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.

Showing 1-10 of 32 results. Next

A174480 Rectangular array of coefficients in successive iterations of x*exp(x), as read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 6, 1, 1, 4, 15, 23, 1, 1, 5, 28, 102, 104, 1, 1, 6, 45, 274, 861, 537, 1, 1, 7, 66, 575, 3400, 8598, 3100, 1, 1, 8, 91, 1041, 9425, 50734, 98547, 19693, 1, 1, 9, 120, 1708, 21216, 187455, 880312, 1270160, 136064, 1, 1, 10, 153, 2612, 41629
Offset: 1

Views

Author

Paul D. Hanna, Apr 17 2010

Keywords

Comments

Triangle A174485 forms a matrix that transforms a diagonal into an adjacent diagonal in this array.

Examples

			Form an array of coefficients in the iterations of x*exp(x), which begin:
n=1: [1, 1, 1/2!, 1/3!, 1/4!, 1/5!, 1/6!, ...];
n=2: [1, 2, 6/2!, 23/3!, 104/4!, 537/5!, 3100/6!, ...];
n=3: [1, 3, 15/2!, 102/3!, 861/4!, 8598/5!, 98547/6!, ...];
n=4: [1, 4, 28/2!, 274/3!, 3400/4!, 50734/5!, 880312/6!, ...];
n=5: [1, 5, 45/2!, 575/3!, 9425/4!, 187455/5!, 4367245/6!, ...];
n=6: [1, 6, 66/2!, 1041/3!, 21216/4!, 527631/5!, 15441636/6!, ...];
n=7: [1, 7, 91/2!, 1708/3!, 41629/4!, 1242892/5!, 43806175/6!, ...];
n=8: [1, 8, 120/2!, 2612/3!, 74096/4!, 2582028/5!, 106459312/6!, ...];
n=9: [1, 9, 153/2!, 3789/3!, 122625/4!, 4885389/5!, 230689017/6!, ...];
n=10:[1, 10, 190/2!, 5275/3!, 191800/4!, 8599285/5!, 457584940/6!,...];
...
This array begins with the above unreduced numerators for n >= 1, k >= 1.
		

Crossrefs

Cf. A174485, diagonals: A174481, A174482, A174483, A174484.

Programs

  • PARI
    {T(n, k)=local(F=x, xEx=x*exp(x+x*O(x^(k+1)))); for(i=1,n,F=subst(F, x, xEx));(k-1)!*polcoeff(F, k)}

Formula

T(n,k) = [x^k/(k-1)! ] G_{n}(x) where G_{n}(x) = G_{n-1}(x*exp(x)) with G_0(x)=x, for n>=1, k>=1.

A262671 Number of pointed multiset partitions of normal pointed multisets of weight n.

Original entry on oeis.org

1, 6, 42, 335, 2956, 28468, 296540
Offset: 1

Views

Author

Gus Wiseman, Sep 26 2015

Keywords

Comments

A pointed multiset k[1...k...n] with point k is normal if its entries [1...k...n] span an initial interval of positive integers. Pointed multiset partitions are triangles (or compositions) in the multiorder of pointed multisets.

Examples

			The a(2) = 6 pointed multiset partitions are:
1[1[11]],1[1[1]1[1]],
1[1[12]],1[1[1]2[2]],
2[2[12]],2[1[1]2[2]].
The a(3) = 42 pointed multiset partitions are:
1[1[111]],1[1[1]1[11]],1[1[11]1[1]],1[1[1]1[1]1[1]],
1[1[122]],1[1[1]2[22]],1[1[12]2[2]],1[1[1]2[2]2[2]],
2[2[122]],2[1[1]2[22]],2[1[12]2[2]],2[2[2]2[12]],2[2[12]2[2]],2[1[1]2[2]2[2]],
1[1[112]],1[1[1]1[12]],1[1[1]2[12]],1[1[11]2[2]],1[1[12]1[1]],1[1[1]1[1]2[2]],
2[2[112]],2[1[1]2[12]],2[1[11]2[2]],2[1[1]1[1]2[2]],
1[1[123]],1[1[1]2[23]],1[1[1]3[23]],1[1[12]3[3]],1[1[13]2[2]],1[1[1]2[2]3[3]],
2[2[123]],2[1[1]2[23]],2[1[13]2[2]],2[2[2]3[13]],2[2[12]3[3]],2[1[1]2[2]3[3]],
3[3[123]],3[1[1]3[23]],3[1[12]3[3]],3[2[2]3[13]],3[2[12]3[3]],3[1[1]2[2]3[3]].
		

