A060187 -id:A060187 - cp's OEIS frontend

A154420 Maximal coefficient of MacMahon polynomial (cf. A060187) p(x,n)=2^n(1 - x)^(n + 1) LerchPhi[x, -n, 1/2]; that is, a(n) = Max(coefficients(p(x,n))).

1, 1, 6, 23, 230, 1682, 23548, 259723, 4675014, 69413294, 1527092468, 28588019814, 743288515164, 16818059163492, 504541774904760, 13397724585164019, 455522635895576646, 13892023109165902550, 527896878148304296900

Offset: 0

Roger L. Bagula, Jan 09 2009

nonn

Since the center is the maximum in the Pascal, Eulerian and MacMahon triangles, a(n)=MacMahon[n,Floor[n/2]]

Vaclav Kotesovec, Table of n, a(n) for n = 0..300
Peter Luschny, Generalized Eulerian polynomials.

Cf. A060187, A006551.

gf := proc(n, k) local f; f := (x,t) -> x*exp(t*x/k)/(1-x*exp(t*x));
series(f(x,t), t, n+2); ((1-x)/x)^(n+1)*k^n*n!*coeff(%, t, n):
collect(simplify(%), x) end:
seq(coeff(gf(n,1),x,iquo(n,2)),n=0..18); # Middle Eulerian numbers, A006551.
seq(coeff(gf(n,2),x,iquo(n,2)),n=0..18); # Middle midpoint Eulerian numbers.
# Peter Luschny, May 02 2013

p[x_, n_] = 2^n*(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2];
Table[Max[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]], {n, 0, 30}]

a(n) ~ sqrt(3) * 2^(n+1) * n^n / exp(n). - Vaclav Kotesovec, Oct 28 2021

Edited by N. J. A. Sloane, Jan 15 2009

A138076 Triangle read by rows: T(n, k) = (-1)^(n+k) * A060187(n+1, k+1).

Original entry on oeis.org

1, -1, 1, 1, -6, 1, -1, 23, -23, 1, 1, -76, 230, -76, 1, -1, 237, -1682, 1682, -237, 1, 1, -722, 10543, -23548, 10543, -722, 1, -1, 2179, -60657, 259723, -259723, 60657, -2179, 1, 1, -6552, 331612, -2485288, 4675014, -2485288, 331612, -6552, 1, -1, 19673, -1756340, 21707972, -69413294, 69413294, -21707972, 1756340, -19673, 1

Offset: 0

Roger L. Bagula, Nov 26 2009

Former name: A signed version of A060187 obtained by taking the Z-transform of p(t,x) = exp(t*(1+2*x)). - G. C. Greubel, Jul 21 2024

			Triangle begins as:
   1;
  -1,     1;
   1,    -6,      1;
  -1,    23,    -23,        1;
   1,   -76,    230,      -76,       1;
  -1,   237,  -1682,     1682,    -237,        1;
   1,  -722,  10543,   -23548,   10543,     -722,      1;
  -1,  2179, -60657,   259723, -259723,    60657,  -2179,     1;
   1, -6552, 331612, -2485288, 4675014, -2485288, 331612, -6552, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000165, A002436, A060187, A177043, A178118.

A060187:= func< n,k | (&+[(-1)^(k-j)*Binomial(n, k-j)*(2*j-1)^(n-1): j in [1..k]]) >;
A138076:= func< n,k | (-1)^(n+k)*A060187(n+1,k+1) >;
[A138076(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 21 2024

p[t_] = Exp[t]*x/(Exp[2*t] + x);
Table[CoefficientList[(n!*(1+x)^(n+1)/x)*SeriesCoefficient[Series[p[ t], {t,0,30}], n], x], {n,0,12}]//Flatten

@CachedFunction
def t(n,k): # t = A060187
    if k==1 or k==n: return 1
    return (2*(n-k)+1)*t(n-1, k-1) + (2*k-1)*t(n-1, k)
def A138076(n,k): return (-1)^(n+k)*t(n+1,k+1)
flatten([[A138076(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 21 2024

