A181544 -id:A181544 - cp's OEIS frontend

A183204 Central terms of triangle A181544.

1, 4, 48, 760, 13840, 273504, 5703096, 123519792, 2751843600, 62659854400, 1451780950048, 34116354472512, 811208174862904, 19481055861877120, 471822589361293680, 11511531876280913760, 282665135367572129040

Offset: 0

Paul D. Hanna, Dec 30 2010

The g.f. for row n of triangle A181544 is (1-x)^(3n+1)*Sum_{k>=0}C(n+k-1,k)^3*x^k.
This sequence is s_7 in Cooper's paper. - Jason Kimberley, Nov 06 2012
Diagonal of the rational function R(x,y,z,w) = 1/(1 - (w*x*y + w*x*z + w*y*z + x*y + x*z + y + z)). - Gheorghe Coserea, Jul 14 2016
This is one of the Apery-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Every prime eventually divides some term of this sequence. - Amita Malik, Aug 20 2017

			Triangle A181544 begins:
(1);
1, (4), 1;
1, 20, (48), 20, 1;
1, 54, 405, (760), 405, 54, 1;
1, 112, 1828, 8464, (13840), 8464, 1828, 112, 1; ...

Seiichi Manyama, Table of n, a(n) for n = 0..702 (terms 0..499 from Jason Kimberley)
S. Cooper, Sporadic sequences, modular forms and new series for 1/pi, Ramanujan J., December 2012, Volume 29, Issue 1, pp 163-183.
Shaun Cooper, Jesús Guillera, Armin Straub, and Wadim Zudilin, Crouching AGM, Hidden Modularity, arXiv:1604.01106 [math.NT], 5-April-2016.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See s7 p. 3.
Timothy Huber, Daniel Schultz, and Dongxi Ye, Ramanujan-Sato series for 1/pi, Acta Arith. (2023) Vol. 207, 121-160. See p. 11.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5.
Robert Osburn, Armin Straub, and Wadim Zudilin, A modular supercongruence for 6F5: an Apéry-like story, arXiv:1701.04098 [math.NT], 2017.
Wadim Zudilin, A generating function of the squares of Legendre polynomials, preprint arXiv:1210.2493 [math.CA], 2012.

Cf. A108625, A181544, A183205, A245086.
Related to diagonal of rational functions: A268545-A268555.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

P:=PolynomialRing(Integers()); C:=Binomial;
A183204:=func; // or directly:
A183204:=func;
[A183204(n):n in[0..16]]; // Jason Kimberley, Oct 29 2012

Table[Sum[Binomial[n,j]^2 * Binomial[2*j,n] * Binomial[j+n,j],{j,0,n}],{n,0,20}] (* Vaclav Kotesovec, Apr 05 2015 *)

{a(n)=polcoeff((1-x)^(3*n+1)*sum(j=0, 2*n, binomial(n+j, j)^3*x^j), n)}

a(n) = [x^n] (1-x)^(3n+1) * Sum_{k>=0} C(n+k-1,k)^3*x^k.
a(n) = Sum_{j = 0..n} C(n,j)^2 * C(2*j,n) * C(j+n,j). [Formula of Wadim Zudilin provided by Jason Kimberley, Nov 06 2012]
1/Pi = sqrt(7) Sum_{n>=0} (-1)^n a(n) (11895n + 1286)/22^(3n+3). [Cooper, equation (41)] - Jason Kimberley, Nov 06 2012
G.f.: sqrt((1-13*x+(1-26*x-27*x^2)^(1/2))/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2)))*hypergeom([1/12,5/12],[1],13824*x^7/(1-21*x+8*x^2+(1-8*x)*(1-26*x-27*x^2)^(1/2))^3)^2. - Mark van Hoeij, May 07 2013
a(n) ~ 3^(3*n+3/2) / (4 * (Pi*n)^(3/2)). - Vaclav Kotesovec, Apr 05 2015
G.f. A(x) satisfies 1/(1+4*x)^2 * A( x/(1+4*x)^3 ) = 1/(1+2*x)^2 * A( x^2/(1+2*x)^3 ) [see Cooper, Guillera, Straub, Zudilin]. - Joerg Arndt, Apr 08 2016
a(n) = (-1)^n*binomial(3n+1,n)* 4F3({-n,n+1,n+1,n+1};{1,1,2(n+1)}; 1). - M. Lawrence Glasser, May 15 2016
Conjecture D-finite with recurrence: n^3*a(n) - (2*n-1)*(13*n^2-13*n+4)*a(n-1) - 3*(n-1)*(3*n-4)*(3*n-2)*a(n-2) = 0. - R. J. Mathar, May 15 2016
0 = (-x^2+26*x^3+27*x^4)*y''' + (-3*x+117*x^2+162*x^3)*y'' + (-1+86*x+186*x^2)*y' + (4+24*x)*y, where y is g.f. - Gheorghe Coserea, Jul 14 2016
From Jeremy Tan, Mar 14 2024: (Start)
The conjectured D-finite recurrence can be proved by Zeilberger's algorithm.
a(n) = Sum_{k=0..n} binomial(n,k)^2 * binomial(n+k,n) * binomial(2*n-k,n) = [(w*x*y*z)^n] ((w+y)*(x+z)*(y+z)*(w+x+y+z))^n. (End)
a(n) = Sum_{0 <= j, k <= n} binomial(n, k)^2 * binomial(n, j)^2 * binomial(k+j, n) = Sum_{k = 0..n} binomial(n, k)^2 * A108625(n, k). - Peter Bala, Jul 08 2024
From Peter Bala, Sep 18 2024: (Start)
a(n) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n+k, k)^3*binomial(3*n+1, n-k). Cf A245086.
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*A143007(n, k) (verified using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)

