cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A302922 Raw moments of a Fibonacci-geometric probability distribution.

Original entry on oeis.org

1, 6, 58, 822, 15514, 366006, 10361818, 342239862, 12918651034, 548600581686, 25885279045978, 1343513774912502, 76071145660848154, 4666162902628259766, 308236822886732856538, 21815861409181135034742, 1646982315540717414270874, 132109620398598537723816246
Offset: 0

Views

Author

Albert Gordon Smith, Apr 15 2018

Keywords

Comments

If F(k) is the k-th Fibonacci number, where F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2), then p(k)=F(k-1)/2^k is a normalized probability distribution on the positive integers.
For example, it is the probability that k coin tosses are required to get two heads in a row, or the probability that a random series of k bits has its first two consecutive 1's at the end.
The g.f. for this distribution is g(x) = x^2/(4-2x-x^2) = (1/4)x^2 + (1/8)x^3 + (1/8)x^4 + (3/32)x^5 + ....
The n-th moments about zero of this distribution, known as raw moments, are defined by a(n) = Sum_{k>=1} (k^n)p(k). They appear to be integers and form this sequence.
The e.g.f. for the raw moments is g(e^x) = 1 + 6x + 58x^2/2! + 822x^3/3! + ....
For n >= 1, a(n) appears to be even.
Dividing these terms by 2 gives sequence A302923.
The central moments (i.e., the moments about the mean) also appear to be integers. They form sequence A302924.
The central moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302925.
The cumulants of this distribution, defined by the cumulant e.g.f. log(g(e^x)), also appear to be integers. They form sequence A302926.
The cumulants also appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.
Note: Another probability distribution on the positive integers that has integral moments and cumulants is the geometric distribution p(k)=1/2^k. The sequences related to these moments are A000629, A000670, A052841, and A091346.

Examples

			a(0)=1 is the 0th raw moment of the distribution, which is the total probability.
a(1)=6 is the 1st raw moment, known as the mean of the distribution. It is the arithmetic average of integers following the distribution.
a(2)=58 is the 2nd raw moment. It is the arithmetic average of the squares of integers following the distribution.
		

Crossrefs

Raw half-moments: A302923.
Central moments: A302924.
Central half-moments: A302925.
Cumulants: A302926.
Half-cumulants: A302927.

Programs

  • Mathematica
    Module[{max, r, g},
      max = 17;
      r = Range[0, max];
      g[x_] := x^2/(4 - 2 x - x^2);
      r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x]
    ]
  • PARI
    Vec(serlaplace(exp(2*x)/(4-2*exp(x)-exp(2*x)))) \\ Michel Marcus, Apr 17 2018

Formula

In the following,
F(k) is the k-th Fibonacci number, as defined in the Comments.
phi=(1+sqrt(5))/2 is the golden ratio, and psi=(1-sqrt(5))/2.
Li(s,z) is the polylogarithm of order s and argument z.
When s is a negative integer as it is here, Li(s,z) is a rational function of z: Li(-n,z) = (z(d/dz))^n(z/(1-z)).
For n>=0:
a(n) = Sum_{k>=1} ((k^n)(F(k-1)/2^k));
a(n) = Sum_{k>=1} ((k^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
a(n) = (Li(-n,phi/2)/phi-Li(-n,psi/2)/psi)/sqrt(5).
E.g.f.: g(e^x) where g(x) = x^2/(4-2x-x^2) is the g.f. for the probability distribution.
a(n) ~ n! * (5 - sqrt(5)) / (10 * (log(sqrt(5) - 1))^(n+1)). - Vaclav Kotesovec, Apr 13 2022

A306417 Number of self-conjugate set partitions of {1, ..., n}.

Original entry on oeis.org

1, 1, 0, 1, 1, 2, 7, 7, 46, 39, 321
Offset: 0

Views

Author

Gus Wiseman, Feb 14 2019

Keywords

Comments

This sequence counts set partitions fixed under Callan's conjugation operation.