Crossrefs

Programs

  • Mathematica
    ReplaceListRepeated[forms_List, rerules_List] :=
    Union[Flatten[
       FixedPointList[
        Function[preforms,
         Union[Flatten[ReplaceList[#, rerules] & /@ preforms, 1]]],
        forms], 1]]
    pointedPartitions[JIX[r_, b_List?OrderedQ]] /; MemberQ[b, r] :=
      Cases[ReplaceListRepeated[{Z[Y[JIX[r, {r}]],
          Y @@ DeleteCases[b, r, 1, 1]]}, {Z[Y[sof___, JIX[w_, t_]],
            Y[for___, x_, aft___]] /; OrderedQ[{w, x}] :>
          Z[Y[sof, JIX[w, t], JIX[x, {x}]], Y[for, aft]],
         Z[Y[JIX[w_, t_], soa___], Y[for___, x_, aft___]] /;
           OrderedQ[{x, w}] :>
          Z[Y[JIX[x, {x}], JIX[w, t], soa], Y[for, aft]],
         Z[Y[sof___, JIX[w_, {tof__}]], Y[for___, x_, aft___]] :>
          Z[Y[sof, JIX[w, Sort[{tof, x}]]], Y[for, aft]],
         Z[Y[JIX[w_, {tof__}], soa___], Y[for___, x_, aft___]] :>
          Z[Y[JIX[w, Sort[{tof, x}]], soa], Y[for, aft]]}],
       Z[Y[pts__], Y[]] :> JIX[r, {pts}]];
    allnormpms[n_Integer] :=
      Join @@ Function[s,
         JIX[#, Array[Count[s, y_ /; y <= #] + 1 &, n]] & /@
          Range[Length[s] + 1]] /@ Subsets[Range[n - 1] + 1];
    Join @@ pointedPartitions /@ allnormpms[3] /.
    JIX -> Apply(* to construct the example *)
    Array[Plus @@ (Length[pointedPartitions[#]] & /@
         allnormpms[#]) &, 7](* to compute the sequence *)

A240165 E.g.f.: exp( x*(1 + exp(2*x)) ).

Original entry on oeis.org

1, 2, 8, 44, 288, 2192, 18976, 182912, 1934848, 22231808, 275203584, 3645178880, 51370694656, 766634946560, 12066538676224, 199607631945728, 3459736006950912, 62662715180515328, 1183139425871331328, 23237689444403511296, 473852525131782946816, 10014501808427774246912
Offset: 0

Views

Author

Paul D. Hanna, Aug 02 2014

Keywords

Examples

			E.g.f.: E(x) = 1 + 2*x + 8*x^2/2! + 44*x^3/3! + 288*x^4/4! + 2192*x^5/5! +...
where E(x) = exp(x) * exp(x*exp(2*x)).
O.g.f.: A(x) = 1 + 2*x + 8*x^2 + 44*x^3 + 288*x^4 + 2192*x^5 +...
where
A(x) = 1/(1-x) + x/(1-3*x)^2 + x^2/(1-5*x)^3 + x^3/(1-7*x)^4 + x^4/(1-9*x)^5 +...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k] *(2*k+1)^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Aug 06 2014 *)
    With[{nn=30},CoefficientList[Series[Exp[x(1+Exp[2x])],{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Jul 17 2016 *)
  • PARI
    {a(n)=local(A=1);A=exp( x*(1 + exp(2*x +x*O(x^n))) );n!*polcoeff(A, n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1);A=sum(k=0, n, x^k/(1 - (2*k+1)*x +x*O(x^n))^(k+1));polcoeff(A, n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=sum(k=0,n, binomial(n,k) * (2*k+1)^(n-k) )}
    for(n=0,30,print1(a(n),", "))