A357407 a(n) = coefficient of x^n, n >= 0, in A(x) = exp( Sum_{n>=1} A183204(n)*x^n/n ), where A183204 equals the central terms of triangle A181544.

Original entry on oeis.org

1, 4, 32, 360, 4964, 78064, 1344020, 24708928, 477282794, 9580852360, 198322047840, 4209371498256, 91221481924426, 2011834246746792, 45039165331725264, 1021419638492387856, 23426910170090512779, 542666070296546760492, 12681393784980089971368

Offset: 0

Paul D. Hanna, Oct 19 2022

nonn

A183204(n) = Sum_{k=floor(n/2)..n} C(n,k)^2 * C(2*k,n) * C(n+k,k).

A183204(n) equals the coefficient of x^n in (1-x)^(3*n+1) * Sum_{k>=0} binomial(n+k-1,k)^3 * x^k, which is the central term of row n of triangle A181544.

			G.f.: A(x) = 1 + 4*x + 32*x^2 + 360*x^3 + 4964*x^4 + 78064*x^5 + 1344020*x^6 + 24708928*x^7 + 477282794*x^8 + 9580852360*x^9 + 198322047840*x^10 + ...
where
log(A(x)) = 4*x + 48*x^2/2 + 760*x^3/3 + 13840*x^4/4 + 273504*x^5/5 + 5703096*x^6/6 + 123519792*x^7/7 + 2751843600*x^8/8 + 62659854400*x^9/9 + ... + A183204(n)*x^n/n + ...

Paul D. Hanna, Table of n, a(n) for n = 0..500

Cf. A183204, A181544.

{A183204(n) = sum(k=n\2,n, binomial(n,k)^2 * binomial(2*k,n) * binomial(n+k,k) )}
{a(n) = polcoeff( exp( sum(m=1,n, A183204(m)*x^m/m ) + x*O(x^n) ),n)}
for(n=0,20,print1(a(n),", "))

a(n) ~ c * 3^(3*n) / n^(5/2), where c = 0.289447274610263555814082139782101227837126089347468995035938970190651243... - Vaclav Kotesovec, Mar 14 2023

A293286 a(n) = A181544(2n, 2n-1).

Original entry on oeis.org

20, 8464, 4050864, 2116980800, 1173644492800, 678353946298560, 404352269157205152, 246796318508780847360, 153477802845690943118400, 96903346351876187722368000, 61954834924471706682462940800, 40029904663914104968204952365824, 26096917229103772343967114415006304

Offset: 1

Eric M. Schmidt, Oct 04 2017

nonn

David J. Gross and Vladimir Rosenhaus, The Bulk Dual of SYK: Cubic Couplings, arXiv:1702.08016 [hep-th], 2017, p. 33.

t[n_, k_] := SeriesCoefficient[Sum[Binomial[n + j, j]^3 x^j, {j, 0, n + k}] (1 - x)^(3n + 1), {x, 0, k}];
a[n_] := t[2n, 2n - 1];
Array[a, 13] (* Jean-François Alcover, Feb 14 2019 *)

def a(n) :
    R. = QQ[]; p = 2*n; q = 2*n-1
    return ((1-x)^(3*p+1) * sum(binomial(p+r,r)^3 * x^r for r in [0..p+q]))[q]