T(n, k) = (-1)^(n+k) * A060187(n+1, k+1).
From G. C. Greubel, Jul 21 2024: (Start)
T(2*n, n) = (-1)^n * A177043(n).
Sum_{k=0..n} T(n, k) = (1/2)*(1 + (-1)^n)*(-1)^floor((n+ 1)/2) * A002436(floor(n/2)).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)^n * A000165(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = (-1)^n * A178118(n+1). (End)

A176291 A symmetrical triangle based on Narayana numbers and Eulerian numbers of type B: T(n, k) = 2 + A060187(n, k) - 2binomial(n, k)binomial(n+1, k)/(k+1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 13, 13, 1, 1, 58, 192, 58, 1, 1, 209, 1584, 1584, 209, 1, 1, 682, 10335, 23200, 10335, 682, 1, 1, 2125, 60267, 258745, 258745, 60267, 2125, 1, 1, 6482, 330942, 2482938, 4671488, 2482938, 330942, 6482, 1, 1, 19585, 1755262, 21702934, 69402712, 69402712, 21702934, 1755262, 19585, 1

Offset: 0

Roger L. Bagula, Apr 14 2010

Row sums are: {1, 2, 4, 28, 310, 3588, 45236, 642276, 10312214, 185760988, 3715773650, ...}.

			Triangle begins as:
  1;
  1,    1;
  1,    2,      1;
  1,   13,     13,       1;
  1,   58,    192,      58,       1;
  1,  209,   1584,    1584,     209,       1;
  1,  682,  10335,   23200,   10335,     682,      1;
  1, 2125,  60267,  258745,  258745,   60267,   2125,    1;
  1, 6482, 330942, 2482938, 4671488, 2482938, 330942, 6482, 1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A007318, A060187.

B:=Binomial;; Flat(List([0..10], n-> List([0..n], k-> 2*(1 - B(n,k)*B(n+1,k)/(k+1)) + Sum([0..k], j-> (-1)^j*B(n+1,j)*(2*(k-j)+1)^n) ))); # G. C. Greubel, Nov 23 2019

B:=Binomial; [2*(1 - B(n,k)*B(n+1,k)/(k+1)) + (&+[(-1)^j*B(n+1,j) *(2*(k-j)+1)^n: j in [0..k]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Nov 23 2019

b:= binomial; T:= 2 + sum((-1)^j*b(n+1,j)*(2*(k-j)+1)^n, j=0..k) - 2*b(n, k)*b(n+1, k)/(k+1); seq(seq(T(n, k), k = 0 .. n), n = 0 .. 10); # G. C. Greubel, Nov 23 2019

(* First program *)
p[x_, n_]= (1 - x)^(n+1)*Sum[(2*k+1)^n*x^k, {k, 0, Infinity}];(*A060187*)
f[n_, m_]:= CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m+1]];
T[n_, m_]:= 2 -(-f[n, m] +2*Binomial[n, m]*Binomial[n+1, m]/(m+1));
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten
(* Second program *)
B:=Binomial; T[n_,k_]:= T[n,k]= 2 +Sum[(-1)^j*B[n+1,j]*(2*(k-j)+1)^n, {j, 0,k}] -2*B[n,k]*B[n+1,k]/(k+1); Table[T[n,k], {n,0,10}, {k,0,n} ]//Flatten (* G. C. Greubel, Nov 23 2019 *)

T(n,k) = b=binomial; 2 + sum(j=0,k, (-1)^j*b(n+1,j)*(2*(k-j)+1)^n) - 2*b(n, k)*b(n+1, k)/(k+1); \\ G. C. Greubel, Nov 23 2019

b=binomial; [[2 + sum( (-1)^j*b(n+1,j)*(2*(k-j)+1)^n for j in (0..k)) - 2*b(n, k)*b(n+1, k)/(k+1) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Nov 23 2019

T(n, k) = 2 + A060187(n, k) - 2*binomial(n, k)*binomial(n+1, k)/(k+1), where A060187(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1, k-j)* (2*j+1)^n.

Edited by G. C. Greubel, Nov 23 2019

A177429 Triangle read by rows: T(n,m)=A060187(1+n,1+m) *n! / (n-m)!

