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 11-20 of 32 results. Next

A346922 Expansion of e.g.f. 1 / (1 + log(1 - x)^3 / 3!).

Original entry on oeis.org

1, 0, 0, 1, 6, 35, 245, 2044, 19572, 210524, 2513760, 33012276, 472963876, 7340889192, 122703087416, 2197496734224, 41979155247520, 852063971170960, 18312093589455440, 415420659953439840, 9920128280950954080, 248735658391768241280, 6533773435848445617600
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 07 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[1/(1 + Log[1 - x]^3/3!), {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k] Abs[StirlingS1[k, 3]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
  • PARI
    my(x='x+O('x^25)); Vec(serlaplace(1/(1+log(1-x)^3/3!))) \\ Michel Marcus, Aug 07 2021
    
  • PARI
    a(n) = sum(k=0, n\3, (3*k)!*abs(stirling(n, 3*k, 1))/6^k); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n,k) * |Stirling1(k,3)| * a(n-k).
a(n) ~ n! * 6^(1/3) / (3 * exp(6^(1/3)) * (1 - exp(-6^(1/3)))^(n+1)). - Vaclav Kotesovec, Aug 08 2021
a(n) = Sum_{k=0..floor(n/3)} (3*k)! * |Stirling1(n,3*k)|/6^k. - Seiichi Manyama, May 06 2022

A347002 Expansion of e.g.f. exp( -log(1 - x)^3 / 3! ).

Original entry on oeis.org

1, 0, 0, 1, 6, 35, 235, 1834, 16352, 163764, 1818030, 22143726, 293476326, 4203311892, 64682865156, 1064154324024, 18636296872320, 346103784493560, 6793394350116600, 140508244952179200, 3054120126193160280, 69596730438090806880, 1659041650323705102840
Offset: 0

Views

Author

Ilya Gutkovskiy, Aug 10 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[Exp[-Log[1 - x]^3/3!], {x, 0, nmax}], x] Range[0, nmax]!
    a[0] = 1; a[n_] := a[n] = Sum[Binomial[n - 1, k - 1] Abs[StirlingS1[k, 3]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 22}]
  • PARI
    a(n) = sum(k=0, n\3, (3*k)!*abs(stirling(n, 3*k, 1))/(6^k*k!)); \\ Seiichi Manyama, May 06 2022

Formula

a(0) = 1; a(n) = Sum_{k=1..n} binomial(n-1,k-1) * |Stirling1(k,3)| * a(n-k).
a(n) = Sum_{k=0..floor(n/3)} (3*k)! * |Stirling1(n,3*k)|/(6^k * k!). - Seiichi Manyama, May 06 2022

A243569 Unsigned Stirling numbers of the first kind s(n,8).

Original entry on oeis.org

1, 36, 870, 18150, 357423, 6926634, 135036473, 2681453775, 54631129553, 1146901283528, 24871845297936, 557921681547048, 12953636989943896, 311333643161390640, 7744654310169576800, 199321978221066137360
Offset: 8

Views

Author

Stanislav Sykora, Jun 06 2014

Keywords

Crossrefs

Programs

  • Mathematica
    Drop[Table[Abs[StirlingS1[n, 8]], {n, 0, 20}], 8] (* Vaclav Kotesovec, Jun 06 2014 *)
  • PARI
    abs(stirling(n,8))

A243570 Unsigned Stirling numbers of the first kind s(n,9).

Original entry on oeis.org

1, 45, 1320, 32670, 749463, 16669653, 368411615, 8207628000, 185953177553, 4308105301929, 102417740732658, 2503858755467550, 63030812099294896, 1634980697246583456, 43714229649594412832
Offset: 9

Views

Author

Stanislav Sykora, Jun 06 2014

Keywords

Crossrefs

Programs

  • Mathematica
    Drop[Table[Abs[StirlingS1[n, 9]], {n, 0, 20}], 9] (* Vaclav Kotesovec, Jun 06 2014 *)
  • PARI
    abs(stirling(n,9))

A163939 Triangle related to the o.g.f.s. of the right hand columns of A163934 (E(x,m=4,n)).