A181543 Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 27, 27, 1, 1, 64, 216, 64, 1, 1, 125, 1000, 1000, 125, 1, 1, 216, 3375, 8000, 3375, 216, 1, 1, 343, 9261, 42875, 42875, 9261, 343, 1, 1, 512, 21952, 175616, 343000, 175616, 21952, 512, 1, 1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1

Offset: 0

Paul D. Hanna, Oct 30 2010

Diagonal of rational function R(x,y,z,t) = 1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z) with respect to x, y, z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t). - Gheorghe Coserea, Jul 01 2018

			Triangle begins:
  1;
  1,   1;
  1,   8,     1;
  1,  27,    27,      1;
  1,  64,   216,     64,       1;
  1, 125,  1000,   1000,     125,       1;
  1, 216,  3375,   8000,    3375,     216,      1;
  1, 343,  9261,  42875,   42875,    9261,    343,     1;
  1, 512, 21952, 175616,  343000,  175616,  21952,   512,   1;
  1, 729, 46656, 592704, 2000376, 2000376, 592704, 46656, 729, 1;
  ...

Indranil Ghosh, Rows 0..120 of triangle, flattened
Jeffrey S. Geronimo, Hugo J. Woerdeman, and Chung Y. Wong, The autoregressive filter problem for multivariable degree one symmetric polynomials, arXiv:2101.00525 [math.CA], 2021.

Cf. A000172 (row sums), A181545 (antidiagonal sums), A002897, A181544, A248658.

Variants: A008459, A007318.

T:= (n, k)-> binomial(n, k)^3:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Jan 06 2021

Flatten[Table[Binomial[n,k]^3,{n,0,10},{k,0,n}]] (* Harvey P. Dale, May 23 2011 *)

T(n,k)=binomial(n,k)^3
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print())

T(n,k)=polcoeff(polcoeff(sum(m=0,n,(3*m)!/m!^3*x^(2*m)*y^m/(1-x-x*y+x*O(x^n))^(3*m+1)),n,x),k,y)
for(n=0,10,for(k=0,n,print1(T(n,k),", "));print()) \\ Paul D. Hanna, Nov 04 2010

diag(expr, N=22, var=variables(expr)) = {
  my(a = vector(N));
  for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
  for (n = 1, N, a[n] = expr;
    for (k = 1, #var, a[n] = polcoeff(a[n], n-1)));
  return(a);
};
x='x; y='y; z='z; t='t;
concat(apply(Vec, diag(1/(1 + y + z + x*y + y*z + t*x*z + (t+1)*x*y*z), 10, [x, y, z]))) \\ Gheorghe Coserea, Jul 01 2018

Row sums equal A000172, the Franel numbers.
Central terms are A002897(n) = C(2n,n)^3.
Antidiagonal sums equal A181545;
The g.f. of the antidiagonal sums is Sum_{n>=0} (3n)!/(n!)^3 * x^(3n)/(1-x-x^2)^(3n+1).
G.f. for column k: [Sum_{j=0..2k} A181544(k,j)*x^j]/(1-x)^(3k+1), where the row sums of A181544 equals De Bruijn's s(3,n) = (3n)!/(n!)^3.
G.f.: A(x,y) = Sum_{n>=0} (3n)!/n!^3 * x^(2n)*y^n/(1-x-x*y)^(3n+1). - Paul D. Hanna, Nov 04 2010

A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4n+1,i) binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

Original entry on oeis.org

1, 1, 16, 36, 16, 1, 1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1, 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1, 1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680

Offset: 1

Yahia Kahloune, Feb 01 2014

In general, define 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,4,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 4 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 08 2020