Original entry on oeis.org

1, 1, 1, 1, 12, 2, 1, 69, 138, 6, 1, 304, 2760, 1824, 24, 1, 1185, 33640, 100920, 28440, 120, 1, 4332, 316290, 2825760, 3795480, 519840, 720, 1, 15253, 2547594, 54541830, 218167320, 152855640, 10982160, 5040, 1, 52416, 18570272, 835056768

Offset: 0

Roger L. Bagula, May 08 2010

Row sums are: 1, 2, 15, 214, 4913, 164306, 7462423, 439114838, 32358353217, 2909210035042, 312597121198751,...

			1;
1, 1;
1, 12, 2;
1, 69, 138, 6;
1, 304, 2760, 1824, 24;
1, 1185, 33640, 100920, 28440, 120;
1, 4332, 316290, 2825760, 3795480, 519840, 720;
1, 15253, 2547594, 54541830, 218167320, 152855640, 10982160, 5040;
1, 52416, 18570272, 835056768, 7854023520, 16701135360, 6685297920, 264176640, 40320;
1, 177057, 126456480, 10940817888, 209905801056, 1049529005280, 1312898146560, 318670329600, 7138938240, 362880;

Cf. A060187

A177429 := proc(n,k)
    A060187(n+1,k+1)*n!/(n-k)! ;
end proc: # R. J. Mathar, Jun 16 2015

(*A060187*);
p[x_, n_] = (1 - x)^(n + 1)*Sum[(2*k + 1)^n*x^k, {k, 0, Infinity}];
f[n_, m_] := CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m + 1]];
t[n_, m_] := f[n, m]*n!/(n - m)!;
Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[%]

A178118 Antidiagonal sums of the triangle A060187.

Original entry on oeis.org

0, 1, 1, 2, 7, 25, 100, 469, 2481, 14406, 90995, 621553, 4561112, 35736921, 297435521, 2618575194, 24297706927, 236870849417, 2419213831452, 25820011544781, 287327296473585, 3326999636488190, 40011485288491131

Offset: 0

Roger L. Bagula, May 20 2010

nonn

This sequence is an analog to the Lucas formula which obtains A000045 as the antidiagonal sums of the Pascal triangle A007318.

David M. Burton, Elementary number theory, McGraw Hill (2002), page 286

Cf. A000800, A060187, A000045

p[x_, n_] = (1 - x)^(n + 1)*Sum[(2*k + 1)^n*x^k, {k, 0, Infinity}];
f[n_, m_] := CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m + 1]]
a[n_] := Sum[f[n - m - 1, m], {m, 0, Floor[(n - 1)/2]}]
Table[a[n], {n, 0, 30}]

a(n) = sum_{m=0.. floor[(n-1)/2]} A060187(n-m-1,m).

Exact definition moved to formula - the Assoc. Eds. of the OEIS, Aug 20 2010

A178122 Triangle T(n,m) = A060187(n+1,m+1) + 2*binomial(n,m) - 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 27, 27, 1, 1, 82, 240, 82, 1, 1, 245, 1700, 1700, 245, 1, 1, 732, 10571, 23586, 10571, 732, 1, 1, 2191, 60697, 259791, 259791, 60697, 2191, 1, 1, 6566, 331666, 2485398, 4675152, 2485398, 331666, 6566, 1, 1, 19689, 1756410, 21708138, 69413544, 69413544, 21708138, 1756410, 19689, 1

Offset: 0

Roger L. Bagula, May 20 2010

			Rows n>=0 and columns 0<=m<=n start as:
  1;
  1,     1;
  1,     8,       1;
  1,    27,      27,        1;
  1,    82,     240,       82,        1;
  1,   245,    1700,     1700,      245,        1;
  1,   732,   10571,    23586,    10571,      732,        1;
  1,  2191,   60697,   259791,   259791,    60697,     2191,       1;
  1,  6566,  331666,  2485398,  4675152,  2485398,   331666,    6566,     1;
  1, 19689, 1756410, 21708138, 69413544, 69413544, 21708138, 1756410, 19689, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000165, A007318, A060187, A142458.

