A061981 -id:A061981 - cp's OEIS frontend

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2n+1,i) binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

1, 1, 4, 1, 1, 20, 48, 20, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 232, 5158, 27664, 47290, 27664, 5158, 232, 1, 1, 716, 37257, 450048, 1822014, 2864328, 1822014, 450048, 37257, 716, 1, 1, 2172, 247236, 6030140, 49258935, 163809288, 242384856, 163809288, 49258935, 6030140, 247236, 2172, 1

Offset: 1

Roger L. Bagula, Jan 06 2009

From Yahia Kahloune, Jan 30 2014: (Start)
In general, let b(k,e,p) = Sum_{i=0..k} (-1)^i*binomial(e*p+1,i)*binomial(k+e-i,e)^p. Then T(n,k) = b(k,2,n).
With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{k=0..e*(p-1)} b(k,e,p)*binomial(n+e+k,e*p+k).
For example, A085438(n) = Sum_{i=1..n} binomial(1+i,2)^3 = T(3,0)*binomial(2+n,7) + T(3,1)*binomial(3+n,7) + T(3,2)*binomial(4+n,7) + T(3,3)*binomial(5+n,7) + T(3,4)*binomial(6+n,7) = (1/5040)*(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n).
(End)
T(n,k) is the number of permutations of 2 indistinguishable copies of 1..n with exactly k descents. A descent is a pair of adjacent elements with the second element less than the first. - Andrew Howroyd, May 06 2020

			Triangle begins:
  1;
  1,     4,       1;
  1,    20,      48,        20,           1;
  1,    72,     603,      1168,         603,           72,           1;
  1,   232,    5158,     27664,       47290,        27664,        5158,  232, 1;
  1,   716,   37257,    450048,     1822014,      2864328,     1822014, ...;
  1,  2172,  247236,   6030140,    49258935,    163809288,   242384856, ...;
  1,  6544, 1568215,  72338144,  1086859301,   6727188848, 19323413187, ...;
  1, 19664, 9703890, 811888600, 21147576440, 225167210712, ... ;
  ...
The T(2,1) = 4 permutations of 1122 with 1 descent are 1212, 1221, 2112, 2211. - _Andrew Howroyd_, May 15 2020

Andrew Howroyd, Table of n, a(n) for n = 1..1600 (rows 1..40)
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 12.
Helmut Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.

Columns k=0..9 are A000012, A061981, A151624, A151625, A151626, A151627, A151628, A151629, A151630, A151631.
Row sums are A000680.
Similar triangles for e=1..6: A173018 (or A008292), this sequence, A174266, A236463, A237202, A237252.
Cf. A000680, A005799, A085438, A334781.

[(&+[(-1)^j*Binomial(2*n+1,j)*Binomial(k-j+2,2)^n: j in [0..k]]): k in [0..2*n-2], n in [1..12]]; // G. C. Greubel, Jun 13 2022

A154283 := proc(n,k)
        (1-x)^(2*n+1)*add( (l*(l+1)/2)^n*x^(l-1),l=0..k+1) ;
        coeftayl(%,x=0,k) ;
end proc: # R. J. Mathar, Feb 01 2013

p[x_, n_]= (1-x)^(2*n+1)*Sum[(k*(k+1)/2)^n*x^k, {k, 0, Infinity}]/x;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n,10}]//Flatten

T(n,k)={sum(i=0, k, (-1)^i*binomial(2*n+1, i)*binomial(k+2-i, 2)^n)} \\ Andrew Howroyd, May 09 2020

def A154283(n,k): return sum((-1)^j*binomial(2*n+1, j)*binomial(k-j+2, 2)^n for j in (0..k))
flatten([[A154283(n,k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jun 13 2022

T(n,k) = (-1) times coefficient of x^k in (x-1)^(2*n+1) * Sum_{k>=0} (k*(k+1)/2)^n *x^(k-1).
From Yahia Kahloune, Jan 29 2014: (Start)
Sum_{i=1..n} binomial(1+i,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+2+k,2*p+1).
binomial(n,2)^p = Sum_{k=0..2*p-2} T(p,k)*binomial(n+k,2*p). (End)
From Peter Bala, Dec 21 2019: (Start)
E.g.f. as a continued fraction: (1-x)/(1-x + ( 1-exp((1-x)^2*t))*x/(1-x + (1-exp(2*(1-x)^2*t))*x/(1-x + (1-exp(3*(1-x)^2*t))*x/(1-x + ... )))) =  1 + x*t + x*(x^2 + 4*x + 1)*t^2/2! + x*(x^4 + 20*x^3 + 48*x^2 + 20*x + 1)*t^3/3! + ... (use Prodinger equation 1.1).
The sequence of alternating row sums (unsigned) [1, 1, 2, 10, 104, 1816,...] appears to be A005799. (End)

Edited by N. J. A. Sloane, Jan 30 2014 following suggestions from Yahia Kahloune (among other things, the signs of all terms have been reversed).

Edited by Andrew Howroyd, May 09 2020

A061980 Square array A(n,k) = A(n-1,k) + A(n-1, floor(k/2)) + A(n-1, floor(k/3)), with A(0,0) = 1, read by antidiagonals.