Formula

O.g.f.: Sum_{n>=0} x^n / (1 - (2*n+1)*x)^(n+1).
a(n) = Sum_{k=0..n} binomial(n,k) * (2*k+1)^(n-k) for n>=0.
From Vaclav Kotesovec, Aug 06 2014: (Start)
a(n) ~ exp((1+exp(2*r))*r - n) * n^(n+1/2) / (r^n * sqrt(r + exp(2*r)*r*(1 + 6*r + 4*r^2))), where r is the root of the equation r*(1 + exp(2*r) + 2*r*exp(2*r)) = n.
(a(n)/n!)^(1/n) ~ exp(1/(2*LambertW(sqrt(n/2)))) / LambertW(sqrt(n/2)).
(End)

A240989 Expansion of e.g.f. exp(x^2 * (exp(x) - 1)).

Original entry on oeis.org

1, 0, 0, 6, 12, 20, 390, 2562, 11816, 105912, 1063530, 8815070, 81342492, 895185876, 9971185406, 112642410090, 1372455608400, 17750397057392, 236950003516626, 3286258330135734, 47688868443593540, 719345273005797900, 11222288509573985382, 181168865439054099266
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 06 2014

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[E^(x^2*(E^x-1)), {x, 0, 20}], x] * Range[0, 20]!
  • PARI
    x='x+O('x^30); Vec(serlaplace(exp(x^2*(exp(x) - 1)))) \\ G. C. Greubel, Nov 21 2017
    
  • PARI
    a(n) = n!*sum(k=0, n\3, stirling(n-2*k, k, 2)/(n-2*k)!); \\ Seiichi Manyama, Jul 09 2022
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=(i-1)!*sum(j=3, i, j/(j-2)!*v[i-j+1]/(i-j)!)); v; \\ Seiichi Manyama, Jul 09 2022

Formula

a(n) ~ exp((n-r^3)/(2+r)-n) * n^(n+1/2) / (r^n * sqrt((2*r^3*(3+r) + n*(1+r)*(4+r))/(2+r))), where r is the root of the equation r^2*((2+r) * exp(r) - 2) = n.
(a(n)/n!)^(1/n) ~ exp(1/(3*LambertW(n^(1/3)/3))) / (3*LambertW(n^(1/3)/3)).
From Seiichi Manyama, Jul 09 2022: (Start)
a(n) = n! * Sum_{k=0..floor(n/3)} Stirling2(n-2*k,k)/(n-2*k)!.
a(0) = 1; a(n) = (n-1)! * Sum_{k=3..n} k/(k-2)! * a(n-k)/(n-k)!. (End)

A185298 Expansion of e.g.f. x*exp(x)*exp(x*exp(x)).

Original entry on oeis.org

0, 1, 4, 18, 92, 520, 3222, 21700, 157544, 1224576, 10133450, 88843084, 821832156, 7992373168, 81458868974, 867700216380, 9636146477648, 111323478770560, 1335253363581330, 16598183219157772, 213488758730421380, 2837046652845555696, 38899888173340835894
Offset: 0