			T(n,0) = 1;
T(n,1) = 5^n - (4*n+1);
T(n,2) = 15^n - (4*n+1)*5^n + C(4*n+1,2);
T(n,3) = 35^n - (4*n+1)*15^n + C(4*n+1,2)*5^n - C(4*n+1,3);
T(n,4) = 70^n - (4*n+1)*35^n + C(4*n+1,2)*15^n - C(4*n+1,3)*5^n + C(4*n+1,4).
Triangle T(n,k) begins:
1,
1, 16, 36, 16, 1;
1, 112, 1828, 8464, 13840, 8464, 1828, 112, 1;
1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1;
1, 3104, 693960, 37229920, 733059110, 6501577152, 29066972368, 69830127680, 93200908410, 69830127680, 29066972368, 6501577152, 733059110, 37229920, 693960, 3104, 1;
1, 15600, 11000300, 1558185200, 75073622025, 1585757994496, 16938467955200, 99825129369600, 342907451401150, 710228619472800, 903546399077256, 710228619472800, 342907451401150, 99825129369600, 16938467955200, 1585757994496, 75073622025, 1558185200, 11000300, 15600, 1;
  ...
Example:
Sum_{i=1..n} C(3+i,4)^3 = C(n+4,13) + 112*C(n+5,13) + 1828*C(n+6,13) + 8464*C(n+7,13) + 13840*C(n+8,13) + 8464*C(n+9,13) + 1828*C(n+10,13) + 112*C(n+11,13) + C(+12,13).
C(n,4)^3 = C(n,12) + 112*C(n+1,12) + 1828*C(n+2,12) + 8464*C(n+3,12) + 13840*C(n+4,12) + 8464*C(n+5,12) + 1828*C(n+6,12) + 112*C(n+7,12) + C(n+8,12).

Vincenzo Librandi, Table of n, a(n) for n = 1..780
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.

Columns k=2..8 are A151640, A151641, A151642, A151643, A151644, A151645, A151646.
Row sums are A014608.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, this sequence, A237202, A237252.
Sum_{i=1..n} binomial(3+i,4)^p for p=2..3 gives: A086023, A086024.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 4, p_] := Sum[(-1)^i*Binomial[4*p+1, i]*Binomial[k-i, 4]^p /. k -> 4+i, {i, 0, k-4}]; row[p_] := Table[b[k, 4, p], {k, 4, 4*p}]; Table[row[p], {p, 1, 6}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(4*n+1, i)*binomial(k+4-i, 4)^n)} \\ Andrew Howroyd, May 08 2020

Sum_{i=1..n} binomial(3+i,4)^p = Sum{k=0..4*(p-1)} T(p,k) * binomial(n+4+k, 4*p+1).

binomial(n,4)^p = Sum_{k=0..4*(p-1)} T(p,k) * binomial(n+k, 4*p).

a(36) corrected by Vincenzo Librandi, Feb 14 2014

Edited by Andrew Howroyd, May 08 2020

A126086 Number of paths from (0,0,0) to (n,n,n) such that at each step (i) at least one coordinate increases, (ii) no coordinate decreases, (iii) no coordinate increases by more than 1 and (iv) all coordinates are integers.

Original entry on oeis.org

1, 13, 409, 16081, 699121, 32193253, 1538743249, 75494983297, 3776339263873, 191731486403293, 9850349744182729, 510958871182549297, 26716694098174738321, 1406361374518034383189, 74456543501901550262689, 3961518532003397434536961, 211689479080817606497324033

Offset: 0

Nick Hobson, Mar 03 2007

nonn

Also, the number of alignments for 3 sequences of length n each (Slowinski 1998).

This is a 3-dimensional generalization of A001850.

			Illustrating a(1) = 13:
000 -> 001 -> 011 -> 111
000 -> 001 -> 101 -> 111
000 -> 001 -> 111
000 -> 010 -> 011 -> 111
000 -> 010 -> 110 -> 111
000 -> 010 -> 111
000 -> 100 -> 101 -> 111
000 -> 100 -> 110 -> 111
000 -> 100 -> 111
000 -> 011 -> 111
000 -> 101 -> 111
000 -> 110 -> 111
000 -> 111

Alois P. Heinz, Table of n, a(n) for n = 0..500 (first 101 terms from Nick Hobson)
A. Bostan, S. Boukraa, J.-M. Maillard, and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
E. Duchi and R. A. Sulanke, The 2^{n-1} Factor for Multi-Dimensional Lattice Paths with Diagonal Steps, Séminaire Lotharingien de Combinatoire, B51c (2004).
Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.
Steffen Eger, The Combinatorics of Weighted Vector Compositions, arXiv:1704.04964 [math.CO], 2017.
Nick Hobson, Python program
L. Reid, Problem #8: How Many Paths from A to B?, Missouri State University's Challenge Problem Set from the 2005-2006 academic year.
J. B. Slowinski, The Number of Multiple Alignments, Molecular Phylogenetics and Evolution 10:2 (1998), 264-266. doi:10.1006/mpev.1998.0522

Cf. A001850, A115866, A263064, A263065, A181544.