Examples

			The  a(3) = 1 through a(7) = 7 self-conjugate set partitions:
  {{12}{3}}  {{13}{24}}  {{123}{4}{5}}  {{135}{246}}    {{13}{246}{57}}
                         {{13}{2}{45}}  {{124}{35}{6}}  {{15}{246}{37}}
                                        {{13}{25}{46}}  {{1234}{5}{6}{7}}
                                        {{14}{2}{356}}  {{124}{3}{56}{7}}
                                        {{14}{236}{5}}  {{134}{2}{5}{67}}
                                        {{14}{25}{36}}  {{14}{2}{3}{567}}
                                        {{145}{26}{3}}  {{14}{23}{57}{6}}
		

Crossrefs

A316747 Stirling transform of (2*n)!.

Original entry on oeis.org

1, 2, 26, 794, 44810, 4050362, 536119946, 97759687034, 23495075990090, 7197163489723322, 2737224615568742666, 1265459307754418362874, 698926543187678223962570, 454516898016585094157146682, 343753040265700944173260034186, 299168865461564926143049346952314
Offset: 0

Views

Author

Vaclav Kotesovec, Jul 12 2018

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[StirlingS2[n, k]*(2*k)!, {k, 0, n}], {n, 0, 20}]
  • PARI
    my(N=20, x='x+O('x^N)); Vec(serlaplace(sum(k=0, N, (2*k)!*(exp(x)-1)^k/k!))) \\ Seiichi Manyama, May 20 2022

Formula

a(n) = Sum_{k=0..n} Stirling2(n,k) * (2*k)!.
a(n) ~ exp(1/8) * (2*n)!.
a(n) ~ sqrt(Pi) * 2^(2*n + 1) * n^(2*n + 1/2) / exp(2*n - 1/8).
E.g.f.: Sum_{k>=0} (2*k)! * (exp(x) - 1)^k / k!. - Seiichi Manyama, May 20 2022

A330149 Expansion of e.g.f. exp(-x) / (1 + log(1 - x)).

Original entry on oeis.org

1, 0, 2, 7, 47, 368, 3494, 38673, 489341, 6966344, 110199090, 1917589771, 36402276107, 748629861016, 16580304397942, 393443385034069, 9958671117295737, 267824225078212336, 7626444798009902530, 229232204568273395919, 7252798333599466521575, 240948882537990850397536
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 03 2019

Keywords

Comments

Inverse binomial transform of A007840.

Crossrefs

Programs

  • Mathematica
    nmax = 21; CoefficientList[Series[Exp[-x]/(1 + Log[1 - x]), {x, 0, nmax}], x] Range[0, nmax]!

Formula

a(n) = Sum_{k=0..n} (-1)^(n - k) * binomial(n,k) * A007840(k).
a(n) ~ n! * exp(n + exp(-1) - 1) / (exp(1) - 1)^(n+1). - Vaclav Kotesovec, Dec 15 2019
a(n) = (-1)^n + Sum_{k=1..n} (k-1)! * binomial(n,k) * a(n-k). - Seiichi Manyama, Dec 19 2023

A337564 Number of sequences of length 2*n covering an initial interval of positive integers and splitting into n maximal runs.

Original entry on oeis.org

1, 1, 6, 80, 1540, 38808, 1206744, 44595408, 1908389340, 92780281880, 5050066185736, 304196411024688, 20087958167374552, 1442953024024996400, 112007566256683719600, 9342904053303870936480, 833388624898522799682780, 79159669418651567937733080
Offset: 0

Views

Author

Gus Wiseman, Sep 03 2020

Keywords

Comments

Sequences covering an initial interval of positive integers are counted by A000670 and ranked by A333217.