Views

Author

Geoffrey Critzer, Feb 20 2011

Keywords

Comments

a(n) is the number of ways to designate an element in each block of the set partitions of {1,2,...,n} and then designate a block.
Inverse binomial transform: b(n) = Sum (-1)^(n-k)*C(n,k)*a(k), k=0..n of A052512. - Alexander R. Povolotsky, Oct 01 2011
Number of pointed set partitions of pointed sets k[1...k...n] for any point k. - Gus Wiseman, Sep 27 2015
Exponential series reversal gives A207833 with alternating signs: 1, -4, 30, -332, 4880, ... . - Vladimir Reshetnikov, Aug 04 2019

Examples

			The a(2) = 4 pointed set partitions are 1[1[12]], 1[1[1]2[2]], 2[1[1]2[2]], 2[2[12]].
The a(3) = 18 pointed set partitions are 1[1[123]], 1[1[1]2[23]], 1[1[1]3[23]], 1[1[12]3[3]], 1[1[13]2[2]], 1[1[1]2[2]3[3]], 2[2[123]], 2[1[1]2[23]], 2[1[13]2[2]], 2[2[2]3[13]], 2[2[12]3[3]], 2[1[1]2[2]3[3]], 3[3[123]], 3[1[1]3[23]], 3[1[12]3[3]], 3[2[2]3[13]], 3[2[12]3[3]], 3[1[1]2[2]3[3]].
		

Crossrefs

Programs

  • Mathematica
    nn=30; a=x Exp[x]; Range[0,nn]! CoefficientList[Series[a Exp[a], {x,0,nn}],x]
  • PARI
    x='x+O('x^33); concat([0],Vec(serlaplace(x*exp(x)*exp(x*exp(x))))) \\ Joerg Arndt, Oct 04 2015

Formula

E.g.f.: A(A(x)) where A(x) = x*exp(x).
a(n) = Sum_{k=1..n} binomial(n,k)*k^(n-k+1). - Vladimir Kruchinin, Sep 23 2011
O.g.f.: Sum_{k>=1} k*x^k/(1 - k*x)^(k+1). - Ilya Gutkovskiy, Oct 09 2018
a(n) ~ exp(r*exp(r) + r - n) * n^(n + 1/2) / (r^(n - 1/2) * sqrt(1 + exp(r)*(1 + 3*r + r^2))), where r = 2*LambertW(exp(1/4)*sqrt(n)/2) - 1/2 + 1/(16*LambertW(exp(1/4)*sqrt(n)/2)^2 + LambertW(exp(1/4)*sqrt(n)/2) - 1). - Vaclav Kotesovec, Mar 21 2023

A292893 Expansion of e.g.f. exp(x * (1 - exp(x))).

Original entry on oeis.org

1, 0, -2, -3, 8, 55, 84, -637, -4992, -10593, 92060, 1012099, 3642000, -18733585, -354606084, -2157876645, 2003383424, 175455790399, 1766183783868, 5436448194707, -96997103373360, -1770215099996721, -13073420293290148, 22275369715313131
Offset: 0

Views

Author

Seiichi Manyama, Sep 26 2017

Keywords

Crossrefs

Column k=1 of A292894.

Programs

  • PARI
    x='x+O('x^66); Vec(serlaplace(exp(x*(1-exp(x)))))
    
  • PARI
    a(n) = n!*sum(k=0, n\2, (-1)^k*stirling(n-k, k, 2)/(n-k)!); \\ Seiichi Manyama, Jul 09 2022
    
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=-(i-1)!*sum(j=2, i, j/(j-1)!*v[i-j+1]/(i-j)!)); v; \\ Seiichi Manyama, Jul 09 2022
    
  • PARI
    a(n) = sum(k=0, n, (-1)^k*(k+1)^(n-k)*binomial(n, k)); \\ Seiichi Manyama, Aug 29 2022
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, (-x)^k/(1-(k+1)*x)^(k+1))) \\ Seiichi Manyama, Aug 29 2022

Formula

From Seiichi Manyama, Jul 09 2022: (Start)
a(n) = n! * Sum_{k=0..floor(n/2)} (-1)^k * Stirling2(n-k,k)/(n-k)!.
a(0) = 1; a(n) = -(n-1)! * Sum_{k=2..n} k/(k-1)! * a(n-k)/(n-k)!. (End)
From Seiichi Manyama, Aug 29 2022: (Start)
a(n) = Sum_{k=0..n} (-1)^k * (k+1)^(n-k) * binomial(n,k).
G.f.: Sum_{k>=0} (-x)^k / (1 - (k+1)*x)^(k+1). (End)

A358738 Expansion of Sum_{k>=0} k! * ( x/(1 - k*x) )^k.