Column k=3 of A262809.

f := proc(n) local i,k; add(add((-1)^k*binomial(k,i)*(-1)^i*binomial(i,n)^3,i=n..k),k=n..3*n) end: # Brendan McKay, Mar 03 2007
seq(sum(binomial(k,n)^3/2^(k+1),k=n..infinity),n=0..10); # Vladeta Jovovic, Mar 01 2008

m = 14; se = Series[1/(1 - x - y - z - x*y - x*z - y*z - x*y*z), {x, 0, m}, {y, 0, m}, {z, 0, m}]; a[n_] := Coefficient[se, (x*y*z)^n]; a[0] = 1; Table[a[n], {n, 0, m}] (* Jean-François Alcover, Sep 27 2011, after Max Alekseyev *)
Table[Sum[Sum[(-1)^k*Binomial[k,i]*(-1)^i*Binomial[i,n]^3,{i,n,k}],{k,n,3*n}],{n,0,20}] (* Vaclav Kotesovec, Mar 15 2014, after Brendan McKay *)

# Naive version - see link for better version.
def f(a, b):
    if a == 0 or b == 0:
        return 1
    return f(a, b - 1) + f(a - 1, b) + f(a - 1, b - 1)
def g(a, b, c):
    if a == 0:
        return f(b, c)
    if b == 0:
        return f(c, a)
    if c == 0:
        return f(a, b)
    return (
        g(a, b, c - 1)
        + g(a, b - 1, c)
        + g(a - 1, b, c)
        + g(a, b - 1, c - 1)
        + g(a - 1, b, c - 1)
        + g(a - 1, b - 1, c)
        + g(a - 1, b - 1, c - 1)
    )
for n in range(6):
    print(g(n, n, n), end=", ")

a(n) can be computed as the coefficient of (xyz)^n in the expansion of 1 / (1-x-y-z-xy-xz-yz-xyz). Also, - 2*(n+1)^2 * a(n) + (n+1)*(5n+8) * a(n+1) - 3*(37*n^2 + 146*n + 139) * a(n+2) - (55*n^2 + 389*n + 685) * a(n+3) + (n+4)^2 * a(n+4) = 0. - Max Alekseyev, Mar 03 2007
For Brendan McKay's explicit formula see the Maple code.
From Dan Dima, Mar 03 2007: (Start)
I found a very simple (although infinite) sum for the number of paths from (0,0,...,0) to (a(1),a(2),...,a(k)) using "nonzero" (2^k-1) steps of the form (x(1),x(2),...,x(k)) where x(i) is in {0,1} for 1<=i<=k, k-dimensions.
f(a(1),a(2),...,a(k)) = Sum( (C(n;a(1)) * C(n;a(2)) * ... C(n;a(k))) / 2^{n+1}, {n, max(a(1),a(2),...,a(k)), infinity}), Sum( (C(n;a(1)) * C(n;a(2)) * ... C(n;a(k))) / 2^{n+1}, {n, 0, infinity}), C(n;a)=n!/a!(n-a)! & we assumed C(n;a)=0 if nAlso f(a(1),a(2),...,a(k)) can be computed as the coefficient of x(1)^a(1)...x(k)^a(k) in the expansion: 1/2 * 1/(1 - (1+x(1))*...*(1+x(k))/2). (End)
From David W. Cantrell (DWCantrell(AT)sigmaxi.net), Mar 03 2007: (Start)
Using pseudo-Mathematica-style notation, f(a(1),a(2),...,a(k)) is 2^(-1 - a(1)) (a(1)!)^(k-1)/(a(2)! a(3)! ... a(k)!) * HypergeometricPFQRegularized[{1, 1 + a(1), 1 + a(1),..., 1 + a(1)}, {1, 1 + a(1) - a(2), 1 + a(1) - a(3),..., 1 + a(1) - a(k)}, 1/2]
Although it should be obvious from the above that there are k denominatorial parameters, it is not obvious that there are to be (k+1) numeratorial parameters [one of which is 1 and the other k of them are 1 + a(1)]. In other words, we have P = k + 1 and Q = k.
For information about HypergeometricPFQRegularized, see http://functions.wolfram.com/HypergeometricFunctions/HypergeometricPFQRegularized/ . (End)
G.f.: hypergeom([1/3,2/3],[1], 54*x/(1-x)^3)/(1-x). - Mark van Hoeij, Mar 25 2012.
Recurrence (of order 3): n^2*(3*n-4)*a(n) = (3*n-2)*(57*n^2 - 95*n + 25)*a(n-1) - (9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) + (n-2)^2*(3*n-1)*a(n-3). - Vaclav Kotesovec, Mar 15 2014
a(n) ~ c * d^n / n, where d = 12*2^(2/3)+15*2^(1/3)+19 = 56.947628372041491... and c = 0.2805916350775843477992461458421909485724690193829181355064... = sqrt((6 + 5*2^(1/3) + 4*2^(2/3))/6)/(2*Pi). - Vaclav Kotesovec, Mar 15 2014, updated Mar 22 2016
From Peter Bala, Jan 16 2020: (Start)
a(n) = Sum_{0 <= j, k <= n} C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). Cf. A001850 and A274668.
a(n) = Sum_{0 <= j, k <= n} (-2)^(j+k)*C(n,k)*C(n,j)*C(n+k,k)*C(n+k+j,k+j). (End)

