A174266 Irregular triangle read by rows: T(n,k) = Sum_{i=0..k} (-1)^i * binomial(3*n+1,i) * binomial(k+3-i,3)^n, 0 <= k <= 3*(n-1).

1, 1, 9, 9, 1, 1, 54, 405, 760, 405, 54, 1, 1, 243, 6750, 49682, 128124, 128124, 49682, 6750, 243, 1, 1, 1008, 83736, 1722320, 12750255, 40241088, 58571184, 40241088, 12750255, 1722320, 83736, 1008, 1, 1, 4077, 922347, 45699447, 789300477, 5904797049, 21475242671, 40396577931, 40396577931, 21475242671, 5904797049, 789300477, 45699447, 922347, 4077, 1

Offset: 1

Author

Roger L. Bagula, Mar 14 2010

Keywords

• nonn
• tabf

Comments

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,3,n).

With these coefficients we can calculate: Sum_{i=1..n} binomial(i+e-1,e)^p = Sum_{i=0..e*(p-1)} b(i,e,p)*binomial(n+e+i,e*p+1).

For example, A086020(n) = Sum_{i=1..n} binomial(2+i, 3)^2 = T(2,0)*binomial(n+3, 7) + T(2,1)*binomial(n+4,7) + T(2,2)*binomial(n+5,7) + T(2,3)*binomial(n+6,7) = (1/5040)*(20*n^7 + 210*n^6 + 854*n^5 + 1680*n^4 + 1610*n^3 + 630*n^2 + 36*n). (End)

T(n,k) is the number of permutations of 3 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

Examples

			Triangle begins:
  1;
  1,      9,         9,            1;
  1,     54,       405,          760,            405,       54,        1;
  1,    243,      6750,        49682,         128124,   128124,    49682, ... ;
  1,   1008,     83736,      1722320,       12750255, 40241088, 58571184, ... ;
  1,   4077,    922347,     45699447,      789300477, ... ;
  1,  16362,   9639783,   1063783164,    38464072830, ... ;
  1,  65511,  98361900,  23119658500,  1641724670475, ... ;
  1, 262116, 992660346, 484099087156, 64856779908606, ... ;
...
The T(2,1) = 9 permutations of 111222 with 1 descent are: 112221, 112212, 112122, 122211, 122112, 121122, 222111, 221112, 211122. - _Andrew Howroyd_, May 07 2020
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1335
• 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.

Crossrefs

Columns k=0..9 are A000012, A289254, A151632, A151633, A151634, A151635, A151636, A151637, A151638, A151639.

Row sums are A014606.

Similar triangles for e=1..6: A173018 (or A008292), A154283, this sequence, A236463, A237202, A237252.

Cf. A060187, A174264.

Programs

Mathematica

(* First program *)
p[n_, x_]:= p[n,x]= (1-x)^(3*n+1)*Sum[(Binomial[k+1, 3])^n*x^k, {k, 0, Infinity}]/x^2;
Table[CoefficientList[p[x, n], x], {n,10}]//Flatten (* corrected by G. C. Greubel, Mar 26 2022 *)
(* Second program *)
T[n_, k_]:= T[n, k]= Sum[(-1)^(k-j+1)*Binomial[3*n+1, k-j+1]*(j*(j^2-1)/2)^n, {j, 0, k+1}]/(3^n);
Table[T[n, k], {n,10}, {k,3*n-2}]//Flatten (* G. C. Greubel, Mar 26 2022 *)

PARI

T(n,k)={sum(i=0, k, (-1)^i*binomial(3*n+1, i)*binomial(k+3-i, 3)^n)} \\ Andrew Howroyd, May 06 2020

Sage

@CachedFunction
def T(n, k): return (1/3^n)*sum( (-1)^(k-j+1)*binomial(3*n+1, k-j+1)*(j*(j^2-1)/2)^n for j in (0..k+1) )
flatten([[T(n, k) for k in (1..3*n-2)] for n in (1..10)]) # G. C. Greubel, Mar 26 2022

Formula

T(n,k) = [x^k] (1-x)^(3*n+1)*(Sum_{k>=0} (k*(k+1)*(k-1)/2)^n*x^k)/(3^n*x^2).

T(n,k) = T(n, 3*n-k).

From Yahia Kahloune, Jan 30 2014: (Start)

Sum_{i=1..n} binomial(2+i,3)^p = Sum_{i=0..3*p-3} T(p,i)*binomial(n+3+i,3*p+1).

binomial(n,3)^p = Sum_{i=0..3*p-3} T(p,i)*binomial(n+i,3*p). (End)

From Sergii Voloshyn, Dec 18 2024: (Start)

Let E be the operator (x^2)D*(1/x)*D*(x^2)*D, where D denotes the derivative operator d/dx. Then (1/6^n)* E^n(x^2/(1 - x)^4) = (row n generating polynomial)/(1 - x)^(3*n+4) = Sum_{i=0..k} (-1)^i * binomial(3*n+1,i) * binomial(k+3-i,3)^n.

For example, when n = 3 we have 1/216*E^3(x^2/(1 - x)^4) = x^2 (1 + 243x + 6750x^2 + 49682x^3 + 128124x^4 + 128124x^5 + 49682x^6 + 6750x^7 + 243x^8 + x^9)/(1 - x)^13. (End)

Extensions

Edited by Andrew Howroyd, May 06 2020