Examples

			The a(0) = 1 through a(2) = 6 sequences:
  ()  (1,1)  (1,1,1,2)
             (1,1,2,2)
             (1,2,2,2)
             (2,1,1,1)
             (2,2,1,1)
             (2,2,2,1)
The a(3) = 80 sequences:
  212222  111121  122233  333112  211133
  221222  111211  133222  333211  233111
  222122  112111  222133  112233  331112
  222212  121111  222331  113322  332111
  122221  123333  331222  221133  111223
  211222  133332  332221  223311  111322
  221122  213333  122223  331122  221113
  222112  233331  132222  332211  223111
  112221  333312  222213  112223  311122
  122211  333321  222231  113222  322111
  211122  122333  312222  222113  111123
  221112  133322  322221  222311  111132
  111221  221333  112333  311222  211113
  112211  223331  113332  322211  231111
  122111  333122  211333  111233  311112
  211112  333221  233311  111332  321111
		

Crossrefs

A335461 has this as main diagonal n = 2*k.
A336108 is the version for compositions.
A337504 is the version for compositions and anti-runs.
A337505 is the version for anti-runs.
A000670 counts sequences covering an initial interval.
A005649 counts anti-runs covering an initial interval.
A124767 counts maximal runs in standard compositions.
A333769 gives run lengths in standard compositions.
A337504 counts compositions of 2*n with n maximal anti-runs.
A337565 gives anti-run lengths in standard compositions.

Programs

  • Mathematica
    allnorm[n_]:=If[n<=0,{{}},Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1]];
    Table[Length[Select[Join@@Permutations/@allnorm[2*n],Length[Split[#]]==n&]],{n,0,3}]
  • PARI
    \\ here b(n) is A005649.
    b(n) = {sum(k=0, n, stirling(n,k,2)*(k + 1)!)}
    a(n) = {if(n==0, 1, b(n-1)*binomial(2*n-1,n-1))} \\ Andrew Howroyd, Dec 31 2020

Formula

a(n) = A005649(n-1)*binomial(2*n-1,n-1) = A005649(n-1)*A001700(n-1) for n > 0. - Andrew Howroyd, Dec 31 2020

Extensions

Terms a(5) and beyond from Andrew Howroyd, Dec 31 2020

A353774 Expansion of e.g.f. 1/(1 - (exp(x) - 1)^3).

Original entry on oeis.org

1, 0, 0, 6, 36, 150, 1260, 16926, 197316, 2286150, 32821020, 548528046, 9515702196, 174531124950, 3521913283980, 76969474578366, 1777400236160676, 43405229295464550, 1126972561394470140, 30949983774936839886, 893095888222540548756, 27035433957000465352950
Offset: 0

Views

Author

Seiichi Manyama, May 07 2022

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(1/(1-(exp(x)-1)^3)))
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, (3*k)!*x^(3*k)/prod(j=1, 3*k, 1-j*x)))
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=6*sum(j=1, i, binomial(i, j)*stirling(j, 3, 2)*v[i-j+1])); v;
    
  • PARI
    a(n) = sum(k=0, n\3, (3*k)!*stirling(n, 3*k, 2));

Formula

G.f.: Sum_{k>=0} (3*k)! * x^(3*k)/Product_{j=1..3*k} (1 - j * x).
a(0) = 1; a(n) = 6 * Sum_{k=1..n} binomial(n,k) * Stirling2(k,3) * a(n-k).
a(n) = Sum_{k=0..floor(n/3)} (3*k)! * Stirling2(n,3*k).
a(n) ~ n! / (6 * log(2)^(n+1)). - Vaclav Kotesovec, May 08 2022

A302923 Raw half-moments of a Fibonacci-geometric probability distribution.

Original entry on oeis.org

3, 29, 411, 7757, 183003, 5180909, 171119931, 6459325517, 274300290843, 12942639522989, 671756887456251, 38035572830424077, 2333081451314129883, 154118411443366428269, 10907930704590567517371, 823491157770358707135437, 66054810199299268861908123
Offset: 1

