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(* Alternative: *) Needs["Combinatorica`"]

Flatten[Table[k*Eulerian[n+1, k], {n, 1, 9}, {k, 1, n}]] (* Peter Luschny, Feb 07 2023 *)

Let E(n, x) = Sum_{j=0..k+1}} (-1)^j binomialA173018(n + 1, j)*(, k + 1 - j)*x^n k and E'(n, x) = (d/dx) E(x, n). Then T(n, k) = [x^(k-1)] E'(n+1, x).

T[n_, k_] := Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; Table[D[Sum[T[n, k]*x^k, {k, 0, n - 1}], x], {n, 1, 10}]; Table[CoefficientList[D[Sum[t[n, k]*x^k, {k, 0, n - 1}], x], x], {n, 1, 10}]; Flatten[%]

T[n_, k_] := Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}];

Table[D[Sum[T[n, k]*x^k, {k, 0, n - 1}], x], {n, 1, 10}];

Table[CoefficientList[D[Sum[T[n, k]*x^k, {k, 0, n - 1}], x], x], {n, 1, 10}];

Flatten[%]

Coefficients of the derivatives of the Eulerian polynomials (with indexing as in A008292A173018).

Let E(n, x) = Sum_{j=0..k+1}} (-1)^j binomial(n + 1, j)*(k + 1 - j)^n and E'(n, x) = (d/dx) E(x, n). Then T(n, k) = [x^(k-1)] E'(n, +1, x).

Triangel Triangle T(n, k) starts:

Cf. A008292, A173018, A001286 (row sums).

Coefficients of derivatives of Eulerian number polynomials (A008292): p(x,n)=Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; p'(x,n)=(d/dx)p{x,n).

Coefficients of the derivatives of the Eulerian polynomials (with indexing as in A008292).

Row sums are: A001286 ( Lah numbers: (n-1)*n!/2. );

{0, 1, 6, 36, 240, 1800, 15120, 141120, 1451520, 16329600}.

pLet E(x,n, x) = Sum[_{j=0..k+1}} (-1)^j Binomial[binomial(n + 1, j])*(k + 1 - j)^n, {j, 0, k + 1}]; p and E'(x,n, x) = (d/dx)p{ E(x, n); t. Then T(n,m k) =Coefficients(p'( [x,^k] E'(n), x).

{1},

Triangel T(n, k) starts:

{ 1};

{ 4, 2},;

{ 11, 22, 3},;

{ 26, 132, 78, 4},;

{ 57, 604, 906, 228, 5},;

{ 120, 2382, 7248, 4764, 600, 6},;

{ 247, 8586, 46857, 62476, 21465, 1482, 7},;

{ 502, 29216, 264702, 624760, 441170, 87648, 3514, 8},;

{1013, 95680, 1365576, 5241416, 6551770, 2731152, 334880, 8104, 9}.

T := (n, k) -> k * combinat:-eulerian1(n+1, k):

for n from 1 to 9 do seq(T(n, k), k = 1..n) od; # Peter Luschny, Feb 07 2023

tT[n_, k_] := Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; Table[D[Sum[tT[n, k]*x^k, {k, 0, n - 1}], x], {n, 1, 10}]; Table[CoefficientList[D[Sum[t[n, k]*x^k, {k, 0, n - 1}], x], x], {n, 1, 10}]; Flatten[%]

Cf. A008292, A001286 (row sums).

nonn,tabl,uned

Edited by Peter Luschny, Feb 07 2023

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nonn,tabl,uned

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_Roger L. Bagula _ and _Gary W. Adamson (rlbagulatftn(AT)yahoo.com), _, Sep 24 2008

Coefficients of derivatives of Eulerian number polynomials (A008292): p(x,n)=Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; p'(x,n)=(d/dx)p{x,n).

1, 4, 2, 11, 22, 3, 26, 132, 78, 4, 57, 604, 906, 228, 5, 120, 2382, 7248, 4764, 600, 6, 247, 8586, 46857, 62476, 21465, 1482, 7, 502, 29216, 264702, 624760, 441170, 87648, 3514, 8, 1013, 95680, 1365576, 5241416, 6551770, 2731152, 334880, 8104, 9

1,2

Row sums are: A001286 ( Lah numbers: (n-1)*n!/2. );

{0, 1, 6, 36, 240, 1800, 15120, 141120, 1451520, 16329600}.

p(x,n)=Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; p'(x,n)=(d/dx)p{x,n); t(n,m)=Coefficients(p'(x,n)).

{1},

{4, 2},

{11, 22, 3},

{26, 132, 78, 4},

{57, 604, 906, 228, 5},

{120, 2382, 7248, 4764, 600, 6},

{247, 8586, 46857, 62476, 21465, 1482, 7},

{502, 29216, 264702, 624760, 441170, 87648, 3514, 8},

{1013, 95680, 1365576, 5241416, 6551770, 2731152, 334880, 8104, 9}

t[n_, k_] := Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]; Table[D[Sum[t[n, k]*x^k, {k, 0, n - 1}], x], {n, 1, 10}]; Table[CoefficientList[D[Sum[t[n, k]*x^k, {k, 0, n - 1}], x], x], {n, 1, 10}]; Flatten[%]

Cf. A008292, A001286.

nonn,uned

Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Sep 24 2008

approved