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).

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

1,3

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,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

Links

Andrew Howroyd, Table of n, a(n) for n = 1..1600 (rows 1..40)

H. Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.

Formula

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)

Example

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

Maple

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

Mathematica

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

Prog

(PARI) 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
(Magma) [(&+[(-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
(SageMath)
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

Crossrefs

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.

Sequence in context: A176422 A156586 A181544 * A185946 A015113 A016519

Adjacent sequences: A154280 A154281 A154282 * A154284 A154285 A154286

Keyword

nonn,easy,tabf

Author

Roger L. Bagula, Jan 06 2009

Extensions

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

Status

approved