Original entry on oeis.org

1, 6, 4, 35, 60, 10, 225, 690, 325, 20, 1624, 7588, 6762, 1316, 35, 13132, 85288, 120358, 46928, 4508, 56, 118124, 1004736, 2028660, 1298860, 265365, 13896, 84, 1172700, 12529400, 33896400, 31862400, 11077255, 1313610, 39915, 120
Offset: 1

Views

Author

Johannes W. Meijer, Aug 13 2009

Keywords

Comments

The asymptotic expansions of the higher order exponential integral E(x,m=4,n) lead to triangle A163934, see A163931 for information on the E(x,m,n). The o.g.f.s. of the right hand columns of triangle A163934 have a nice structure Gf(p) = W4(z,p)/(1-z)^(2*p+2) with p = 1 for the first right hand column, p = 2 for the second right hand column, etc.. The coefficients of the W4(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. The row sums of this triangle lead to A000457, see A163936 for more information.

Examples

			The first few W4(z,p) polynomials are:
W4(z,p=1) = 1/(1-z)^4
W4(z,p=2) = (6+4*z)/(1-z)^6
W4(z,p=3) = (35+60*z+10*z^2)/(1-z)^8
W4(z,p=4) = (225+690*z+325*z^2+20*z^3)/(1-z)^10
		

Crossrefs

Row sums equal A000457.
A000399 equals the first left hand column.
A000292 equals the first right hand column.
Cf. A163931 (E(x,m,n)) and A163934.
Cf. A163936 (E(x,m=1,n)), A163937 (E(x,m=2,n)) and A163938 (E(x,m=3,n)).

Programs

  • Maple
    with(combinat): a := proc(n, m): add((-1)^(n+k+1)*((m-k+2)*(m-k+1)*(m-k)/3!)*binomial(2*n+2, k)*stirling1(m+n-k+1, m-k+2), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..8); # Johannes W. Meijer, revised Nov 27 2012
  • Mathematica
    Table[Sum[(-1)^(n + k + 1)*Binomial[m - k + 2, 3]*Binomial[2*n + 2, k]*StirlingS1[m + n - k + 1, m - k + 2], {k, 0, m - 1}], {n, 1, 50}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
  • PARI
    for(n=1,10, for(m=1,n, print1(sum(k=0, m-1, (-1)^(n+k+1)* binomial(m-k+2,3)* binomial(2*n+2,k)*stirling(m+n-k+1,m-k+2,1)), ", "))) \\ G. C. Greubel, Aug 13 2017

Formula

a(n,m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(m-k+2,3)* binomial(2*n+2,k)*stirling1(m+n-k+1,m-k+2), for 1<= m <=n.

A062254 3rd level triangle related to Eulerian numbers and binomial transforms (A062253 is second level, triangle of Eulerian numbers is first level and triangle with Z(0,0)=1 and Z(n,k)=0 otherwise is 0th level).

Original entry on oeis.org

1, 6, 0, 25, 10, 0, 90, 120, 15, 0, 301, 896, 406, 21, 0, 966, 5376, 5586, 1176, 28, 0, 3025, 28470, 55560, 27910, 3123, 36, 0, 9330, 139320, 456525, 437100, 122520, 7860, 45, 0, 28501, 646492, 3312078, 5339719, 2912833, 494802, 19096, 55, 0
Offset: 0

Views

Author

Henry Bottomley, Jun 14 2001

Keywords

Comments

Binomial transform of n^3*k^n is ((kn)^3 + 3(kn)^2 + (1 - k)(kn))*(k + 1)^(n - 3); of n^4*k^n is ((kn)^4 + 6(kn)^3 + (7 - 4k)(kn)^2 + (1 - 4k + k^2)(kn))*(k + 1)^(n - 4); of n^5*k^n is ((kn)^5 + 10(kn)^4 + (25 - 10k)(kn)^3 + (15 - 30k + 5k^2)(kn)^2 + (1 - 11k + 11k^2 - k^3)(kn))*(k + 1)^(n - 5); of n^6*k^n is ((kn)^6 + 15(kn)^5 + (65 - 20k)(kn)^4 + (90 - 120k + 15k^2)(kn)^3 + (31 - 146k + 91k^2 - 6k^3)(kn)^2 + (1 - 26k + 66k^2 - 26k^3 + k^4)(kn))*(k + 1)^(n - 6). This sequence gives the (unsigned) polynomial coefficients of (kn)^3.

Examples

			Rows start:
 (1),
 (6,0),
 (25,10,0),
 (90,120,15,0),
 ...
		