A060187:= func< n,k | (&+[(-1)^(k-j)*Binomial(n, k-j)*(2*j-1)^(n-1): j in [1..k]]) >;
A178122:= func< n,k | A060187(n+1, k+1) + 2*Binomial(n, k) - 2 >;
[A178122(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 18 2022

p[x_, n_] = (1 - x)^(n + 1)*Sum[(2*k + 1)^n*x^k, {k, 0, Infinity}];
f[n_, m_] := CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m + 1]];
t[n_, m_] := f[n, m] + 2*Binomial[n, m] - 2 ;
Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[%]

def A060187(n,k): return sum( (-1)^(k-j)*binomial(n, k-j)*(2*j-1)^(n-1) for j in (1..k) )
def A178122(n,k): return A060187(n+1, k+1) + 2*binomial(n, k) - 2
flatten([[A178122(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 18 2022

T(n, m) = A060187(n+1,m+1) + 2*A007318(n,m) - 2.
T(n, m) = T(n, n-m).
Sum_{k=0..n} T(n, k) = A000165(n) + 2*(2^n -(n+1)).

Indices in definition corrected, row sum formula added by the Assoc. Eds. of the OEIS - Aug 20 2010

A142240 A triangular sequence from the pattern in row sums of Pascal's triangle A007318, Eulerian numbers A008292 and A060187: Delta_diagonal=m; m={0,1,2,3,...k}.

Original entry on oeis.org

1, 2, 2, 2, 3, 2, 2, 4, 4, 2, 2, 5, 6, 5, 2, 2, 6, 8, 8, 6, 2, 2, 7, 10, 11, 10, 7, 2, 2, 8, 12, 14, 14, 12, 8, 2, 2, 9, 14, 17, 18, 17, 14, 9, 2, 2, 10, 16, 20, 22, 22, 20, 16, 10, 2

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 17 2008

Row sums are:
{1, 4, 7, 12, 20, 32, 49, 72, 102, 140}.
The triangle is calculated by hand.
Row sums are:
1) Pascal A007318:ratio =2: delta row zero:a(n)=2*a(n-1);a(1)=1;
b(n)->0
1,2,4,8,16,32,64,128,256,512
2) Eulerian numbers A008292: ratio =n: delta=1:a(n)=n*a(n-1)
1,2,6,24,120,720,...n!
b(n)->1,2,3,4,...
3) A060187:ratio=2*n: delta=2:a(n)=2*n*a(n-1)
b(n)->2,4,6,8,...
{1, 2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200}.
4) hypothetical next level sums:delta=3:b(n)=b(n-1)+3;a(n)=a(n-1)*b(n);
b(n)->{2, 5, 8, 11, 14, 17, 20, 23, 26, 29} diagonal
{1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600}
5)hypothetical next level sums:delta=3:b(n)=b(n-1)+4;a(n)=a(n-1)*b(n);
b(n)->{2, 6, 10, 14, 18, 22, 26, 30, 34, 38} diagonal
{1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800}
The conjecture that goes with this triangular sequence is that there on n levels like these for Pascal combinatorial quantum levels.

			{1},
{2, 2},
{2, 3, 2},
{2, 4, 4, 2},
{2, 5, 6, 5, 2},
{2, 6, 8, 8, 6, 2},
{2, 7, 10, 11, 10, 7, 2},
{2, 8, 12, 14, 14, 12, 8, 2},
{2, 9, 14, 17, 18, 17, 14, 9, 2},
{2, 10, 16, 20, 22, 22, 20, 16, 10, 2}

Cf. A060187, A007318, A008292.

a={{1},{2,2},{2,3,2},{2,4,4,2}, {2,5,6,5,2},{2,6,8,8,6,2},{2,7,10,11,10,7,2},{2,8,12,14,14, 12,8,2},{2,9,14,17,18,17,14,9,2},{2,10,16,20,22,22,20,16,10,2}} Flatten[a] Table[Apply[Plus,a[[n]]],{n,1,10}]

b(n,m)=b(n-1,m]+m; Delta_diagonal=m; m={0,1,2,3,...k}.

A142707 Coefficients of derivatives of MacMahon polynomials (A060187): p(x,n)=2^n(1 - x)^(1 + n)LerchPhi[x, -n, 1/2]; p'(x,n)=(d/dx)p{x,n).

