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.

Previous Showing 31-37 of 37 results.

A078921 Signed variant of A077012.

Original entry on oeis.org

1, -1, 2, 2, -3, 6, -6, 8, -12, 24, 24, -30, 40, -60, 120, -120, 144, -180, 240, -360, 720, 720, -840, 1008, -1260, 1680, -2520, 5040, -5040, 5760, -6720, 8064, -10080, 13440, -20160, 40320, 40320, -45360, 51840, -60480, 72576, -90720, 120960, -181440, 362880, -362880, 403200, -453600, 518400, -604800, 725760, -907200, 1209600, -1814400, 3628800
Offset: 1

Views

Author

Wouter Meeussen, Dec 14 2002

Keywords

Comments

Row sums give A024167.

Examples

			Triangle starts:
     1
    -1,    2
     2,   -3,    6
    -6,    8,  -12,    24
    24,  -30,   40,   -60,  120
  -120,  144, -180,   240, -360,   720
   720, -840, 1008, -1260, 1680, -2520, 5040
  ...
		

Crossrefs

Programs

  • Mathematica
    Table[Table[ -(-1)^(n-k+1) n/(n-k+1), {k, 1, n}] (n-1)!, {n, 1, 12}]

Formula

T(n, k) = -(-1)^(n-k+1)*(n/(n-k+1))*(n-1)!.
E.g.f.: log(1+x)/(1-y*x). - Vladeta Jovovic, Feb 07 2003
Sum_{n>=1} Sum_{k=1..n} 1/T(n, k) = (e^2+1)/(4*e). - Amiram Eldar, Jun 29 2025

A087301 a(n) = n!*Sum_{i=1..n-1} (-1)^(i+1)/i.

Original entry on oeis.org

2, 3, 20, 70, 564, 3108, 30624, 230256, 2705760, 25771680, 352805760, 4067556480, 63651813120, 861371884800, 15176802816000, 235775183616000, 4620563523072000, 81032645804544000, 1748700390205440000
Offset: 2

Views

Author

Vladeta Jovovic, Oct 20 2003

Keywords

Comments

Stirling transform of A052882(n)=[0,2,9,52,375,...] is a(n+1)=[0,2,3,20,...]. - Michael Somos, Mar 04 2004

Crossrefs

Programs

  • Mathematica
    Rest[Table[n!Sum[(-1)^(i+1)/i,{i,n-1}],{n,20}]] (* Harvey P. Dale, Oct 24 2011 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(log(1+x+x*O(x^n))*x/(1-x),n))

Formula

E.g.f.: x*log(1+x)/(1-x). a(n) = 1/2*(-1)^n*n!*(2*(-1)^n*log(2)+Psi(1/2+1/2*n)-Psi(1/2*n)).
a(n) ~ n! * log(2). - Vaclav Kotesovec, Jul 01 2018

A346943 a(n) = a(n-1) + n*(n+1)*a(n-2) with a(0)=1, a(1)=1.

Original entry on oeis.org

1, 1, 7, 19, 159, 729, 7407, 48231, 581535, 4922325, 68891175, 718638075, 11465661375, 142257791025, 2550046679775, 36691916525775, 730304613424575, 11958031070311725, 261722208861516375, 4805774015579971875, 114729101737416849375, 2334996696935363855625
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 08 2021, following a suggestion from John M. Campbell

Keywords

Comments

From Peter Bala, Dec 09 2024: (Start)
b(n) := A000246(n+2) = (n+2)!/2^(n+1) * binomial(n+1, floor((n+1)/2)) satisfies the same second-order recurrence as a(n) with the initial conditions b(0) = 1 and b(1) = 3. This leads to the finite continued fraction a(n)/b(n) = 1/(1 + 2/(1 + 6/(1 + ... + n*(n+1)/1). Letting n tend to infinity gives the continued fraction representation 1/(1 + 2/(1 + 6/(1 + ... + n*(n+1)/(1 + ...) = Pi/2 - 1, due to Euler - see paragraph 31, p. 48. (End)

Crossrefs

Programs

  • Mathematica
    RecurrenceTable[{a[n] == a[n-1] + n*(n+1)*a[n-2], a[0]==1, a[1]==1}, a, {n,0,20}]
    nmax = 20; CoefficientList[Series[(-2 + Pi + 2*Pi*x + 4*Sqrt[1 - x^2] + 2*x*(-2 + Sqrt[1 - x^2]) - 4*(1 + 2*x) * ArcSin[Sqrt[1 - x]/Sqrt[2]]) / (2*(1 - x)^(5/2) * (1 + x)^(3/2)), {x, 0, nmax}], x] * Range[0, nmax]!

Formula

a(n) ~ n! * (Pi - 2) * n^(3/2) / sqrt(2*Pi).
a(n) ~ (Pi - 2) * n^(n+2) / exp(n).
E.g.f. A(x) satisfies the differential equation -6*A(x) - (6*x + 1)*A'(x) + (1 - x^2)*A''(x) = 0, A(0)=1, A'(0)=1.
E.g.f.: (-2 + Pi + 2*Pi*x + 4*sqrt(1-x^2) + 2*x*(-2+sqrt(1-x^2)) - 4*(1+2*x) * arcsin(sqrt(1-x)/sqrt(2))) / (2*(1-x)^(5/2) * (1+x)^(3/2)).