Views

Author

Albert Gordon Smith, Apr 15 2018

Keywords

Comments

If F(k) is the k-th Fibonacci number, where F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2), then p(k)=F(k-1)/2^k is a normalized probability distribution on the positive integers.
For example, it is the probability that k coin tosses are required to get two heads in a row, or the probability that a random series of k bits has its first two consecutive 1's at the end.
The g.f. for this distribution is g(x) = x^2/(4-2x-x^2) = (1/4)x^2 + (1/8)x^3 + (1/8)x^4 + (3/32)x^5 + ....
The n-th moments about zero of this distribution, known as raw moments, are defined by a(n) = Sum_{k>=1} (k^n)*p(k). They appear to be integers and form A302922.
The e.g.f. for the raw moments is g(e^x) = 1 + 6x + 58x^2/2! + 822x^3/3! + ....
For n >= 1, the raw moments appear to be even. Dividing them by 2 gives this sequence of raw half-moments.
The central moments (i.e., the moments about the mean) also appear to be integers. They form sequence A302924.
The central moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302925.
The cumulants of this distribution, defined by the cumulant e.g.f. log(g(e^x)), also appear to be integers. They form sequence A302926.
The cumulants also appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.
Note: Another probability distribution on the positive integers that has integral moments and cumulants is the geometric distribution p(k)=1/2^k. The sequences related to these moments are A000629, A000670, A052841, A091346.

Examples

			a(1)=3 is half the first raw moment of the distribution. It is half the arithmetic average of integers following the distribution.
a(2)=29 is half the second raw moment. It is half the arithmetic average of the squares of integers following the distribution.
		

Crossrefs

Raw moments: A302922.
Central moments: A302924.
Central half-moments: A302925.
Cumulants: A302926.
Half-cumulants: A302927.

Programs

  • Mathematica
    Module[{max, r, g, rawMoments},
      max = 17;
      r = Range[0, max];
      g[x_] := x^2/(4 - 2 x - x^2);
      rawMoments = r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x];
      Rest[rawMoments]/2
    ]

Formula

In the following,
F(k) is the k-th Fibonacci number, as defined in the Comments.
phi=(1+sqrt(5))/2 is the golden ratio, and psi=(1-sqrt(5))/2.
Li(s,z) is the polylogarithm of order s and argument z.
When s is a negative integer as it is here, Li(s,z) is a rational function of z: Li(-n,z) = (z(d/dz))^n(z/(1-z)).
For n>=1:
a(n) = (1/2)A302922(n);
a(n) = (1/2)Sum_{k>=1} ((k^n)(F(k-1)/2^k));
a(n) = (1/2)Sum_{k>=1} ((k^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
a(n) = (1/2)(Li(-n,phi/2)/phi-Li(-n,psi/2)/psi)/sqrt(5).
E.g.f.: (1/2)g(e^x) where g(x) = x^2/(4-2x-x^2) is the g.f. for the probability distribution.

A302924 Central moments of a Fibonacci-geometric probability distribution.

Original entry on oeis.org

1, 0, 22, 210, 4426, 102330, 2906362, 95952570, 3622138906, 153816150810, 7257695358202, 376693381614330, 21328770664314586, 1308295248437904090, 86423208789970618042, 6116714829331037666490, 461779664078480243085466, 37040796099362864616022170
Offset: 0

Views

Author

Albert Gordon Smith, Apr 15 2018

Keywords

Comments

If F(k) is the k-th Fibonacci number, where F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2), then p(k)=F(k-1)/2^k is a normalized probability distribution on the positive integers.
For example, it is the probability that k coin tosses are required to get two heads in a row, or the probability that a random series of k bits has its first two consecutive 1's at the end.
The g.f. for this distribution is g(x) = x^2/(4-2x-x^2) = (1/4)x^2 + (1/8)x^3 + (1/8)x^4 + (3/32)x^5 + ....
The mean of this distribution is 6. (See A302922.)
The n-th moments about the mean, known as central moments, are defined by a(n) = Sum_{k>=1} ((k-6)^n)p(k). They appear to be integers and form this sequence.
For n >= 1, a(n) appears to be even. Dividing these terms by 2 gives sequence A302925.
The raw moments (i.e., the moments about zero) also appear to be integers. This is sequence A302922.
The raw moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302923.
The cumulants of this distribution, defined by the cumulant e.g.f. log(g(e^x)), also appear to be integers. They form sequence A302926.
The cumulants also appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.
Note: Another probability distribution on the positive integers that has integral moments and cumulants is the geometric distribution p(k)=1/2^k. The sequences related to these moments are A000629, A000670, A052841, and A091346.