Original entry on oeis.org

1, 1, 3, 15, 103, 893, 9341, 114355, 1603155, 25318137, 444689497, 8597568671, 181430298479, 4149361409077, 102229328244837, 2699254206069387, 76038064580742091, 2276259442660623857, 72160287650141753777, 2414950992007231422007
Offset: 0

Views

Author

Seiichi Manyama, Nov 29 2022

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Sum[k! * (x/(1 - k*x))^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Feb 18 2023 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(sum(k=0, N, k!*(x/(1-k*x))^k))
    
  • PARI
    a(n) = if(n==0, 1, sum(k=1, n, k!*k^(n-k)*binomial(n-1, k-1)));

Formula

a(n) = Sum_{k=1..n} k! * k^(n-k) * binomial(n-1,k-1) for n > 0.
a(n) ~ n! / ((1 + LambertW(1))^2 * LambertW(1)^n). - Vaclav Kotesovec, Feb 18 2023

A245834 E.g.f.: exp( x*(1 + exp(3*x)) ).

Original entry on oeis.org

1, 2, 10, 71, 592, 5777, 64792, 814025, 11264176, 169871633, 2768582104, 48412950929, 902831609368, 17865749820089, 373564063839376, 8223263706957713, 189960800250512608, 4591950749700004385, 115866075506169417256, 3044877330738661504625, 83169542349597382767496, 2356949307613191494567561
Offset: 0

Views

Author

Paul D. Hanna, Aug 02 2014

Keywords

Examples

			E.g.f.: E(x) = 1 + 2*x + 10*x^2/2! + 71*x^3/3! + 592*x^4/4! + 5777*x^5/5! +...
where E(x) = exp(x) * exp(x*exp(3*x)).
O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 71*x^3 + 592*x^4 + 5777*x^5 + 64792*x^6 +...
where
A(x) = 1/(1-x) + x/(1-4*x)^2 + x^2/(1-7*x)^3 + x^3/(1-10*x)^4 + x^4/(1-13*x)^5 +...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Binomial[n,k] *(3*k+1)^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Aug 06 2014 *)
    With[{nn=30},CoefficientList[Series[Exp[x(1+Exp[3x])],{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Jun 09 2019 *)
  • PARI
    {a(n)=local(A=1);A=exp( x*(1 + exp(3*x +x*O(x^n))) );n!*polcoeff(A, n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=local(A=1);A=sum(k=0, n, x^k/(1 - (3*k+1)*x +x*O(x^n))^(k+1));polcoeff(A, n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {a(n)=sum(k=0,n,(3*k+1)^(n-k)*binomial(n,k))}
    for(n=0,30,print1(a(n),", "))

Formula

O.g.f.: Sum_{n>=0} x^n / (1 - (3*n+1)*x)^(n+1).
a(n) = Sum_{k=0..n} binomial(n,k) * (3*k+1)^(n-k) for n>=0.
From Vaclav Kotesovec, Aug 06 2014: (Start)
a(n) ~ exp((1+exp(3*r))*r - n) * n^(n+1/2) / (r^n * sqrt(r + exp(3*r)*r* (1+9*r*(1+r)))), where r is the root of the equation r*(1 + exp(3*r) + 3*r*exp(3*r)) = n.
(a(n)/n!)^(1/n) ~ 3*exp(1/(2*LambertW(sqrt(3*n)/2))) / (2*LambertW(sqrt(3*n)/2)).
(End)

A355463 Expansion of Sum_{k>=0} (x/(1 - k^k * x))^k.