Crossrefs

First column is A000392. Diagonals include A000007 and all but the start of A000217. Row sums are A000399.
Taking all the levels together to create a pyramid, one face would be A010054 as a triangle with a parallel face which is Pascal's triangle (A007318) with two columns removed, another face would be a triangle of Stirling numbers of the second kind (A008277) and a third face would be A000007 as a triangle, (cont.)
(cont.) with a triangle of Eulerian numbers (A008292), A062253, A062254 and A062255 as faces parallel to it. The row sums of this last group would provide a triangle of unsigned Stirling numbers of the first kind (A008275).

Programs

  • PARI
    E(n, k) = if ((n<0) || (k<0), 0, if ((n==0) && (k==0), 1, (k+1)*E(n-1, k)+(n-k)*E(n-1, k-1)));
    A2(n, k) = if ((n<0) || (k<0), 0, (k+2)*A2(n-1, k)+(n-k)*A2(n-1, k-1)+E(n, k));
    A3(n, k) = if ((n<0) || (k<0), 0, (k+3)*A3(n-1, k)+(n-k)*A3(n-1, k-1) + A2(n, k));
    row3(n) = vector(n+1, k, A3(n,k-1)); \\ Michel Marcus, Jan 27 2025

Formula

A(n, k) = (k+3)*A(n-1, k) + (n-k)*A(n-1, k-1) + A062253(n, k).

A341588 E.g.f.: -log(1 + log(1 - x))^3 / 6.

Original entry on oeis.org

1, 12, 130, 1485, 18508, 253400, 3805723, 62437500, 1113510409, 21479997957, 446094038806, 9930796412082, 236037249893092, 5968192832899412, 160007282538148508, 4534905316824903144, 135500246340709682692, 4257646241716404353684, 140366073694357927723936, 4845119946789226304526392
Offset: 3

Views

Author

Ilya Gutkovskiy, Feb 15 2021

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 22; CoefficientList[Series[-Log[1 + Log[1 - x]]^3/6, {x, 0, nmax}], x] Range[0, nmax]! // Drop[#, 3] &
    Table[Sum[Abs[StirlingS1[n, k] StirlingS1[k, 3]], {k, 3, n}], {n, 3, 22}]

Formula

a(n) = Sum_{k=3..n} |Stirling1(n, k) * Stirling1(k, 3)|.
a(n) ~ (n-1)! * log(n)^2 / (2 * (1 - exp(-1))^n) * (1 + (2*gamma - 2*log(exp(1) - 1)) / log(n) + (gamma^2 - Pi^2/6 - 2*log(exp(1) - 1)*gamma + log(exp(1)-1)^2) / log(n)^2). - Vaclav Kotesovec, Jun 04 2022

A052748 Expansion of e.g.f.: -(log(1-x))^3.

Original entry on oeis.org

0, 0, 0, 6, 36, 210, 1350, 9744, 78792, 708744, 7036200, 76521456, 905507856, 11589357312, 159580302336, 2352940786944, 36994905688320, 617953469022720, 10929614667747840, 204073497562936320, 4011658382046919680, 82822558521844224000, 1791791417179298304000
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

Original name: A simple grammar.

Crossrefs

Column k=3 of A225479.

Programs

  • Maple
    spec := [S,{B=Cycle(Z),S=Prod(B,B,B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
    with(combinat):seq(stirling1(j, 3)*3!*(-1)^(j+1), j=0..50); # Leonid Bedratyuk, Aug 07 2012
  • PARI
    a(n) = {3!*stirling(n,3,1)*(-1)^(n+1)} \\ Andrew Howroyd, Jul 27 2020

Formula

E.g.f.: log(1/(1-x))^3.
Recurrence: {a(1)=0, a(0)=0, a(2)=0, a(3)=6, (-1 - 3*n - 3*n^2 - n^3)*a(n+1) + (9*n + 7 + 3*n^2)*a(n+2) + (-6 - 3*n)*a(n+3) + a(n+4)}.
a(n) = stirling1(n, 3)*3!*(-1)^(n+1). - Leonid Bedratyuk, Aug 07 2012
a(n) = 6*A000399(n). - Andrew Howroyd, Jul 27 2020

Extensions

Name changed and terms a(20) and beyond from Andrew Howroyd, Jul 27 2020

A127054 Rectangular table, read by antidiagonals, defined by the following rule: start with all 1's in row zero; from then on, row n+1 equals the partial sums of row n excluding terms in columns k = m*(m+1)/2 (m>=1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 10, 9, 4, 1, 1, 34, 33, 15, 5, 1, 1, 154, 153, 65, 23, 6, 1, 1, 874, 873, 339, 119, 32, 7, 1, 1, 5914, 5913, 2103, 719, 186, 42, 8, 1, 1, 46234, 46233, 15171, 5039, 1230, 267, 54, 9, 1, 1, 409114, 409113, 124755, 40319, 9258, 1891, 380
Offset: 0

Views

Author

Paul D. Hanna, Jan 04 2007

Keywords

Comments

Variant of table A125781. Generated by a method similar to Moessner's factorial triangle (A125714).