Examples

			a(0)=1 is the 0th central moment of the distribution, which is the total probability.
a(1)=0 is the 1st central moment, or the "mean about the mean". It is zero by definition of central moments.
a(2)=22 is the 2nd central moment, known as the variance or the square of the standard deviation. It measures how far integers following the distribution are from the mean by averaging the squares of their differences from the mean.
		

Crossrefs

Central half-moments: A302925.
Raw moments: A302922.
Raw half-moments: A302923.
Cumulants: A302926.
Half-cumulants: A302927.

Programs

  • Mathematica
    Module[{max, r, g, moments},
      max = 17;
      r = Range[0, max];
      g[x_] := x^2/(4 - 2 x - x^2);
      moments = r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x];
      Table[Sum[Binomial[n, k] moments[[k + 1]] (-6)^(n - k), {k, 0, n}], {n, 0, max}]
    ]

Formula

In the following,
F(k) is the k-th Fibonacci number, as defined in the Comments.
phi=(1+sqrt(5))/2 is the golden ratio, and psi=(1-sqrt(5))/2.
LerchPhi(z,s,a) = Sum_{k>=0} z^k/(a+k)^s is the Lerch transcendant.
For n >= 0:
a(n) = Sum_{k>=1} (((k-6)^n)(F(k-1)/2^k));
a(n) = Sum_{k>=1} (((k-6)^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
a(n) = (LerchPhi(phi/2,-n,-5)-LerchPhi(psi/2,-n,-5))/(2 sqrt(5));
a(n) = Sum_{k=0..n} (binomial(n,k)*A302922(k)*(-6)^(n-k)).

A302925 Central half-moments of a Fibonacci-geometric probability distribution.

Original entry on oeis.org

0, 11, 105, 2213, 51165, 1453181, 47976285, 1811069453, 76908075405, 3628847679101, 188346690807165, 10664385332157293, 654147624218952045, 43211604394985309021, 3058357414665518833245, 230889832039240121542733, 18520398049681432308011085
Offset: 1

Views

Author

Albert Gordon Smith, Apr 15 2018

Keywords

Comments

If F(k) is the k-th Fibonacci number, where F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2), then p(k)=F(k-1)/2^k is a normalized probability distribution on the positive integers.
For example, it is the probability that k coin tosses are required to get two heads in a row, or the probability that a random series of k bits has its first two consecutive 1's at the end.
The g.f. for this distribution is g(x) = x^2/(4-2x-x^2) = (1/4)x^2 + (1/8)x^3 + (1/8)x^4 + (3/32)x^5 + ....
The mean of this distribution is 6. (See A302922.)
The n-th moments about the mean, known as central moments, are defined by a(n) = Sum_{k>=1} ((k-6)^n)p(k). They appear to be integers and form A302924.
For n >= 1, that sequence appears to be even. Dividing those terms by 2 gives this sequence.
The raw moments (i.e., the moments about zero) also appear to be integers. This is sequence A302922.
The raw moments also appear to be even for n >= 1. Dividing them by 2 gives sequence A302923.
The cumulants of this distribution, defined by the cumulant e.g.f. log(g(e^x)), also appear to be integers. They form sequence A302926.
The cumulants also appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.
Note: Another probability distribution on the positive integers that has integral moments is the geometric distribution p(k)=1/2^k. The sequences related to these moments are A000629, A000670, A052841, and A091346.

Examples

			a(1)=0 is half the 1st central moment of the distribution, or half the "mean about the mean". It is zero by definition of central moments.
a(2)=11 is half the 2nd central moment, or half the variance, or half the square of the standard deviation.
		

Crossrefs

Central moments: A302924.
Raw moments: A302922.
Raw half-moments: A302923.
Cumulants: A302926.
Half-cumulants: A302927.