Original entry on oeis.org

1, 1, 2, 10, 131, 5656, 869097, 490286392, 1264458639313, 12443651667592768, 681538604797281047489, 153070077563816488157872384, 205935348854901274982393017521537, 1352544986573612111579941739713633174912
Offset: 0

Views

Author

Seiichi Manyama, Jul 03 2022

Keywords

Crossrefs

Programs

  • Mathematica
    Flatten[{1, Table[Sum[Binomial[n-1,k-1] * k^(k*(n-k)), {k,1,n}], {n,1,20}]}] (* Vaclav Kotesovec, Feb 16 2023 *)
  • PARI
    my(N=20, x='x+O('x^N)); Vec(sum(k=0, N, (x/(1-k^k*x))^k))
    
  • PARI
    a(n) = if(n==0, 1, sum(k=1, n, k^(k*(n-k))*binomial(n-1, k-1)));

Formula

a(n) = Sum_{k=1..n} k^(k*(n-k)) * binomial(n-1,k-1) for n > 0.

A116071 Triangle T, read by rows, equal to Pascal's triangle to the matrix power of Pascal's triangle, so that T = C^C, where C(n,k) = binomial(n,k) and T(n,k) = A000248(n-k)*C(n,k).

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 10, 9, 3, 1, 41, 40, 18, 4, 1, 196, 205, 100, 30, 5, 1, 1057, 1176, 615, 200, 45, 6, 1, 6322, 7399, 4116, 1435, 350, 63, 7, 1, 41393, 50576, 29596, 10976, 2870, 560, 84, 8, 1, 293608, 372537, 227592, 88788, 24696, 5166, 840, 108, 9, 1
Offset: 0

Views

Author

Paul D. Hanna, Feb 03 2006

Keywords

Comments

Column 0 = A000248 (Number of forests with n nodes and height at most 1).
Column 1 = A052512 (Number of labeled trees of height 2).
Row sums = A080108 (Sum_{k=1..n} k^(n-k) * C(n-1,k-1)).
Central terms = A116072(n) = (n+1) * A000108(n) * A000248(n).
From Peter Bala, Sep 13 2012: (Start)
For commuting lower unitriangular matrix A and lower triangular matrix B we define A raised to the matrix power B, denoted by A^B, to be the lower unitriangular matrix Exp(B*Log(A)). Here Exp denotes the matrix exponential defined by the power series
Exp(A) = 1 + A + A^2/2! + A^3/3! + ...
and the matrix logarithm Log(A) is defined by the series
Log(A) = (A-1) - 1/2*(A-1)^2/2 + 1/3*(A-1)^3 - ....
Let A = [f(x),x] and B = [g(x),x] be exponential Riordan arrays in the Appell subgroup and suppose f(0) = 1. Then A and B commute and A^B is the exponential Riordan array [exp(g(x)*log(f(x))),x], also belonging to the Appell group. In the present case we are taking A = B = [exp(x),x], equal to the Pascal triangle A007318.
For any lower unitriangular matrix A (with, say, rational entries) the infinite tower of powers A^(A^(A^(...))) is well-defined (and also has rational entries). An example is given in the Formula section. (End)