More terms from Max Alekseyev, Mar 03 2007

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

Original entry on oeis.org

1, 1, 36, 225, 400, 225, 36, 1, 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1, 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1

Offset: 1

Yahia Kahloune, Feb 05 2014

In general, define 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,6,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 6 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

			For example :
  T(n,0) = 1;
  T(n,1) = 7^n - (6*n+1);
  T(n,2) = 28^n - (6*n+1)*7^n + C(6*n+1,2);
  T(n,3) = 84^n - (6*n+1)*28^n + C(6*n+1,2)*7^n + C(6*n+1,3);
  T(n,4) = 210^n - (6*n+1)*84^n + C(6*n+1,2)*28^n - C(6*n+1,3)*7^n + C(6*n+1,4).
Triangle T(n,k) begins:
 1;
 1, 36, 225, 400, 225, 36, 1;
 1, 324, 15606, 233300, 1424925, 4050864, 5703096, 4050864, 1424925, 233300, 15606, 324, 1;
 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1;
 1, 16776, 16689816, 3656408776, 286691702976, 10255094095176, 192698692565176, 2080037792142216, 13690633212385551, 57229721552316976, 156200093827061616, 283397584598631216, 345271537321293856, 283397584598631216, 156200093827061616, 57229721552316976,13690633212385551, 2080037792142216, 192698692565176, 10255094095176, 286691702976, 3656408776, 16689816, 16776, 1;
...
Example:
Sum_{i=1..n} C(5+i,6)^2 = A086027(n) = C(n+6,13) + 36*C(n+7,13) + 225*C(n+8,13) + 400*C(n+9,13) + 225*C(n+10,13) + 36*C(n+11,13) + C(n+12,13).
binomial(n,6)^2 = C(n,12) + 36*C(n+1,12) + 225*C(n+2,12) + 400*C(n+3,12) + 225*C(n+4,12) + 36*C(n+5,12) + C(n+6,12).

G. C. Greubel, Table of n, a(n) for the first 25 rows, flattened
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.

Columns k=2..6 are A151651, A151652, A151653, A151654, A151655.
Row sums are A248814.
Similar triangles for e=1..5: A173018 (or A008292), A154283, A174266, A236463, A237202.
Sum_{i=1..n} binomial(5+i,6)^p for p=1..3 gives: A000580, A086027, A086028.
Cf. A087127, A086023, A086024, A086025, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 6, p_] := Sum[(-1)^i*Binomial[6*p+1, i]*Binomial[k-i, 6]^p /. k -> 6+i, {i, 0, k-6}]; row[p_] := Table[b[k, 6, p], {k, 6, 6*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(6*n+1, i)*binomial(k+6-i, 6)^n)} \\ Andrew Howroyd, May 06 2020

Sum_{i=1..n} binomial(5+i,6)^p = Sum{k=0..6*(p-1)} T(p,k) * binomial(n+6+k, 6*p+1).

binomial(n,6)^p = Sum_{k=0..6*(p-1)} T(p,k) * binomial(n+k, 6*p).

Edited by Andrew Howroyd, May 06 2020

A237202 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(5n+1,i) binomial(k+5-i,5)^n, 0 <= k <= 5*(n-1).

