3, 1, 1, 1, 2, 5, 1, 1, 1, 2, 3, 1, 1, 4, 1, 2, 6, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 7, 1, 1, 2, 3, 4, 5, 1, 1, 1, 2, 1, 1, 1, 2, 3, 8, 1, 1, 1, 2, 1, 4, 1, 1, 2, 3, 6, 1, 1, 1, 2, 5, 9, 1, 1, 1, 2, 3, 1, 1, 4, 1, 2, 1, 1, 1, 2, 3, 1, 10

Offset: 1

Jwalin Bhatt, May 12 2025

nonn

The frequencies of the terms follow the Yule-Simon distribution with parameter value 1. The geometric mean approaches A245254 in the limit.

			After every ((2*3)/2=3) ones we see a 2,
after every ((3*4)/2=6) ones we see a 3,
after every ((4*5)/2=10) ones we see a 4 and so on.

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

Cf. A245254, A383899.

from itertools import islice
def beta_distribution_generator():
    num_ones, num_reached = 0, 1
    while num_ones := num_ones+1:
        yield 1
        for num in range(2, num_reached+2):
            if num_ones % (num*(num+1)//2) == 0:
                yield num
                num_reached += num == num_reached+1
A383855 = list(islice(beta_distribution_generator(), 120))

A383899 A sequence constructed by greedily sampling the Yule-Simon distribution for parameter value 1, to minimize discrepancy selecting the smallest value in case of ties.

Original entry on oeis.org

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

Offset: 1

Jwalin Bhatt, May 14 2025

nonn

The geometric mean approaches A245254 in the limit.

The probability mass function of the Yule-Simon distribution with parameter 1 is given by p(k) = 1/(k*(k+1)) for k >= 1.

			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 |     0.5       |     0.166     |     0.083     |   1    |
| 2 |     0         |     0.333     |     0.166     |   2    |
| 3 |     0.5       |    -0.5       |     0.25      |   1    |
| 4 |     0         |    -0.333     |     0.333     |   3    |
| 5 |     0.5       |    -0.166     |    -0.583     |   1    |

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

Cf. A245254, A383855.

probCountDiff[j_, k_, count_]:=k/(j*(j+1))-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]
A383899=samplePDF[120]

cp's OEIS Frontend

User: Jwalin Bhatt

A386904 A sequence constructed by greedily sampling the Borel distribution for parameter value 1/2 to minimize discrepancy.

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A386950 A sequence constructed by greedily sampling the pdf (H(10)-H(i))/10, where H(n) is the n-th Harmonic number and the support is [0,9], 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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A386016 A sequence constructed by greedily sampling the Borel distribution for parameter value 1/2 to minimize ratio discrepancy.

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A386009 Decimal expansion of Product_{k>=2} k^( ((k/2)^(k-1)) / (exp(k/2)*k!) ).

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Formula

A385686 Decimal expansion of exp((Sum_{k>=2} log(k)/k!)/(e-1)).

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A385685 Sequence where k is appended after every k! occurrences of 1, with multiple values following a 1 listed in order.

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A385170 a(n) is the integer part of the reciprocal of the distance of x_n from its nearest integer, where x_n is the n-th extrema of gamma(x).

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