Examples

			E.g.f.: E(x,y) = 1 + (1 + y)*x + (3 + 2*y + y^2)*x^2/2!
  + (10 + 9*y + 3*y^2 + y^3)*x^3/3!
  + (41 + 40*y + 18*y^2 + 4*y^3 + y^4)*x^4/4!
  + (196 + 205*y + 100*y^2 + 30*y^3 + 5*y^4 + y^5)*x^5/5! +...
where E(x,y) = exp(x*y) * exp(x*exp(x)).
O.g.f.: A(x,y) = 1 + (1 + y)*x + (3 + 2*y + y^2)*x^2
  + (10 + 9*y + 3*y^2 + y^3)*x^3
  + (41 + 40*y + 18*y^2 + 4*y^3 + y^4)*x^4
  + (196 + 205*y + 100*y^2 + 30*y^3 + 5*y^4 + y^5)*x^5 +...
where
A(x,y) = 1/(1-x*y) + x/(1-x*(y+1))^2 + x^2/(1-x*(y+2))^3 + x^3/(1-x*(y+3))^4 + x^4/(1-x*(y+4))^5 + x^5/(1-x*(y+5))^6 + x^6/(1-x*(y+6))^7 + x^7/(1-x*(y+7))^8 +...
Triangle begins:
  1;
  1, 1;
  3, 2, 1;
  10, 9, 3, 1;
  41, 40, 18, 4, 1;
  196, 205, 100, 30, 5, 1;
  1057, 1176, 615, 200, 45, 6, 1;
  6322, 7399, 4116, 1435, 350, 63, 7, 1;
  41393, 50576, 29596, 10976, 2870, 560, 84, 8, 1;
  293608, 372537, 227592, 88788, 24696, 5166, 840, 108, 9, 1; ...
		

Crossrefs

Programs

  • Mathematica
    (* The function RiordanArray is defined in A256893. *)
    RiordanArray[Exp[# Exp[#]]&, #&, 10, True] // Flatten (* Jean-François Alcover, Jul 19 2019 *)
  • PARI
    /* By definition C^C: */
    {T(n,k)=local(A, C=matrix(n+1,n+1,r,c,binomial(r-1,c-1)), L=matrix(n+1,n+1,r,c,if(r==c+1,c))); A=sum(m=0,n,L^m*C^m/m!); A[n+1,k+1]}
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • PARI
    /* From e.g.f.: */
    {T(n,k)=local(A=1);A=exp( x*y + x*exp(x +x*O(x^n)) );n!*polcoeff(polcoeff(A, n,x),k,y)}
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • PARI
    /* From o.g.f. (Paul D. Hanna, Aug 03 2014): */
    {T(n,k)=local(A=1);A=sum(k=0, n, x^k/(1 - x*(k+y) +x*O(x^n))^(k+1));polcoeff(polcoeff(A, n,x),k,y)}
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • PARI
    /* From row polynomials (Paul D. Hanna, Aug 03 2014): */
    {T(n,k)=local(R);R=sum(k=0,n,(k+y)^(n-k)*binomial(n,k));polcoeff(R,k,y)}
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))
    
  • PARI
    /* From formula for T(n,k) (Paul D. Hanna, Aug 03 2014): */
    {T(n,k) = sum(j=0,n-k, binomial(n,j) * binomial(n-j,k) * j^(n-k-j))}
    for(n=0,10,for(k=0,n,print1(T(n,k),", "));print(""))

Formula

E.g.f.: exp( x*exp(x) + x*y ).
From Peter Bala, Sep 13 2012: (Start)
Exponential Riordan array [exp(x*exp(x)),x] belonging to the Appell group. Thus the e.g.f. for the k-th column of the triangle is x^k/k!*exp(x*exp(x)).
The inverse array, denote it by X, is a signed version of A215652. The infinite tower of matrix powers X^(X^(X^(...))) equals the inverse of Pascal's triangle. (End)
O.g.f.: Sum_{n>=0} x^n / (1 - x*(n+y))^(n+1). - Paul D. Hanna, Aug 03 2014
G.f. for row n: Sum_{k=0..n} binomial(n,k) * (k + y)^(n-k) for n>=0. - Paul D. Hanna, Aug 03 2014
T(n,k) = Sum_{j=0..n-k} C(n,j) * C(n-j,k) * j^(n-k-j) = A000248(n-k)*C(n,k). - Paul D. Hanna, Aug 03 2014
Infinitesimal generator is A216973. - Peter Bala, Feb 13 2017
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