Programs

  • Mathematica
    Module[{max, r, g, moments, centralMoments},
      max = 17;
      r = Range[0, max];
      g[x_] := x^2/(4 - 2 x - x^2);
      moments = r! CoefficientList[Normal[Series[g[Exp[x]], {x, 0, max}]], x];
      centralMoments = Table[Sum[Binomial[n, k] moments[[k + 1]] (-6)^(n - k), {k, 0, n}], {n, 0, max}];
      Rest[centralMoments]/2
    ]

Formula

In the following,
F(k) is the k-th Fibonacci number, as defined in the Comments.
phi=(1+sqrt(5))/2 is the golden ratio, and psi=(1-sqrt(5))/2.
LerchPhi(z,s,a) = Sum_{k>=0} z^k/(a+k)^s is the Lerch transcendant.
For n>=1:
a(n) = (1/2)A302924(n);
a(n) = (1/2)Sum_{k>=1} (((k-6)^n)(F(k-1)/2^k));
a(n) = (1/2)Sum_{k>=1} (((k-6)^n)(((phi^(k-1)-psi^(k-1))/sqrt(5))/2^k));
a(n) = (1/2)(LerchPhi(phi/2,-n,-5)-LerchPhi(psi/2,-n,-5))/(2 sqrt(5));
a(n) = (1/2)Sum_{k=0..n} (binomial(n,k)A302922(k)(-6)^(n-k)).

A302926 Cumulants of a Fibonacci-geometric probability distribution.

Original entry on oeis.org

0, 6, 22, 210, 2974, 56130, 1324222, 37489410, 1238235454, 46740118530, 1984855550782, 93653819396610, 4860878501987134, 275227990564092930, 16882335978752910142, 1115211301788480951810, 78930528072274523870014, 5958837996496319756259330
Offset: 0

Views

Author

Albert Gordon Smith, Apr 15 2018

Keywords

Comments

If F(k) is the k-th Fibonacci number, where F(0)=0, F(1)=1, and F(n)=F(n-1)+F(n-2), then p(k)=F(k-1)/2^k is a normalized probability distribution on the positive integers.
For example, it is the probability that k coin tosses are required to get two heads in a row, or the probability that a random series of k bits has its first two consecutive 1's at the end.
The g.f. for this distribution is g(x) = x^2/(4-2x-x^2) = (1/4)x^2 + (1/8)x^3 + (1/8)x^4 + (3/32)x^5 + ....
The cumulants of this distribution, defined by the cumulant e.g.f. log(g(e^x)), appear to be integers and form this sequence.
The cumulants appear to be even for n >= 0. Dividing them by 2 gives sequence A302927.
The n-th moments about zero of this distribution, known as raw moments, are defined by a(n) = Sum_{k>=1} (k^n)p(k). They also appear to be integers and form sequence A302922.
For n >= 1, the raw moments also appear to be even. Dividing them by 2 gives sequence A302923.
The central moments (i.e., the moments about the mean) also appear to be integers. They form sequence A302924.
For n >= 1, the central moments also appear to be even. Dividing them by 2 gives sequence A302925.
Note: Another probability distribution on the positive integers that has integral moments and cumulants is the geometric distribution p(k)=1/2^k. The sequences related to these moments are A000629, A000670, A052841, and A091346.
Variant of A103437. - R. J. Mathar, Jun 09 2018

Examples

			a(0)=0 is the 0th cumulant of the distribution. The 0th cumulant is always zero.
a(1)=6 is the 1st cumulant, which is always the mean.
a(2)=22 is the 2nd cumulant, which is always the variance.
		

Crossrefs

Half-cumulants: A302927.
Raw moments: A302922.
Raw half-moments: A302923.
Central moments: A302924.
Central half-moments: A302925.

Programs

  • Mathematica
    Module[{max, r, g},
      max = 17;
      r = Range[0, max];
      g[x_] := x^2/(4 - 2 x - x^2);
      r! CoefficientList[Normal[Series[Log[g[Exp[x]]], {x, 0, max}]], x]
    ]

Formula

E.g.f.: log(g(e^x)) where g(x) = x^2/(4-2x-x^2) is the g.f. for the probability distribution.
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