Original entry on oeis.org

1, 0, 3, 0, 2, 9, 0, 1, 8, 27, 0, 0, 6, 26, 81, 0, 0, 4, 23, 80, 243, 0, 0, 3, 20, 76, 242, 729, 0, 0, 3, 17, 72, 237, 728, 2187, 0, 0, 1, 17, 66, 232, 722, 2186, 6561, 0, 0, 1, 11, 66, 222, 716, 2179, 6560, 19683, 0, 0, 1, 11, 54, 222, 701, 2172, 6552, 19682, 59049

Offset: 0

Henry Bottomley, May 24 2001

			Array begins as:
    1,   0,   0,   0,   0,   0,   0, ...;
    3,   2,   1,   0,   0,   0,   0, ...;
    9,   8,   6,   4,   3,   3,   1, ...;
   27,  26,  23,  20,  17,  17,  11, ...;
   81,  80,  76,  72,  66,  66,  54, ...;
  243, 242, 237, 232, 222, 222, 202, ...;
  729, 728, 722, 716, 701, 701, 671, ...;
Antidiagonal rows begin as:
  1;
  0, 3;
  0, 2, 9;
  0, 1, 8, 27;
  0, 0, 6, 26, 81;
  0, 0, 4, 23, 80, 243;
  0, 0, 3, 20, 76, 242, 729;
  0, 0, 3, 17, 72, 237, 728, 2187;
  0, 0, 1, 17, 66, 232, 722, 2186, 6561;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Row sums are 6^n: A000400.
Columns are A000244, A024023, A060188, A061981, A061982 twice, A061983 twice, etc.
Cf. A000244, A061290, A061930, A061979, A061984, A061987.

A[n_, k_]:= A[n, k]= If[n==0, Boole[k==0], A[n-1,k] +A[n-1,Floor[k/2]] +A[n-1, Floor[k/3]]];
T[n_, k_]:= A[k, n-k];
Table[A[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 18 2022 *)

@CachedFunction
def A(n,k):
    if (n==0): return 0^k
    else: return A(n-1, k) + A(n-1, (k//2)) + A(n-1, (k//3))
def T(n, k): return A(k, n-k)
flatten([[T(n, k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 18 2022

A(n,k) = A(n-1,k) + A(n-1, floor(k/2)) + A(n-1, floor(k/3)), with A(0,0) = 1.
T(n, k) = A(k, n-k).
Sum_{k=0..n} A(n, k) = A000400(n).
T(n, n) = A(n, 0) = A000244(n). - G. C. Greubel, Jun 18 2022

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.

A146745 Coefficients of Pascal's triangle polynomial minus MacMahon polynomial A060187 with minus the first and last row terms and powers of x divided out: f(n)=3^n - 2n - 1; q(x,n)=2^n(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2]; p(x,n)=((q[x, n] - (x + 1)^n)/x - f[n] - f[n]*x^(n - 2))/x.

Original entry on oeis.org

224, 1672, 1672, 10528, 23528, 10528, 60636, 259688, 259688, 60636, 331584, 2485232, 4674944, 2485232, 331584, 1756304, 21707888, 69413168, 69413168, 21707888, 1756304, 9116096, 178300784, 906923072, 1527092216, 906923072

Offset: 2

Roger L. Bagula, Nov 01 2008

Row sums starting with n=4 are {224, 3344, 44584, 640648, 10308576, 185754720, 3715772120}. First elements in each row are {224, 1672, 1672, 10528, 60636, 331584, 1756304, 9116096}. Subtracting out the row terms gives the middle elements of the difference.

			Triangle starts
{224},
{1672, 1672},
{10528, 23528, 10528},
{60636, 259688, 259688, 60636},
{331584, 2485232, 4674944, 2485232, 331584},
{1756304, 21707888, 69413168, 69413168, 21707888, 1756304},
{9116096, 178300784, 906923072, 1527092216, 906923072, 178300784, 9116096}

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 - f[n] - f[n]*x^(n - 2))/x; Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 4, 10}]; Flatten[%]

f(n) = 3^n - 2*n - 1;
q(x,n) = 2^n*(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2];
p(x,n) = ((q[x, n] - (x + 1)^n)/x - f[n] - f[n]*x^(n - 2))/x;
t(n,m) = Coefficients(p(x,n)) with n starting at 4.

cp's OEIS Frontend

A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2*n+1,i) * binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

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A061980 Square array A(n,k) = A(n-1,k) + A(n-1, floor(k/2)) + A(n-1, floor(k/3)), with A(0,0) = 1, read by antidiagonals.

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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.

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A146745 Coefficients of Pascal's triangle polynomial minus MacMahon polynomial A060187 with minus the first and last row terms and powers of x divided out: f(n)=3^n - 2*n - 1; q(x,n)=2^n*(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2]; p(x,n)=((q[x, n] - (x + 1)^n)/x - f[n] - f[n]*x^(n - 2))/x.

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A154283 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(2n+1,i) binomial(k+2-i,2)^n, 0 <= k <= 2*(n-1).

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.

A146745 Coefficients of Pascal's triangle polynomial minus MacMahon polynomial A060187 with minus the first and last row terms and powers of x divided out: f(n)=3^n - 2n - 1; q(x,n)=2^n(1 - x)^(n + 1)* LerchPhi[x, -n, 1/2]; p(x,n)=((q[x, n] - (x + 1)^n)/x - f[n] - f[n]*x^(n - 2))/x.