Original entry on oeis.org

1, 6, 2, 23, 46, 3, 76, 460, 228, 4, 237, 3364, 5046, 948, 5, 722, 21086, 70644, 42172, 3610, 6, 2179, 121314, 779169, 1038892, 303285, 13074, 7, 6552, 663224, 7455864, 18700056, 12426440, 1989672, 45864, 8, 19673, 3512680, 65123916, 277653176

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 24 2008

Row sums are:A014479

0, 1, 8, 72, 768, 9600, 138240, 2257920, 41287680, 836075520, 18579456000.

			{1},
{6, 2},
{23, 46, 3},
{76, 460, 228, 4},
{237, 3364, 5046, 948, 5},
{722, 21086, 70644, 42172, 3610, 6},
{2179, 121314, 779169, 1038892, 303285, 13074, 7},
{6552, 663224, 7455864, 18700056, 12426440, 1989672, 45864, 8},
{19673, 3512680, 65123916, 277653176, 347066470, 130247832, 12294380, 157384, 9},
{59038, 18232282, 534902712, 3627693128, 7635462340, 5441539692, 1248106328, 72929128, 531342, 10}

Cf. A060187, A014479.

Clear[p, x, n, a]; p[x_, n_] = 2^n*(1 - x)^(1 + n)*LerchPhi[x, -n, 1/2]; Table[FullSimplify[Expand[D[p[x, n], x]]], {n, 0, 10}]; Table[CoefficientList[FullSimplify[Expand[D[p[x, n], x]]], x], {n, 0, 10}]; Flatten[%]

p(x,n)=2^n*(1 - x)^(1 + n)*LerchPhi[x, -n, 1/2]; p'(x,n)=(d/dx)p{x,n); t(n,m)=Coefficients(p'(x,n)).

A146543 The LerchPhi functional part of A060187 MacMahon numbers is treated/ solved for as a curvature to give a set of polynomial triangle sequence coefficients: p(x,n)=Sum[A060187(n,m)*x^(m-1),{m,0,n}]; q(x,n)=k from Solve[FullSimplify[ExpandAll[p[x, n]/(x - 1)^n]] - (1 + k/x^2) == 0, k].

Original entry on oeis.org

2, 0, 8, 2, 20, 26, 0, 80, 224, 80, 2, 232, 1692, 1672, 242, 0, 728, 10528, 23568, 10528, 728, 2, 2172, 60678, 259688, 259758, 60636, 2186, 0, 6560, 331584, 2485344, 4674944, 2485344, 331584, 6560, 2, 19664, 1756376, 21707888, 69413420, 69413168

Offset: 0

Roger L. Bagula, Oct 31 2008

nonn

The concept here is that the increase in curvature causes transformation of Pascal's triangle into the Eulerian numbers and the MacMahon numbers, while leaving the numerical Modulo 2 Sierpinski Self -Similarity intact. The resulting polynomials have a finite Blaschke elliptical structure. The row sums are: {0, 2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200}.

			{}, {2}, {0, 8}, {2, 20, 26}, {0, 80, 224, 80}, {2, 232, 1692, 1672, 242}, {0, 728, 10528, 23568, 10528, 728}, {2, 2172, 60678, 259688, 259758, 60636, 2186}, {0, 6560, 331584, 2485344, 4674944, 2485344, 331584, 6560}, {2, 19664, 1756376, 21707888, 69413420, 69413168, 21708056, 1756304, 19682}, {0, 59048, 9116096, 178301024, 906923072, 1527092720, 906923072, 178301024,9116096, 59048}

Kenneth Hoffman, Banach Spaces of Analytic Functions, Dover, New York, 1962, page 66, page 132.
Lennart Carleson and Theodore W. Gamelin, Complex Dynamics, Springer,New York,1993,pp 103 ( Herman's Rings as Finite Blaschke sets)

A060187, A008292

Clear[q, p, x, n, a]; p[x_, n_] = p[x_, n_] = (1 - x)^(n + 1)*PolyLog[ -n, x]/x; q[x_, n_] := ((x - 1)^n/x^2)*k /. Solve[FullSimplify[ExpandAll[p[x, n]/(x - 1)^n]] - (1 + k/x^2) == 0, k]; Table[FullSimplify[Expand[q[x, n]]], {n, 0, 10}]; Table[Flatten[CoefficientList[FullSimplify[Expand[q[x, n]]], x]], {n, 0, 10}]; Flatten[%]

p(x,n)=Sum[A060187(n,m)*x^(m-1),{m,0,n}]; q(x,n)=k from Solve[FullSimplify[ExpandAll[p[x, n]/(x - 1)^n]] - (1 + k/x^2) == 0, k]; t(n,m)=Coefficients(((x - 1)^n/x^2)*q(n,x)).