Original entry on oeis.org

1, 1, 25, 100, 100, 25, 1, 1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1, 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275

Offset: 1

Yahia Kahloune, Feb 05 2014

In general, define 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,5,n).
Using these coefficients we can obtain formulas for binomial(n,e)^p and for Sum_{i=1..n} binomial(e-1+i,e)^p.
In particular:
  binomial(n, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+k, e*p).
  Sum_{i=1..n} binomial(e-1+i, e)^p = Sum_{k=0..e*(p-1)} b(k,e,p) * binomial(n+e+k, e*p+1).
T(n,k) is the number of permutations of 5 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 08 2020

			T(n,0) = 1;
T(n,1) = 6^n - (5*n+1);
T(n,2) = 21^n - (5*n+1)*6^n + C(5*n+1,2);
T(n,3) = 56^n - (5*n+1)*21^n + C(5*n+1,2)*6^n - C(5*n+1,3) ;
T(n,4) = 126^n - (5*n+1)*56^n + C(5*n+1,2)*21^n - C(5*n+1,3)*6^n  + C(5*n+1,4).
Triangle T(n,k) begins:
1;
1, 25, 100, 100, 25, 1;
1, 200, 5925, 52800, 182700, 273504, 182700, 52800, 5925, 200, 1;
1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 125, 1;
1, 7750, 3882250, 447069750, 18746073375, 359033166276, 3575306548500, 20052364456500, 66640122159000, 135424590593500, 171219515211316, 135424590593500, 66640122159000, 20052364456500, 3575306548500, 359033166276, 18746073375, 447069750, 3882250, 7750, 1;
...
Example:
Sum_{i=1..n} C(4+i,5)^3 = C(n+5,16) + 200*C(n+6,16) + 5925*(n+7,16) + 52800*C(n+8,16) + 182700*C(n+9,16) + 273504*C(n+10,16) + 182700*C(n+11,16) + 52800*C(n+12,16) + 5925*C(n+13,16) + 200*C(n+14,16) + C(n+15,16).
C(n,5)^3 = C(n,15) + 200*C(n+1,15) + 5925*C(n+2,15) + 52800*C(n+3,15) + 182700*C(n+4,15) + 273504*C(n+5,15) + 182700*C(n+6,15) + 52800*C(n+7,15) + 5925*C(n+8,15) + 200*C(n+9,15) + C(n+10,15).

G. C. Greubel, Table of n, a(n) for the first 25 rows, flattened
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.

Columns k=2..5 are A151647, A151648, A151649, A151650.
Row sums are A014609.
Similar triangles for e=1..6: A173018 (or A008292), A154283, A174266, A236463, this sequence, A237252.
Sum_{i=1..n} binomial(4+i,5)^p for p=2..3 gives: A086025, A086026.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A181544.

b[k_, 5, p_] := Sum[(-1)^i*Binomial[5*p+1, i]*Binomial[k-i, 5]^p /. k -> 5+i, {i, 0, k-5}]; row[p_] := Table[b[k, 5, p], {k, 5, 5*p}]; Table[row[p], {p, 1, 5}] // Flatten (* Jean-François Alcover, Feb 05 2014 *)

T(n,k)={sum(i=0, k, (-1)^i*binomial(5*n+1, i)*binomial(k+5-i, 5)^n)} \\ Andrew Howroyd, May 08 2020

Sum_{i=1..n} binomial(4+i,5)^p = Sum{k=0..5*(p-1)} T(p,k) * binomial(n+5+k, 5*p+1).

binomial(n,5)^p = Sum_{k=0..5*(p-1)} T(p,k) * binomial(n+k, 5*p).

Edited by Andrew Howroyd, May 08 2020

A183205 a(n) = [x^n] (1-x)^(3n+1)/(n+1) * Sum_{k>=0} C(n+k-1,k)^3*x^k.

Original entry on oeis.org

1, 2, 16, 190, 2768, 45584, 814728, 15439974, 305760400, 6265985440, 131980086368, 2843029539376, 62400628835608, 1391503990134080, 31454839290752912, 719470742267557110, 16627360903974831120, 387786053931422003360

Offset: 0

Paul D. Hanna, Dec 30 2010

nonn

Cf. A181544, A183204.

{a(n)=polcoeff((1-x)^(3*n+1)/(n+1)*sum(j=0, 2*n, binomial(n+j, j)^3*x^j), n)}

a(n) = A183204(n)/(n+1), where A183204 equals the central terms of triangle A181544.

A262014 Triangle in which the g.f. for row n is (1-x)^(4n+1) Sum_{j>=0} binomial(n+j-1,j)^4 * x^j, read by rows of k=0..3*n terms.