Examples

			Rows are partial sums excluding terms in columns k = {1,3,6,10,...}:
row 2 = partial sums of [1, 3, 5,6, 8,9,10, 12,13,14,15, ...];
row 3 = partial sums of [1, 9, 23,32, 54,67,81, 113,131,150,170, ...];
row 4 = partial sums of [1, 33, 119,186, 380,511,661, 1045,1283,...].
The terms that are excluded in the partial sums are shown enclosed in
parenthesis in the table below. Rows of this table begin:
1,(1), 1, (1), 1, 1, (1), 1, 1, 1, (1), 1, 1, 1, 1, (1), 1, ...;
1,(2), 3, (4), 5, 6, (7), 8, 9, 10, (11), 12, 13, 14, 15, (16), ...;
1,(4), 9, (15), 23, 32, (42), 54, 67, 81, (96), 113, 131, 150, ...;
1,(10), 33, (65), 119, 186, (267), 380, 511, 661, (831), 1045, ...;
1,(34), 153, (339), 719, 1230, (1891), 2936, 4219, 5765, (7600), ...;
1,(154), 873, (2103), 5039, 9258, (15023), 25148, 38203, 54625, ..;
1,(874), 5913, (15171), 40319, 78522, (133147), 238124, 379339, ...;
1,(5914), 46233, (124755), 362879, 742218, (1305847), 2477468, ...;
1,(46234), 409113, (1151331), 3628799, 7742058, (14059423), ...;
1,(409114), 4037913, (11779971), 39916799, 88369098, (164977399),...;
Columns include:
k=1: A003422 (Left factorials: !n = Sum k!, k=0..n-1);
k=2: A007489 (Sum of k!, k=1..n);
k=3: A097422 (Sum{k=1 to n} H(k) k!, where H(k) = sum{j=1 to k} 1/j);
k=4: A033312 (n! - 1);
k=5: Partial sums of A001705;
k=6: partial sums of A000399 (Stirling numbers of first kind s(n,3)).
		

Crossrefs

Cf. variants: A125781, A125714; antidiagonal sums: A127055; diagonal: A127056; columns: A003422, A007489, A097422, A033312.

Programs

  • Maple
    {T(n,k)=local(A=0,b=2,c=0,d=0);if(n==0,A=1, until(d>k,if(c==b*(b-1)/2,b+=1,A+=T(n-1,c);d+=1);c+=1));A}

A348063 Coefficient of x^2 in expansion of n!* Sum_{k=0..n} binomial(x,k).

Original entry on oeis.org

1, 0, 11, 5, 304, 364, 15980, 34236, 1368936, 4429656, 173699712, 771653376, 30605906304, 175622947200, 7149130156800, 50800930272000, 2137822335475200, 18241636315507200, 796397873127782400, 7971407298921830400, 361615771356450508800, 4168685961862906982400, 196587429737202833817600
Offset: 2

Views

Author

Seiichi Manyama, Sep 26 2021

Keywords

Crossrefs

Programs

  • PARI
    a(n) = n!*polcoef(sum(k=2, n, binomial(x, k)), 2);
    
  • PARI
    a(n) = if(n<2, 0, a(n-1)+(n-1)^2*a(n-2)+(-1)^n*(n-2)!);
    
  • PARI
    N=40; x='x+O('x^N); Vec(serlaplace(log(1+x)^2/(2*(1-x))))
    
  • Python
    from sympy.abc import x
    from sympy import ff, expand
    def A348063(n): return sum(ff(n,n-k)*expand(ff(x,k)).coeff(x**2) for k in range(2,n+1)) # Chai Wah Wu, Sep 27 2021

Formula

a(n) = a(n-1) + (n-1)^2 * a(n-2) + (-1)^n * (n-2)!.
E.g.f.: (log(1 + x))^2/(2 * (1 - x)).
a(n) ~ n! * log(2)^2 / 2 * (1 + (-1)^n*log(n)/(log(2)^2*n)). - Vaclav Kotesovec, Sep 27 2021
Previous Showing 11-20 of 32 results. Next