A146568 Coefficients of Pascal's triangle polynomial minus MacMahon polynomial A060187 with a power of x divided out: q(x,n)=2^n(1 - x)^(n + 1) LerchPhi[x, -n, 1/2]; p(x,n)=((x+1)^n-q(x,n))/x.

Original entry on oeis.org

4, 20, 20, 72, 224, 72, 232, 1672, 1672, 232, 716, 10528, 23528, 10528, 716, 2172, 60636, 259688, 259688, 60636, 2172, 6544, 331584, 2485232, 4674944, 2485232, 331584, 6544, 19664, 1756304, 21707888, 69413168, 69413168, 21707888, 1756304

Offset: 2

Roger L. Bagula, Nov 01 2008

First elements in each row are: 3^n - 2*n - 1 (A061981).

			Triangle starts:
{4},
{20, 20},
{72, 224, 72},
{232, 1672, 1672, 232},
{716, 10528, 23528, 10528, 716},
{2172, 60636, 259688, 259688, 60636, 2172},
{6544, 331584, 2485232, 4674944, 2485232, 331584, 6544},
{19664, 1756304, 21707888, 69413168, 69413168, 21707888, 1756304, 19664},
{59028, 9116096, 178300784, 906923072, 1527092216, 906923072, 178300784, 9116096, 59028}

Cf. A061981, A060187.

q[x_, n_] = 2^n*(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2]; p[x_, n_] = (q[x, n] - (x + 1)^n)/x; Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 2, 10}]; Flatten[%]

q(x,n)=2^n*(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2]; p(x,n)=((x+1)^n-q(x,n))/x; t(n,m)=Coefficients(p(x,n)) with n starting at 2.

cp's OEIS Frontend

A154420 Maximal coefficient of MacMahon polynomial (cf. A060187) p(x,n)=2^n*(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2]; that is, a(n) = Max(coefficients(p(x,n))).

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A138076 Triangle read by rows: T(n, k) = (-1)^(n+k) * A060187(n+1, k+1).

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A176291 A symmetrical triangle based on Narayana numbers and Eulerian numbers of type B: T(n, k) = 2 + A060187(n, k) - 2*binomial(n, k)*binomial(n+1, k)/(k+1).

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A177429 Triangle read by rows: T(n,m)=A060187(1+n,1+m) *n! / (n-m)!

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A178118 Antidiagonal sums of the triangle A060187.

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A178122 Triangle T(n,m) = A060187(n+1,m+1) + 2*binomial(n,m) - 2, read by rows.

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A142240 A triangular sequence from the pattern in row sums of Pascal's triangle A007318, Eulerian numbers A008292 and A060187: Delta_diagonal=m; m={0,1,2,3,...k}.

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A154420 Maximal coefficient of MacMahon polynomial (cf. A060187) p(x,n)=2^n(1 - x)^(n + 1) LerchPhi[x, -n, 1/2]; that is, a(n) = Max(coefficients(p(x,n))).

A176291 A symmetrical triangle based on Narayana numbers and Eulerian numbers of type B: T(n, k) = 2 + A060187(n, k) - 2binomial(n, k)binomial(n+1, k)/(k+1).

A142707 Coefficients of derivatives of MacMahon polynomials (A060187): p(x,n)=2^n(1 - x)^(1 + n)LerchPhi[x, -n, 1/2]; p'(x,n)=(d/dx)p{x,n).

A146568 Coefficients of Pascal's triangle polynomial minus MacMahon polynomial A060187 with a power of x divided out: q(x,n)=2^n(1 - x)^(n + 1) LerchPhi[x, -n, 1/2]; p(x,n)=((x+1)^n-q(x,n))/x.