A371685 Triangle read by rows: T(n, k) = n! * Sum_{j=0..n-1} binomial(k - 1, j) / (j + 1).

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 5, 6, 9, 14, 14, 24, 36, 56, 90, 94, 120, 180, 280, 450, 744, 444, 720, 1080, 1680, 2700, 4464, 7560, 3828, 5040, 7560, 11760, 18900, 31248, 52920, 91440, 25584, 40320, 60480, 94080, 151200, 249984, 423360, 731520, 1285200
Offset: 0

Views

Author

Peter Luschny, Apr 06 2024

Keywords

Examples

			Triangle starts:
[0]    0;
[1]    1,    1;
[2]    1,    2,    3;
[3]    5,    6,    9,    14;
[4]   14,   24,   36,    56,    90;
[5]   94,  120,  180,   280,   450,   744;
[6]  444,  720, 1080,  1680,  2700,  4464,  7560;
[7] 3828, 5040, 7560, 11760, 18900, 31248, 52920, 91440;
		

Crossrefs

Cf. A029767 (main diagonal), A024167 (column 0), A371768 (row sums).

Programs

  • Maple
    T := (n, k) -> local j; n!*add(binomial(k-1, j)/(j + 1), j = 0..n-1):
    T := (n, k) -> local j; n!*ifelse(n = 0, 0, ifelse(k=0, add(-(-1)^j/j, j = 1..n), (2^k - 1) / k)):
    seq(print(seq(T(n, k), k = 0..n)), n = 0..7);

Formula

Restricted to the range 1 <= k <= n: T(n, k) = n!*(2^k - 1)/k.

A302582 a(n) = n! * [x^n] log(1 + x)/(1 - x)^n.

Original entry on oeis.org

0, 1, 3, 29, 386, 6774, 146484, 3762744, 111868560, 3777096240, 142734788640, 5967788097600, 273488036169600, 13631083378617600, 734083968523046400, 42477063883483622400, 2628184745184816384000, 173147202267665649408000, 12100888735302910523904000, 894183767796064712795136000
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 10 2018

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n! SeriesCoefficient[Log[1 + x]/(1 - x)^n, {x, 0, n}], {n, 0, 19}]
    Table[n! Sum[(-1)^(k + 1) Binomial[2 n - k - 1, n - k]/k, {k, 1,  n}], {n, 0, 19}]
    Join[{0}, Table[n^2 (2 (n - 1))! HypergeometricPFQ[{1, 1, 1 - n}, {2, 2 - 2 n}, -1]/n!, {n, 19}]]

Formula

a(n) = n!*Sum_{k=1..n} (-1)^(k+1)*binomial(2*n-k-1,n-k)/k.
a(n) ~ log(3/2) * 2^(2*n - 1/2) * n^n / exp(n). - Vaclav Kotesovec, May 05 2018

A336250 a(n) = (n!)^n * Sum_{k=1..n} (-1)^(k+1) / k^n.

Original entry on oeis.org

0, 1, 3, 197, 313840, 24191662624, 137300308036448256, 81994640912971156525105152, 6958651785463110878359050928999366656, 108902755985567407887534498777329973193771818418176, 395560567918154447056086270973712023435510589158871531520000000000
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 14 2020

Keywords

Crossrefs

Programs

  • Mathematica
    Table[(n!)^n Sum[(-1)^(k + 1)/k^n, {k, 1, n}], {n, 0, 10}]
    Table[(n!)^n SeriesCoefficient[-PolyLog[n, -x]/(1 - x), {x, 0, n}], {n, 0, 10}]
  • PARI
    a(n) = (n!)^n * sum(k=1, n, (-1)^(k+1) / k^n); \\ Michel Marcus, Jul 14 2020

Formula

a(n) = (n!)^n * [x^n] -polylog(n,-x) / (1 - x).

A347978 E.g.f.: 1/(1 + x)^(1/(1 - x)).

Original entry on oeis.org

1, -1, 0, -3, 4, -30, 186, -630, 11600, -26712, 1005480, -2581920, 117196872, -485308824, 17734457664, -131070696120, 3387342915840, -43890398953920, 801577841697216, -17363169328243392, 233460174245351040, -7968629225100337920, 84363134551361043840
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 22 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[1/(1 + x)^(1/(1 - x)), {x, 0, nmax}], x] Range[0, nmax]!
    A024167[n_] := n! Sum[(-1)^(k + 1)/k, {k, 1, n}]; a[0] = 1; a[n_] := a[n] = -Sum[Binomial[n - 1, k - 1] A024167[k] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace(1/(1+x)^(1/(1-x)))) \\ Michel Marcus, Sep 22 2021

Formula

E.g.f.: exp( Sum_{k>=1} x^k * Sum_{j=1..k} (-1)^j / j ).
a(0) = 1; a(n) = -Sum_{k=1..n} binomial(n-1,k-1) * A024167(k) * a(n-k).
a(0) = 1; a(n) = -Sum_{k=1..n} binomial(n,k) * A073478(k) * a(n-k).
Previous Showing 31-37 of 37 results.