Original entry on oeis.org

1, 1, 11, 11, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 243, 6750, 49682, 128124, 128124, 49682, 6750, 243, 1, 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1, 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 1275, 1, 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476

Offset: 0

Paul D. Hanna, Sep 10 2015

			Triangle begins:
 1;
 1, 11, 11, 1;
 1, 72, 603, 1168, 603, 72, 1;
 1, 243, 6750, 49682, 128124, 128124, 49682, 6750, 243, 1;
 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1;
 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 1275, 1;
 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1;
 ...
Row g.f.s begin:
 n=0: (1) = (1-x) * (1 + x + x^2 + x^3 + x^4 +...);
 n=1: (1 + 11*x + 11*x^2 + x^3)  =  (1-x)^5 * (1 + 2^4*x + 3^4*x^2 + 4^4*x^3 + 5^4*x^4 + 6^4*x^5 +...);
 n=2: (1 + 72*x + 603*x^2 + 1168*x^3 + 603*x^4 + 72*x^5 + x^6)  =  (1-x)^9 * (1 + 3^4*x + 6^4*x^2 + 10^4*x^3 + 15^4*x^5 + 21^4*x^6 +...);
 n=3: (1 + 243*x + 6750*x^2 + 49682*x^3 + 128124*x^4 + 128124*x^5 + 49682*x^6 + 6750*x^7 + 243*x^8 + x^9)  =  (1-x)^13 * (1 + 4^4*x + 10^4*x^2 + 20^4*x^3 + 35^4*x^4 + 56^4*x^5 + 84^4*x^6 +...);
 ...

Paul D. Hanna, Table of n, a(n) for n = 0..1425, as a flattened triangle of rows 0..30
Ilia Gaiur, Vladimir Rubtsov, and Duco van Straten, Product formulas for the Higher Bessel functions, arXiv:2405.03015 [math.AG], 2024. See p. 18.
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. 14.

Cf. A008977 (row sums), A262015 (diagonal), A202750, A258402.

Cf. A181544 (triangle variant).

{T(n, k)=polcoeff(sum(j=0, n+k, binomial(n+j, j)^4*x^j)*(1-x)^(4*n+1), k)}
for(n=0, 10, for(k=0, 3*n, print1(T(n, k), ", ")); print(""))

Row sums form A008977(n) = (4*n)!/(n!)^4.
T(n,1) = A258402(n) = (n^2 + 4*n + 6) * n^2.
From Sergii Voloshyn, Dec 17 2024: (Start)
Let E be the operator D*x*D*x*D*x*D, where D denotes the derivative operator d/dx. Then (1/(n)!^4)  * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(4*n+1) = Sum_{j>=0} binomial(n+j,j)^4 * x^j.
For example, when n = 2 we have (1/2!)^4*E^3(1/(1 - x)) = (1 + 243 x + 6750 x^2 + 49682 x^3 + 128124 x^4 + 128124 x^5 + 49682 x^6 + 6750 x^7 + 243 x^8 + x^9)/(1 - x)^13.  (End)

Showing 1-10 of 10 results.

cp's OEIS Frontend

A183204 Central terms of triangle A181544.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A357407 a(n) = coefficient of x^n, n >= 0, in A(x) = exp( Sum_{n>=1} A183204(n)*x^n/n ), where A183204 equals the central terms of triangle A181544.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A293286 a(n) = A181544(2n, 2n-1).

Views

Author

Keywords

Links

Programs

Mathematica

Sage

A181543 Triangle of cubed binomial coefficients, T(n,k) = C(n,k)^3, read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Formula

A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4*n+1,i) * binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A126086 Number of paths from (0,0,0) to (n,n,n) such that at each step (i) at least one coordinate increases, (ii) no coordinate decreases, (iii) no coordinate increases by more than 1 and (iv) all coordinates are integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

Extensions

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6*n+1,i) * binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

Views

Author

Keywords

Comments

A236463 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(4n+1,i) binomial(k+4-i,4)^n, 0 <= k <= 4*(n-1).

A237252 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(6n+1,i) binomial(k+6-i,6)^n, 0 <= k <= 6*(n-1).

A237202 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(5n+1,i) binomial(k+5-i,5)^n, 0 <= k <= 5*(n-1).

A262014 Triangle in which the g.f. for row n is (1-x)^(4n+1) Sum_{j>=0} binomial(n+j-1,j)^4 * x^j, read by rows of k=0..3*n terms.